P1-Allens Made Physics Theory {PART-2}

JEE-Physics TR ANSPARENT GLASS SLA B (Normal shift) When an object is placed infront of a glass slab, it shift the object in the direction of incident light and form a image at a distance x. OI t  1 x  1      SOME ILLUSTR ATIONS OF REFR ACTION  Bending of an object When a point object in a denser medium is seen from a d rarer medium it appears to bend by   Twinkling of stars Due to fluctuations in refractive index of atmosphere the refraction becomes irregular and the light sometimes reaches the eye and sometimes it does not. This gives rise to twinkling of stars. GOLDEN KEY POINTS •  is a scalar and has no units and dimensions. • If  and  are electric permittivity and magnetic permeability respectively of free space while  and  those of a given medium, then according to electromagnetic theory. 1 1 c  c 00 and vm    nm  v   r r 0 0 • As in vacuum or free space, speed of light of all wavelengths is maximum and equal to c so for all wavelengths cc    1 the refractive index of free space is minimum and is vm c Example A ray of light is incident on a transparent glass slab of refractive index 1.62. If the reflected and refracted rays are mutually perpendicular, what is the angle of incidence ? [tan–1 (1.62) = 58.3°] Solution According to given problem : r + 90° + r' = 180° i.e, r' = 90° – r =1 i r r' = (90° – i) [ i = r] and as according to Snell's law: 1 sin i = sin r' O sin i = sin (90 – i) sin i = cos i [sin (90 – i) = cos i] r'  tan i =  i = tan–1 = tan–1 (1.62) = 58.3° Example NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 A 20 cm thick glass slab of refractive index 1.5 is kept infront of a plane mirror. An object is kept in air at a distance 40 cm from the mirror. Find the position of image w.r.t an observer near the object. What is effect of separation between glass slab and the mirror on image. Solution Shifting in object due to glass slab x = d 1  1  20 1  1   20 cm   1.5  3 Slab Distance of object from mirror (as seen by mirror)  40  20  100 cm 3 mM 33 x  =2 Image will be formed at a distance 100 cm from mirror M. 100 100 3 3 3 100 –x 20 O m' 3 Shifting in image due to glass slab = 3 cm M' So distance of image from mirror = 100  20  80 cm 3 33 Distance of image from the actual plane mirror is independent of separation b between glass slab and the mirror. If the distance is more then brightness of image will be less. E 16

JEE-Physics Example If one face of a prism angle 30° and =2 is silvered, the incident ray retraces its initial path. What is the angle of incidence ? A Solution 30° As incident ray retraces its path the ray is incident normally on the silvere face of the prism as shown in figure. D i rE Further, as in AED 30° + 90° + D = 180°  D = 60° Now as by construction, D + r = 90°  r = 90° – 60° = 30° = 2 21  1 1  i  45 C B 22 2  from Snell's law at surface AC, 1 sin i  2 sin 30   sin i  Example An object is placed 21 cm infront of a concave mirror of radius of curvature 20 cm. A glass slab of thickness 3 cm and refractive index 1.5 is placed closed to the mirror in space between the object and the mirror. Find the position of final image formed if distance of nearer surface of the slab from the mirror is 10 cm. 3cm O 10cm 21cm Solution 3cm Shift by slab x = d  1  3 1  1  1cm 1    1.5  for image formed by mirror u = – (21 – 1) cm = – 20 cm. Ox 21cm 111 1 1 1    v = – 20 cm u v f 20 v 10 shift in the direction of light v = – (20 + 1) = – 21 cm. Example NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 f A particle is dropped along the axis from a height 2 on a concave mirror of focal length f as shown in figure. Find the maximum speed of image. t=0 Solution vIM  m2vOM  m2 gt where h=f/2 g f f 2f  2f  2 4f2 gt ///////////////////////////////////////////////// u f  gt2   gt  f  gt2 2  m     v1   2  gt   f  f gt2  f f   2  2  For maximum speed dvI  0  t  f3 3fg dt 3g  v Im ax  4 E 17

JEE-Physics TOTAL INTERNAL REFLECTION When light ray travel from denser to rarer medium it bend away from the normal if the angle of incident is increased, angle of refraction will also increased. At a particular value of angle the refracted ray subtend 900 angle with the normal, this angle of incident is known as critical angle (C). If angle of incident further increase the ray come back in the same medium this phenomenon is known as total internal reflection. r R rare x medium 90° i1<C R i1 x' denser //// //// C i2 troetfalel cintitoenrnal medium i2>C D /// incident ray // CONDITIONS r 2+ h 2  Angle of incident > critical angle [i > c] NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65  Light should travel from denser to rare medium   Glass to air, water to air, Glass to water Snell's Law at boundary xx', D sin C = R sin 90° sin C  R D Graph between angle of deviation () and angle of incidence (i) as rays goes from denser to rare medium  If i <  µ sini = µ sin r; r  sin 1  D  so   r  i  sin 1  D   R sin i  R sin i  i cD R normal  r R  – 2 C rare medium D i 2 i denser medium O C i normal  2 rarer medium  R  If i > c ;  = – 2i D ii denser medium air D Cr A C i  A point object is situated at the bottom of tank filled with a liquid C of refractive index  upto height h. It is found light from the source water i >C  come out of liquid surface through a circular portion above the object r 11 r 1 r2 sin C  & sin C      r2  h2  2  r2  h2 r2  h2 rA  2r2  r2  h2  (2  1)r2  h2   radius of circular portion C h area = r2 h r  2  1 and C 18 E

JEE-Physics SOME ILLUSTR ATIONS OF TOTAL INTERNAL REFLECTION  Sparkling of diamond : The sparkling of diamond is due to total internal reflection inside it. As refractive index for diamond is 2.5 so c = 24°. Now the cutting of diamond are such that i > C. So TIR will take place again and again inside it. The light which beams out from a few places in some specific directions makes it sparkle.  Optical Fibre : In it light through multiple total internal reflections is propagated along the axis of a glass fibre of radius of few microns in which index of refraction of core is greater than that of surroundings.   >  light pipe  Mirage and looming : Mirage is caused by total internal reflection in deserts where due to heating of the earth, refractive index of air near the surface of earth becomes lesser than above it. Light from distant objects reaches the surface of earth with i >  so that TIR will take place and we see the image of an object along with the object C as shown in figure. cold air rare hot air sky hot surface denser NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 Similar to 'mirage' in deserts, in polar regions 'looming' takes place due to TIR. Here  decreases with height and so the image of an object is formed in air if (i> C ) as shown in figure. GOLDEN KEY POINTS • A diver in water at a depth d sees the world outside through a horizontal circle of radius. r = d tan c. • In case of total internal reflection, as all (i.e. 100%) incident light is reflected back into the same medium there is no loss of intensity while in case of reflection from mirror or refraction from lenses there is some loss of intensity as all light can never be reflected or refracted. This is why images formed by TIR are much brighter than formed by mirrors or lenses. Example A rectangular block of glass is placed on a printed page laying on a horizontal surface. Find the minimum value of the refractive index of glass for which the letters on the page are not visible from any of the vertical faces of the block. Solution The situation is depicted in figure. Light will not emerge out from the vertical face BC if at it 1  1 DC i > C or sin i > sin C sin i > as sin C  ... (i)      But from Snell's law at O 1 × sin = sin r And in  OPR, r + 90 + i = 180  r + i = 90°  r = 90 – i R iP sin  r So sin = sin (90 – i) = cos i  cos i  AO B   paper so sin i  1  cos2 i  sin  2 1 ... (ii)     so substituting the value of sin i from equation (ii) in (i), sin2  1 i.e.,2 > 1 + sin2  (sin2 ) = 1 2 > 2 > 2  = 2 1  max min 2  E 19

JEE-Physics REFR ACTION AT TR ANSPARENT CURVED SURFACE 1 = refractive index of the medium in which actual incident ray lies. m = refractive index of the medium in which actual refractive ray lies. 2 O = Object P= pole C= centre of curvature R = PC = radius of curvature NM 1 Refraction from curved surface 1 P 2 2 1sin 1 = 2sin 2 O object  if angle is very small : 11 = 22...(i) C I image But1 =   ...(ii)  = 2 ...(iii) from (i), (ii) and (iii) 1( + ) = 2( – ) 1 + 1 = 2 – 2 1 +2 = (2 – 1)  1PM  2PM  (2  1 )PM 2  1  2  1 u v R vu R SIGN CON VENTION FOR R ADIUS OF CURVATURE light light light light object object object object P C P C C PC P R R R R R = positive R = negative R = negative R = positive These are valid for all single refraction surfaces – convex, concave or plane. In case of plane refracting surface R , µ2  µ1  µ2  µ1  2  1  0 i.e. u  1 or d Ac  1 v u R v u v 2 d Ap 2 FOCAL LENGTH OF A SINGLE SPHERICAL SURFACE A single spherical surface as two principal focus points which are as follows– ( i ) First focus: The first principal focus is the point on the axis where when an object is placed, the image is formed at infinity. That is when u= f, v = , then from  1  2   2  1  F1 NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 u v  R   1  1  2  1  1R  f1 R (2  1 ) We get  f1 (ii) Second focus: Similarly, the second principal focus is the point where parallel rays focus. That is u = –, v = f, then 1 1 2 2 = 2  1 ; f2  2 R F2 f2 R (2  1 )   (iii) Ratio of Focal length: f1   1 f2 2 20 E

JEE-Physics Example An air bubble in glass ( = 1.5) is situated at a distance 3 cm from a spherical surface of diameter 10 cm as shown in Figure. At what distance from the surface will the bubble appear if the surface is (a) convex (b) concave. =1.5 A =1.5  =1 =1 P P C OI CO I 3cm 3cm 5cm 5cm Solution In case of refraction from curved surface 2  1  (2  1 ) vu R 1 (1.5) 1  1.5 (a) 1 = 1.5 , 2 = 1 , R = – 5 cm and u = –3 cm  v  (3)  (5)  v = –2.5 cm the bubble will appear at a distance 2.5 cm from the convex curved surface inside the glass. 1 (1.5) 1  1.5 (b) 1 = 1.5 , 2 = 1 , R = 5 cm and u = –3 cm  v  (3)  (5)  v  1.66 cm the bubble will appear at a distance 1.66 cm from the concave curved surface inside the glass. Note : If the surface is plane then R    case (a) or (b) would yield 1  (1.5)  (1  1.5 )  v = – 2cm v (3)  Example In a thin spherical fish bowl of radius 10 cm filled with water of refractive index (4/3), there is a small fish at a distance 4 cm from the centre C as shown in Figure. Where will the fish appear to be, if seen from (a) E and (b) F 10cm C E F (neglect the thickness of glass) ? 4cm Solution In the case of refraction from curved surface 2  1  (2  1 ) vu R NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 4 F C OI E (a) Seen from E 1  3 ,  2 = 1, R = – 10 cm & u = –(10 – 4) = –6 cm   1 4 14  v  90  5.3cm 5.3cm  3 3 6cm v 6 10 17 i.e., fish will appear at a distance 5.3 cm from E towards F (lesser than actual distance, i.e., 6 cm) 4 (b) Seen from F 1  3 , 2 = 1, R = – 10 cm and u= –(10 + 4) = –14 cm 44 F C O IE 1 v  210  16.154   1  3 3  13 cm 14cm v 14 10 16.154cm so fish will appear at a distance 16.154 cm from F toward E (more than actual distance, i.e., 14 cm) E 21

JEE-Physics LENS A lens is a piece of transparent material with two refracting surfaces such that at least one is curved and refractive index of its material is different from that of the surroundings. A thin spherical lens with refractive index greater than that of surroundings behaves as a convergent or convex lens, i.e., converges parallel rays if its central (i.e. paraxial) portion is thicker than marginal one. However if the central portion of a lens is thinner than marginal, it diverges parallel rays and behaves as divergent or concave lens. This is how wse and classify identify convergent and divergent lenses. R R2 R R  R R R2 R1 R2 RR R1 R2= R1 R2 Bi-convex equi-convex plano-convex cancavo-convex Bi-concave equiconcave plano convexo concave concave  Optical Centre : O is a point for a given lens through which any ray passes undeviated C1 O C2 O diroepcttiecdaltocentre convex lens C1 C2 concave lens  Principal Axis : C C is a line passing through optical centre and perpendicular to the lens. 12  Principal Focus : A lens has two surfaces and hence two focal points. First focal point is an object point on the principal axis for which image is formed at infinity. v= F1 F1 O O convex concave lens lens While second focal point is an image point on the principal axis for which object lies at infinity NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 u= u= F2 F2 O convex O concave lens f f lens  Focal Length f is defined as the distance between optical centre of a lens and the point where the parallel beam of light converges or appears to converge.  Aperture : In reference to a lens, aperture means the effective diameter. Intensity of image formed by a lens which depends on the light passing through the lens will depend on the square of aperture, i.e., Intensity  (Aperture)2 22 E

JEE-Physics LENS–MAKER'S FORMULA In case of image formation by a lens X Image formed by first surface acts as object for the second. M AB M L So, from the formula of refraction at curved surface. 2  1  2  1 O C2 C1 I I1 vu R For first surface A L  M  L  M ...(i) [ 2 = L, 1 = M] v1 u R1 Y For second surface B M  L  M  L   L  M ...(ii) [ 2 = M, 1 = L, 1 = 2, u  v] v v1 R2 R2 1 By adding (i) and (ii) M 1  1  (L   ) 1  1   1  1  L  M 1  1   (   1 ) 1  1     L   v u   R2  v u M  R2   R2  ...(iii) M  M  R   R   R  1 1 1 Now if object is at infinity, Image will be formed at the focus, u=– v=f So 1    1   1  1  ...(iv) f  R1 R2    2F F O F 2F I   h1 O This is known as lens makers formula by equating (iii) and (iv) 1 1  1 this is known as lens formula  vu f Magnification   m  height of image  h1 v f  fv height of object h0  u fu f NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 RULES FOR IMAGE FORMATION  A ray passing through optical centre proceeds undeviated through the lens  A ray passing through first focus or directed towards it, after refraction from the lens, becomes parallel to the principal axis.  A ray passing parallel to the principal axis after refraction through the lens passes or appears to pass through F 2 For Convergent or convex lens Object Im a g e Magnification  F m  1 & m< 0  – 2F F–2F m  1 & m<0 2F 2F m = –1 F – 2F –2 F  m 1 & m<0 Just before F towards C  m < < -1 Just before F towards P  m >>1 F–O In front of lens m >1 E 23

JEE-Physics IMAGE FORMATION FOR CON VEX LENS (CON VERGENT LENS) (i) Object is placed at infinity (ii) Object is placed in between  – 2F Image : Image : at F real inverted very small in size real (F – 2F) inverted small in size (diminished) m  1 & m< 0 m  1 & m<0 (iii) Object is placed at 2F (iv) Object is placed in between 2F – F Image : Image : real (at 2F) inverted equal (of same size) real (2F – ) inverted enlarged (m = – 1) m  1 & m<0  2F F OF 2F  2F F O F 2F I  I O O (v) Object is placed in between F – O Image :  2F F F 2F  virtual (in front of lens) erected enlarge I OO (m > + 1) IMAGE FORMATION FOR CONCAVE LENS (DIVERGENT LENS) Imge is virtual, diminished, erect, towards the object, m = +ve (i) Object is placed at infinity (ii) Object is placed infront of lens Image : Image : At F virtual erected between F and optical centre NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 diminished (m << + 1) virtual erected diminished (m < + 1) u= FF OF I S i g n c o n v e n t io n f o r o b j e c t /im a g e f o r l e n s Real object u – ve Real image v + ve Virtual object u + ve Virtual image v – ve 24 E

POWER OF LENS JEE-Physics Reciprocal of focal length in meter is known as power of lens. f1 f2 P 1 100 f1 f2  SI UNIT : dioptre (D) Power of lens : f(m) f(cm) dioptre [in air] COMBINATION OF LENSES Two thin lens are placed in contact to each other 1 11  power of combination. P = P + P  F f1 f2 1 2 Use sign convention when solve numericals Two thin lens are placed in at a small distance d (provided incident rays are parallel to principal axis). 111 d    F f1 f2 f1 f2  P = P + P – d PP 1 2 12 Use sign convention when solving numericals  Newton's Formula f  x1x2 O F1 F2 I x = distance of object from focus. 1 x = distance of image from focus. 2 SOME SPECIAL CASES ( i ) The focal length of equiconvex lens placed in air refractive index of lens L =  refractive index of medium  M = 1 R1(+) object R = + R, R =–R R2( 1 2 1  (  1)  1    1   Focal length f  R f  R  R   2 (  1)   ( i i ) Focal length of planoconvex lens placed in air R1(+) light 1  1 1  R R2= f  R   ( 1) object  (  1)   Focal length f  NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 If object is placed towards plane surface 1  (  1)  1    1  Focal length f  R f      ( 1)  R  light object (i i i ) If an equiconvex lens of focal length f is cut into equal parts by a horizontal Rf 1= R2=fR plane AB then the focal length of each part will be equal to that of initial lens. B Because  , R and R will remain unchanged. Only intensity will be reduced. 1 2  intensity I  (apertures)2 A  intensity through a single part will be reduced E 25

JEE-Physics ( i v ) If the same lens is cut into equal parts by a vertical plane CD the focal length C 2f of each part will be double of initial value but intensity will remain unchanged. f 2f For equiconvex lens 1  (  1)2 For plano convex lens 1  1 fR  f1 R 12  D So f f1  f = 2f  Focal length of each part = 2 focal length of each 1 part become 2f (focal length of original lens) ( v ) If a lens is made of number of layers of different refractive index for a given wavelength 1 1 2 1 then no. of images is equal to number of refractive index, as  (  1) 2 f 1 In figure number of images = 2 B (vi) Focal length of lens depends on wavelength.  1  ( 1)  1 f   f >f f RV White light v R white light vR FR FV FR FV O vR R v A ( v i i ) If half portion of lens is covered by black paper then intensity of image will be reduced but complete image will be formed. (viii) Sun–goggles : BA radius of curvature of two surfaces is equal with centre on the same side object sun R1 goggles R = R = +R so 1  (  1) 1  1 R2 12 f  R R  R1=R2 P=0  1  0  f   and P = 0 sun goggles have no power NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 f ( i x ) If refractive index of medium < Refractive index of lens M< L If  < L  then f = +ve  Convex lens behave as convex lens. While concave lens behave as concave lens. ( x ) Refractive index of medium = Refractive index of lens (M = L) M L M M L M 1   L  1 1 ; 1 0  f   & P=0 f  M  1  R1  R 2  f Lens will behave as plane glass plate  M= L (A) (B) 26 E

JEE-Physics (xi) Refractive index of medium > Refractive index of lens L M L M M M M > L 1 [ f will be negative ] L  M convex lens will behave as concave lens and concave lens will  M> L behave as convex lens. If a air bubble is formed in water it behaves as concave lens. Example A point source S is placed at distance of 15 cm from a converging lens of focal length 10 cm. Where should a (i) concave mirror (ii) convex mirror of focal length 12 cm be placed so that real image is formed on object itself. Solution 11 1 1 1 1 u = –15cm, f = +10cm; v  u  f  v  (15)  10  v = 30 cm (i) x = v + 2f  30 + 2 × 12 = 54 cm (ii) x = v – 2f = 30 – 2 × 12 = 6 cm OI //////////////////////////////////O I ///////////////////////////////////////////////// Example 1 A convex lens of focal length f is producing real image which is times of the size of the object. Find out n position of the object. Solution v1 u Image is real so m= u = – n  v =– n from lens formula 11 1 1  1  1  (n  1)  1  u  f 1  n   vu f u u f uf  n NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 Example If f = +0.5m, what is the power of the lens ? (a) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 (b) cm. What is the refractive index of glass ? A convex lens has 20 cm focal length in air. What is the focal length in water ? (Refractive index of air– (c) water = 1.33, refractive index for air glass is 1.5) Solution 11 1  1  1  1  (  1)  1 1    1.5 (a) P    2D (b) f (µ  1)  R1 R 2  12  10  15  f(m) 0.5 1 1  1  1 fw   1  fw  1.5  1  20  78.2cm f  (  1)  R1 R2  so f  ( 1) ; fa  w   1  1.5  1 (c)  1.33 E 27

JEE-Physics Example Column I (optical system) Incident ray R1=R2=20cm Incident ray R1=R2=20cm Incident ray R1=R2=20cm Incident ray R1=R2=20cm (A) =1.5 (B) water  water (C) water  air (D) air  water air glass air (=4/3) glass (=1) (=1) glass (=4/3) (=4/3) glass (=4/3) ( =1) ( =1) Column II (focal length) (P) 80 cm (Q) 40 cm (R) 30 cm (S) 20 cm Ans. (A) –S (B) –P (C) –R Solution For (A) : 1    1  1  1  1.5  1  1  1  1  f  20cm f  p1 p2   20 20  20 For (B) : 1   1.5  1  1  1  1 f  4/3  20 20    f  80cm 80 For (C) : 1.5  4/3  1.5  4 1  & 1  1.5  1  1 .5   1  f  30cm v1  3   20  f v1  20  For (D) 1.5  1  1.5  1  1  & 4/3 1.5  4  1 .5    1   f  40cm v1   20  f   3   20  v1 DISPLACEMENT METHOD It is used for determination of focal length of convex lens in laboratory. A thin convex lens of focal length f is placed between an object and a screen fixed at a distance D apart.If D > 4f there are two position of lens at which a sharp image of the object is formed on the screen object screen By lens formula 11 1 1  1 1 NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65D±D(D -4f)    u2 – Du + Df = 0  u = there are three v u f D  u u f 2 possibilities (i) for D < 4f u will be imaginary hence physically no position of lens is possible (ii) for D = 4f D u = = 2f so only one position of lens is possible 2 and since v = D – u = 4f – 2f = u =2f (iii) for D > 4f u = D  D ( D  4 f ) and u = D  D ( D  4 f ) 12 22 So there are two positions of lens for which real image will be formed on the screen.(for two distances u1 and u2 of the objec t from lens) 28 E

JEE-Physics object I2 I1 If the distance between two positions of lens is x then x=u –u = D  D D  4f D  D D  4f D D  4f   x2 = D2 – 4 Df f = D2  x2 21 = 22 4D Distance of image corresponds to two positions of the lens : v1 D  u1 D 11    [D  D (D  4f)] = [D + D (D  4f) ] = u2  v1  u2 2 2 D 11 D (D  4f)] = u1  v2  u1 v2  D  u2   [D + D (D  4f)] = [D  2 2 for two positions of the lens distances of object and image are interchangeable. Now x = u – u and D = v + u = u + u [ v = u ] 21 1121 12 so Dx and v= Dx =u; m = I1  v1  D  x and m = I2  v2  D  x u1 ( v2 )  2 1 2 21 O u1 D  x 2 O u2 D  x Now m1  m2 Dx Dx  I1 I2 1O  I1 I2  O2 Dx Dx Example A convex lens is placed between an object and a screen which are at a fixed distance apart for one position of the lens. The magnification of the image obtained on the screen is m1. When the lens is moved by a distance d the magnification of the image obtained on the same screen is m , Find the focal length of the lens. 2 Solution If D is the distance between the object and the screen, d the separation of the two position of lens throwing two images on the screen then m1  (D  d) and m2  (D  d)  m – m = 4Dd D2  d2 d d (D  d) (D  d) 1 2 D2  d2 but = f so m – m =  f = 4D 1 2f m1  m2 NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 Example In a displacement method using lens, we obtain two images for separation of the lens d. One image is magnified as much as the other is diminished. If m is the magnifications of one image, find the focal length of the lens. Solution d 1 md f m 2  m , so f becomes From above question m1  m2 here if m is taken as m, (m2 1) 1 Example In the displacement method the distance between the object and the screen is 70 cm and the focal length of the lens is 16 cm, find the separations of the magnified and diminished image position of the lens. Solution d  D2  4fd  (70)2  4  16  70  420  20.5cm E 29

JEE-Physics Example An object 25 cm high is placed in front of a convex lens of focal length 30 cm. If the height of image formed is 50 cm, find the distance between the object and the image (real and virtual) ? Solution As object is in front of the lens, it is real and as h = 25 cm, f = 30 cm, h = – 50 cm ; m  h2 50  2 1 2 h1  25 f 30 vv m  f  u  2  30  u  u = – 45 cm  m  u  2  45  v = 90 cm 2F F F 2F 45cm 90cm As in this situation object and image are on opposite sides of lens, the distance between object and image d = u + v = 45 + 90 = 135 cm. If the image is erect (i.e., virtual) 1 m  f f u  2  30 u u = – 15 cm m   v  2  v v = 30 cm  30  u 15 As in the situation both image and object are in front of the lens, the distance between object and image d = v – u = 30 – 15 = 15 cm. 2 COMBINATION OF LENSES AND MIRRORS When several lenses or mirrors are used, the image formation is considered one after another in steps, The image formed by the lens facing the object serves as an object for the next lens or mirror, the image formed by the second lens acts as an object for the third, and so on, The total magnification in such situations will be given by m  I  I1  I2  ...  m = m× m× ... O O I1 1 2 1 Converging lens P = +ve Diverging lens P = –ve Power of Lens [in air] PL  fL L L PL 1   Power For mirror fm Convex mirror P = –ve Concave mirror P = +ve M M SILVERING OF LENS Calculate equivalent focal length of a equiconvex lens silvered at one side. NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 /////////////////////////////////////////////////////=+I1 I2 +I1 I3 /////////////////////////////////////////////////////O P = P + P + P = 2P + P LML LM 1 1 1 1 2 1 2( 1)  2 2 4  4  2 F  R        F f fm f f fm RRR 4  2 30 E

JEE-Physics Example Calculate equivalent focal length of plano convex lens for following case :– (i) When curved surface is silvered. (ii) When plane surface is silvered. Solution (i) P = 2P + P/////////////////////////////////// ///////////////////////////////////////////////////////////////////(ii) P = 2P + P LM//////////////////////////////////// LM O  1  2  1  1  2(  1)  2 O  1  2  1 1  2(  1)  1 F f fm F R R F fL fm F R   1  2  F  R R F R 2  F  2(  1) Example The radius of curvature of the convex face of a plano–convex lens is 12 cm and its refractive index is 1.5. (a) Find the focal length of this lens. The plane surface of the lens is now silvered. (b) At what distance from the lens will parallel rays incident on the convex face converge ? (c) Sketch the ray diagram to locate the image, when a point object is placed on the axis 20 cm from the lens. (d) Calculate the image distance when the object is placed as in (c). Solution L O1 (a) MN PQAs for a lens, by lens–maker's formula  = ( – 1) 11 Here  = 1.5; R = 12 cm and R =   1 2 L O1 R1 R2 MN QPSo  = (1.5 – 1) 1 1 i.e. f = 24 cm i.e., the lens as convergent with focal length 24 cm. 12  R = 12 cm R =  I2 I3 I3 I1 O I2 (A) 12 cm 10cm (B) 30cm (C) (b) As light after passing through the lens will be incident on the mirror which will reflect it back through the lens again, so P = P + P + P = 2P + P But P = 11 and P =– 1 = 0 as M  R   LML LM L L = 0.24 M  2 NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 11 1 So P = 2  0.24 + 0 = 0.12 D. The system is equivalent to a concave mirror of focal length F, P = – F 1 i.e., F = – = – 0.12 m = – 12 cm i.e., the rays will behave as a concave mirror of focal length 12 cm. P 111 we have 111 So as for parallel incident rays u = –  from mirror formula v + u =  + = v  12  v = – 12 cm i.e., parallel incident rays will focus will at a distance of 12 cm in front of the lens as shown in Figure (c) and (d) When object is at 20 cm in front of the given silvered lens which behaves as a concave 111 11 1 mirror of focal length 12 cm, from mirror formula v + u =  we have v + 20 = 12 v= – 30 cm i.e., the silvered lens will form image at a distance of 30 cm in front of it as shown in fig. (C) E 31

JEE-Physics Example A pin is placed 10 cm in front of a convex lens of focal length 20 cm, made of material having refractive index 1.5. The surface of the lens farther away from the pin is silvered and has a radius of curvature 22 cm. Determine the position of the final image. Is the image real or virtual ? Solution As radius of curvature of silvered surface is 22 cm, R 22 IO so M = 2 = 2 = –11 cm = – 0.11 m and hence, M = – 1 11 =– 0.11 = 0.11 D M Further as the focal length of lens is 20 cm, i.e., 0.20 m its power will be given by : P = 1 = 1 D. L  L 0.20 Now as in image formation, light after passing through the lens will be reflected back by the curved mirror 2 1 210 through the lens again P= P + P + P = 2P + P i.e. P    D . LML LM 0.20 0.11 11 So the focal length of equivalent mirror F  1  11 m   110 cm i.e., the silvered lens behave as a concave  P 210 21 11 21 mirror of focal length (110/21) cm. So for object at a distance 10 cm in front of it,    i.e., v = – v 10 110 11cm i.e., image will be 11 cm in front of the silvered lens and will be real as shown in Figure. Example A point object is kept at a distance of 2m from a parabolic reflecting surface y2 = 2x. An equiconvex lens is kept at a distance of 1.80 m from the parabolic surface. The focal length of the lens is 20 cm. Find the position from origin of the image in cm, after reflection from the surface. y object x 0.2m Solution OR y2=2x y2=4ax (x ,y )1 1 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ P (a,0) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2 F  object NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65(x2,0)Cobject \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ Comparing with y2 = 4ax  a = 0.5 PC is a normal so tan     dy 1  y1  final position of image = 0.5 m = 50 cm / dx x1 ,y1  But 2 y1 tan 2  y1  0 & 2 tan   y1  x2  1m x2  x1 tan   1  tan2  x2  x1 1  y 2 2 1 32 E

JEE-Physics Solution Ans. 1.5 For case (a) 1 =  1 1   1 and v = 2R  m=2  1 µ  , v2 v1  R = 2 m v1  R  For case (b) 1   = 1 and v = R  µ=m v2  R 2 m 1 Therefore   2  1 µ  1  µ = 1.5 1  2 Example#27 In figure, L is half part of an equiconvex glass lens ( = 1.5) whose surfaces have radius of curvature R = 40 cm and its right surface is silvered. Normal to its principal axis a plane mirror M is placed on right of the lens. Distance between lens L and mirror M is b. A small object O is placed on left of the lens such that there is no parallax between final images formed by the lens and mirror. If transverse length of final image formed by lens is twice that of image formed by the mirror, calculate distance 'a' in cm between lens and object. \\\\\\\\\\\\\\\\\\\\\\\\\\\\ O M ab Solution Ans. 5 Distance of image of object O from plane mirror = a + b. Since, there is no parallax between the images formed by the silvered lens L and plane mirror M, therefore, two images are formed at the same point. Distance of image = (a + 2b) behind lens. Since, length of image formed by L is twice the length of image formed by the mirror M and length of image formed by a plane mirror is always equal to length of the object, therefore, transverse magnification produced by the lens L is equal to 2.Since, distance of object from L is a, therefore, distance of image from L must be equal to 2a. a  (a + 2b) = 2a  b = 2 The silvered lens L may be assumed as a combination of an equi–convex lens and a concave mirror placed in contact with each other co–axially as shown in figure. \\\\\\\\\\\\\\\\\\\\\\\\\\\\ NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 Focal length of convex lens f is given by 1 = (–1) 1  1   f = 40 cm 1 f1  R1 R2  1 R For concave mirror focal length, f = = – 20 cm m2 The combination L behaves like a mirror whose equivalent focal length F is given by 1 12  F fm f1  F = –10 cm, Hence, for the combination u = –a, v = +2a, F = –10 cm 11 1 Using mirror formula,   a = 5 cm vuF 56 E

Distance of Bird as seen by fish x= d + h JEE-Physics Bf h By differentiating d(x fB )  dh  1 d(d ) d dt dt  dt v 4  dh  5 2  3 cm / s, d(d)  2  ( 2 )  4 cm / s,   4 FB  3 +  = 6 cm/s  dt dt 3  v =  d(d)    dh   4 + 4  8 cm/s BF  dt   dt   3  (3) d(d) 1 dt (for fish image after reflection = 0 )  3 +  (0) = 3 cm/s Similarly speed of image of bird  4 cm/s Example#25  50x In the shown figure the focal length of equivalent system in the form of  13  . Find the value of x. Ans. 2 1  3  1  1  1   1 ; 1   6  1  1  1  3 and 1   8  1  1  1   3 f1  2  10 10  10 f2  5  10 20   f3  5  20 20  50 100 1111  1  3  3 100 = 50x  x=2  10 100 50 = 13 f f1 f2 f3 13 NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 Example#26 Quarter part of a transparent cylinder ABC of radius R is kept on a horizontal floor and a horizontal beam of light falls on the cylinder in the two different arrangement of cylinder as shown in the figure (a) & (b). In arrangement (a) light converges at point D, which is at a distance 2R/m from B. And in arrangement (b) light converges at point E, which is at a distance R/(m – 1) from C. Find out the refractive index of the material. (a) (b) A CB D BCE 2R/m R m-1 E 55

JEE-Physics Example#23 Column-I contains a list of mirrors and position of object. Match this with Column-II describing the nature of image. Column I Column II (A) C F (P) real, inverted, enlarged (B) (Q) virtual, erect, enlarged (C) (R) virtual, erect, diminished (D) F (S) virtual, erect Solution Ans. (A) P (b)RS (C) S (D) QS Example#24 A bird in air is diving vertically over a tank with speed 5 cm/s, base of tank is silvered. A fish in the tank is rising upward along the same line with speed 2 cm/s. Water level is falling at rate of 2 cm/s. [Take : water = 4/3] Column I (cm/s) Column II NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 (A) Speed of the image of fish as seen by the bird directly (P) 8 (B) Speed of the image of fish formed after reflection in (Q) 6 the mirror as seen by the bird (R) 3 (C) Speed of image of bird relative to the fish looking upwards (S) 4 (D) Speed of image of bird relative to the fish looking downwards in the mirror Solution Ans. (A) – (Q) ; (B) – (R) ; (C) – (P) ; (D) – (S) Distance of fish asseen by bird x =h+ d fB  54 E

JEE-Physics Example#19 to 21 There is a spherical glass ball of refractive index 1 and another glass ball of refractive index 2 inside it as shown in figure. The radius of the outer ball is R and that of inner ball is R . A ray is incident on the 12 outer surface of the ball at an angle i . 1 1 i2 B r2 C r1 O A D i1 R2 2 R1 1 9 . Find the value of r 1 (A) sin 1  sin i1  (B) sin 1 1 sin i1  (C) sin 1  1  (D) sin 1  1   1   sin i1   1 sin i1  2 0 . Find the value of i 2 (A) sin 1  R2 sin i1  (B) ] sin 1  R1 sin i1  (C) sin 1  R1 sin i1  (D) sin 1  R2 sin i1   R1 1   R 2 2   R 2 1   R1 2  2 1 . Find the value of r 2 (A) sin 1  R1 sin i1  (B) sin 1  R2 sin i1  (C) sin 1  R1 1  (D) sin 1  R2 sin i1   2R 2   2 R 1   1R 2 sin i1   1R 1  Solution 19. Ans. (A) 1 sin r = sin i  r = sin 1  sin i1  1 1  1  1 20. Ans. (C) Using sine rule sin r1  sin(180  i2 )   sin i = R1 sin r =  R1 sin i1   i2  sin 1  R1 sin i1  R2 R1 2 R2 1  R 2 1   R2 1  21. Ans. (A) 1 sin i = sin r ; 1 R1 sin i1 = sin r; r = sin 1  R1 sin i1  2 22 2 R2 1 2 2  2 R 2  Example#22 Consider a an equilateral prism ABC of glass    3 placed in water    4  2   3  A water  =4/3 NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 glass =3/2 Column-I Column-II (A) FG is parallel to BC (P) Maximum deviation (B) i = 900 (Q) Minimum deviation 1 (C) i= i = sin-1 9 (R) TIR will take place at surface AC  16  1 2 (S) No TIR will take place at surface BC Ans. (A) QS, (B) PS, (C) QS, (D) S (D) EF is perpendicular to AB Solution At maximum deviation i = 900 or i =900 12 At minimum deviation i = i, EF BC; 1 2 For i =0, TIR will not take place at AC 53 1 E

JEE-Physics Solution 13. Ans. (A) distance  Y12  x2    x 2  Y22 Time t = speed = c 14. Ans. (A) dt 2x 2   x   x 2  Y22 = 0  sin 1  sin 2  1   2  For least time dx =0  – Y12  x2 15. Ans. (A) Example#16 to 18 One hard and stormy night you find yourself lost in the forest when you come upon a small hut. Entering it you see a crooked old woman in the corner hunched over a crystal ball. You are about to make a hasty exit when you hear the howl of wolves outside. Taking another look at the gypsy you decide to take your chances with the wolves, but the door is jammed shut. Resigned to a bad situation you approach her slowly, wondering just what is the focal length of that nifty crystal ball. 1 6 . If the crystal ball is 20 cm in diameter with R.I. = 1.5, the gypsy lady is 1.2 m from the ball, where is the image of the gypsy in focus as you walk towards her? (A) 6.9 cm from the crystal ball (B) 7.9 cm from the crystal ball (C) 8.9 cm from the crystal ball (D) None of these 1 7 . The image of old lady is (B) erect, virtual and small (A) real, inverted and enlarged (D) real, inverted and small (C) erect, virtual and magnified 1 8 . The old lady moves the crystal ball closer to her wrinkled old face. At some point you can no longer get an image of her. At what object distance will there be no change of the gypsy formed? (A) 10cm (B) 5 cm (C) 15 cm (D) None of these Solution 16. Ans. (A) 1.5 1  1.5 1 (1) v1  10 For refraction at 1st surface  v1  36cm Lady (2) 120 120cm R=10cm for refraction at 2nd surface 1  1.5  1  1.5  v  80  6.9cm v (36  20) 10 11.5 NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 17. Ans. (D) Total magnification = mm =  1 v1   2 v   v = negative 12  2u   1 v1  u 18. Ans. (B) At this point image will formed at infinity (1) (2) x 20cm 1 1.5 1  1.5  v1  30cm   For refraction at second surface  v1 10 For refraction at first surface 1.5 1  1.5  1  x  5cm  10 x 10 52 E

Solution JEE-Physics Case I Ans.(AC) 10ms1   5ˆi,   5ˆi  vI  0 v om v Im 5ms1 Case II 10ms1   1 5ˆi,   15ˆi  vI  2 0 m s 1 v om v Im 5ms1 Case II 10ms1   15ˆi,  Im  15ˆi  vI  2 0 m s 1 v om v 5ms1 Case IV 10ms1   5ˆi,  Im  15ˆi  vI  0 m s 1 OR v om v 5ms1 vI    2 5  10  20 or 0 m/s 2vm  vo Example#13 to 15 A ray of light travelling with a speed c leaves point 1 shown in figure and is reflected to point 2. The ray strikes the reflecting surface at a distance x from point 1. According to Fermat's principle of least time, among all possible paths between two points, the one actually taken by a ray of light is that for which the time taken is the least (In fact there are some cases in which the time taken by a ray is maximum rather than a minimum). 1 32 Y1    Y2 x  1 3 . Find the time for the ray to reach from point 1 to point 2. NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65  (A) Y12  x2    x 2  Y22  (C) Y1  Y2 2x (B) c cc (D) c c 1 4 . Under what condition is time taken least? (A) 1  2 (B) x    x (C) Y =Y (D) all of these 12 15. Which of the following statement is in accordance with Fermat's principle (A) A ray as it moves from one point to another after reflection takes shortest possible path E (B) A ray as it moves from one point to another after reflection takes longest possible path (C) A ray as it moves from one point to another takes shortest possible time (D) A ray as it moves from one point to another takes longest possible time 51

JEE-Physics Example#10 A ray of light is incident in situation as shown in figure. 2  3=2 =4 300 Which of the following statements is/are true? (B) If 2 = 2.8 then the angle of deviation is 600 (A) If 2 = 3.2 then the angle of deviation is zero (D) If 2 =1.8 then the angle of deviation is 600 (C) If 2 =1.8 then the angle of deviation is 1200 Solution Ans.(BC) 2 3=2  =4  300 1 sin 300 = 2 sin  = 3 sin 1  2= 2 sin  =2 sin 1  1 =900 and 2  2 For 2 <2, TIR will take place at first surface.  3=2   = 1200  2<2  1=4 300 300 Example#11 A fish lies at the bottom of a 4m deep water lake. A bird flies 6 m above the water surface and refractive index NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 of water is 4/3. Then the distance between (A) Bird and image of fish is 9 m (B) Fish and image of bird is 12 m (C) Fish and image of bird is 8m (D) Fish and image of bird is 10m Solution Ans. (AB) For a bird, fish appears 3 m below the water surface and for fish, bird appears 9m above the surface. Example#12 A plane mirror and an object has speeds of 5 m/s and 10 m/s respectively. If the motion of mirror and object is along the normal of the mirror then the speed of image may be :– (A) 0 m/s (B) 10 m/s (C) 20 m/s (D) 25 m/s 50 E

JEE-Physics Example#7 A man of height 2 m stands on a straight road on a hot day. The vertical temperature in the air results in a variation of refractive index with height y as  = 0 (1+ ay) where 0 is the refractive index of air near the road and a=2 × 10–6/m. What is the actual length of the road, man is able to see (A) 2000 m (B) 390 m (C) infinite distance (D) None of these Solution Ans. (A)  sin = 0sin90° = 0  sin= 0  1  1  ay 2m  121  y 2 90° dy  But dx  tan   dx  (dy )  1   x  dy  2 a   2000m dx dy  ay   0 a0 y x Example#8 2  A system of coordinates is drawn in a medium whose refractive index varies as 1 y2 , where 0 y  1 and  =2 for y < 0 as shown in figure. A ray of light is incident at origin at an angle 60° with y–axis as shown in the figure. At point P ray becomes parallel to x-axis. The value of H is :- y HP O x =2 60°  2 1/ 2  2 1/2    1/2  4 1/2  3   1  3  (C) 3  1  1 (A)  (B)   (D)  3 Ans. (A) 1 sin 1  2 sin 2  at origin x =0, y = 0  =2 &  = 60° at point P :  = 90°  2 sin60° sin90°  3  2  y   2  1/ 2 Example#9 y2 1  3  1  NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 A ray of light is incident along a vector ˆi  ˆj  kˆ on a plane mirror lying in y–z plane. The unit vector along the reflected ray can be ˆi  ˆj  kˆ ˆi  ˆj  kˆ ˆi  ˆj  kˆ 3ˆi  ˆj  kˆ (A) (B) (C) (D) 3 3 3 3 Solution Ans. (C,D) According to law of reflection ˆr  eˆ  2 eˆ.nˆ nˆ Here eˆ  ˆi  ˆj  kˆ , nˆ  ˆi 3 so eˆ.nˆ  1  ˆr  ˆi  ˆj  kˆ ± 2ˆi 3ˆi  ˆj  kˆ or ˆi  ˆj  kˆ 3 3 3 = 3 3 E 49

JEE-Physics Example#5 If x and y denote the distances of the object and image from the focus of a concave mirror. The line y= 4x cuts the graph at a point whose abscissa is 20 cm. The focal length of the mirror is y x (A) 20 cm (B) 40 cm (C) 30 cm (D) can't be determined Ans.(B) Solution For x = 20 cm, y = 4 × 20 = 80 cm From Newton's formula xy = f2  (2) (80) = f2  f = 40 cm Example#6 A concave mirror forms an image I corresponding to a point object O. The equation of the circle intercepted by the xy plane on the mirror is y O x (-50,4) I (25,-1) (A) x2 + y2 = 1600 (B) x2 + y2–20x – 1600 = 0 (C) x2 + y2–20x – 1500 = 0 (D) x2 + y2–20x + 1500 = 0 Solution Ans. (C) 11 1 1 11 v  u  f = x1 25 From mirror equation + x1 50 = and f from m= – v1 x1  25  1   4x –100 = x +50  3x = 150  x = 50 unit  ; x1  50 4 1 1 1 1 u4 y NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 O(50,4) x1+50 I x125 x (25,1) 1  1  1    1  1  f = –20 unit  R = – 40 unit f 50  25 50  50  25 100  Centre of circle will be at (10,0) Equation of required circle (x–10)2 + (y–0)2 = (40)2  x2 + y2 – 20 x – 1500 = 0 48 E

JEE-Physics Example#3 On one boundary of a swimming pool, there is a person at point A whose speed of running on ground (boundary) is 10 ms–1 , while that of swimming is 6 ms–1. He has to reach a point B in the swimming pool.The distance covered on the boundary so that the time required to reach the point B in the pool is minimum, is– A Swimming B Pool (A) 10m (B) 6m (C) 7m (D) 116 m Solution Ans. (C) A Required time t  x  42  10  x 2 dt x 10 For minimum time 0  x= 7m B 5 dx Swimming Pool OR Ax 10-x Just like light, he has different speeds on ground and in water, so to minimize the time, c Fermat's principle must hold good. B sin C 6  3  C  370  tan C  3  10  x  x  7m 10 5 4 4 Example#4 A person has D cm wide face and his two eyes are separated by d cm. The minimum width of a mirror required for the person to view his complete face is Dd Dd Dd Dd (A) 2 (B) 4 (C) 4 (D) 2 NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 Solution H1 Ans. (D) E1 According to ray diagram : M'1 M'2 H M' = H1E2 & H M' = H2E1 E2 11 2 22 2 H2 1 Dd 47 M1 H E = D– (D–d) = = HE M2 12 2 2 21 M' M' = D–H M' – H M' = D – D  d 12 11 22   2 Dd = 2 E

JEE-Physics SOME WORKED OUT EXAMPLES Example#1 The correct mirror–image of the figure given is :- (A) (B) (C) (D) Solution Ans. (C) Example#2 A point object O can move along vertical line AB as shown in figure. When image of the object is first visible to D then it is released at t = 0 from rest from A. The time for which image is visible to D is : A O L DL 22 vertical plane B mirror 6L 2L 3L (D) t   (A) g (B) g (C) g Solution Ans. (A) NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 45° Required time is given by 3L 1 gt2  t  6L 45° =2 g D L/2 45° 45° 46 E

JEE-Physics PRESBYOPIA In this case both near and far object are not clearly visible. To remove this defect two separate spectacles one for myopia and other for hypermetropia are used or bifocal lenses are used. ASTIGM ATISM In this defect eye can not see object in two orthogonal direction clearly. It can be removed by using cylindrical lens in particular direction. Example A person can not see clearly an object kept at a distance beyond of 100 cm. Find the nature and the power of lens to be used for seeing clearly the object at infinity. Solution For lens u = –  and and v = – 100 cm 1 1 1 1  1  f  v  100cm concave  Power of lens 11 vu  v P     1D f f f1 Example A far sighted person has a near point of 60 cm. What power lens should be used for eye glasses such that the person can read this book at a distance of 25 cm. Solution Here v = – 60 cm, u = –25 cm 1 1 1  1  1  f  300 cm  Power 11  2.33D f vu 60 25 7 = f in m  3 / 7 NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 E 45

JEE-Physics DEFECTS OF EYES MYOPIA [or Short–sightedness or Near– sightedness] Defective-eye Corrected-eye (A) (B) (i) Distant object are not clearly visible, but near object are clearly visible because image is formed before the retina. (ii) To remove the defect concave lens is used.  The maximum distance. Which a person can see without help of spectacles is known as far point.  If the reference of object is not given then it is taken as infinity.  In this case image of the object is formed at the far point of person. 11 1 P 1  1 1 P vu f distance of far point (in m) distance of object (in m) f 100  100 P distance of far point (in cm) distance of object (in cm) LONG–SIGHTEDNESS OR HYPERMETROPIA Defective-eye IO Corrected-eye (A) (B) (i) Near object are not clearly visible but far object are clearly visible. NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 (ii) The image of near object is formed behind the retina. (iii) To remove this defect convex lens is used. Near Point :– The minimum distance which a person can see without help of spectacles. • In this case image of the object is formed at the near point. • If reference of object is not given it is taken as 25 cm. 11 1 1 11   P   P vu f distance of near point (in m) distance of object (in m) f distance of near point = –ve, distance of object = –ve, P = +ve 44 E

JEE-Physics Example f 1f With diaphragm of the camera lens set at 2 , the correct exposure time is 100 , then with diaphragm set at 4 . Calculate the correct exposure time. Solution 1 11 As exposure time  aperture2  t1  f / 22 and t2  f / 42 here 1 t2  16  4  t2  4 t1 4 t1  100 s then t1 4 s 100 Example A good photographic print is obtained by an exposure of two seconds at a distance of 20 cm from the lamp. Calculate the time of exposure required to get an equally good result at a distance of 40 cm. Solution We know that the intensity of light varies inversely as the (distance)2. When distance is doubled, the intensity becomes one–fourth. So, the time of exposure should be four times. Hence, time of exposure = 2 × 4 = 8 s Example Photograph of the ground are taken from an aircraft, flying at an altitude of 2000 m, by a camera with a lens of focal length 50 cm. The size of the film in the camera is 18 cm  18 cm. What area of the ground can be photographed by this camera at any one time. Solution As here u = – 2000m, f = 0.50m, so from lens formula 1  1  1 , v u f we have 1  1  1  1  1  1  1 as 1  1   v = 0.5m = 50 cm =  v 0.5 v 0.5 2000 0.5 0.5 2000  2000 Now as in case of a lens, m  v  0.5   1  103 So I = (ma) (mb) = m2A [ A = ab] u 2000 4 1 NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 A  I1  18cm  18cm = (720 m  720 m) m2 1 / 4   1 0 3 2  Example The proper exposure time for a photographic print is 20 s at a distance of 0.6 m from a 40 candle power lamp. How long will you expose the same print at a distance of 1.2 m from a 20 candle power lamp ? Solution In case of camera, for proper exposure I D 2t = I D 2t 1 11 2 22 As here D is constant and I = (L/r2); L1  t1  L2  t2 40 20 r12 r22 So 0.62  20  1.22 t  t  160 s E 43

JEE-Physics Example A telescope consisting of an objective of focal length 60 cm and a single–lens eyepiece of focal length 5 cm is focussed at a distant object in such a way that parallel rays emerge from the eye piece. If the object subtends an angle of 2° at the objective, then find the angular width of the image. Solution MP  f0       f0  2  60  24 fe  fe 5 Example The focal lengths of the objective and the eye piece of an astronomical telescope are 60 cm and 5 cm respectively. Calculate the magnifying power and the length of the telescope when the final image is formed at (i) infinity, (ii) least distance of distinct vision (25 cm) Solution (i) When the final image is at infinity, then : f0 60 MP = – fe =– = – 12 and length of the telescope is L = f + f = 60 + 5 = 65 cm 5 0e (ii) For least distance of distinct vision, the magnifying power is : MP   f0 1  fe    60 1  5  12  6  1 4 . 4 fe D  5 25  5 1 1 1 1 1 1 1 1 1        5  Now fe ve ue 5 25 ue ue 25 u = – 4.17 cm  |u | = 4.17 cm e e The length of telescope in this position is L = f + |u | = 60 + 4.17 = 64.17 cm 0e LENS – CAMERA There is a convex lens whose aperture and distance from the film can be adjusted. Object is real and placed between  and 2F, so the image is real, inverted diminished and between F and 2F. O F 2F aperture film 2F F I shutter image If I is the intensity of light, S is the light transmitting area of lens and t is the exposure time, NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 then for proper exposure,I × S × t = constant light transmitting area of a lens is proportional to the square of its aperture D ; I × D2 × t = constant If aperture is kept fixed, for proper exposure, I × t = constant, i.e., I t = I t 11 22 If intensity is kept fixed, for proper exposure, D2 × t = constant 1 Time of exposure  (aperture)2 ... (i) The ratio of focal length to the aperture of lens is called f–number of the camera, focal length 1 f–number = aperture   Aperture  f  number ... (ii) From equation (i) and (ii)   Time of exposure  (f–number)2 42 E

JEE-Physics h' MP  visual angle with instrument ()  MP  f0   f0 A B   h visual angle for unaided eye () h ' ue ue (i) If the final image is at infinity v = –  , u = –ve e e 1 11 So MP   f0 and length of the tube L = f + f   ue  fe  ue  fe . fe 0e (ii) If the final image is at D : v = –D u = –ve e e 1 1 1 1 1 1  1 1  fe  So MP   f0   f0 1  fe  D      fe D  ue fe D  ue fe ue fe D Length of the tube is L  f0  u e S. S. Astronomical – Telescope Compound – Microscope No. No. 1 . It is used to increase visual angle of near 1 . It is used to increase visual angle of distant tiny object. large objects. 2 . In it field and eye lens both are 2 . In it objective lens is of large focal length convergent, of short focal length and and aperture while eye lens of short focal aperture. length and aperture and both are convergent. 3 . Final image is inverted, virtual and 3 . Final image is inverted, virtual and enlarged enlarged and at a distance D to  from at a distance D to  from the eye. the eye. 4 . MP does not change appreciably if 4 . MP becomes (1/m2) times of its initial value objective and eye lens are interchanged if objective and eye–lenses are as [MP ~ (LD / f0 fe)] interchanged as MP ~ [f0 / fe] 5 . MP is increased by decreasing the focal 5 . MP is increased by increasing the focal length of both the lenses. length of objective lens and by decreasing the focal length of eyepiece 6 . RP is increased by decreasing the 6 . RP is increased by increasing the aperture wavelength of light used. of objective.  RP  2 sin   RP  D   1.22  Example A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope ? What is the separation between the objective and the eyepiece? When final image is formed at infinity. Solution NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 Here, f = 144 cm; f = 6.0 cm, MP = ?, L = ? 0e MP = f0  144  24 and L = f + f = 144 + 6.0 = 150.0 cm fe 6.0 0e Example Diameter of the moon is 3.5 × 103 km and its distance from earth is 3.8 × 105 km. It is seen by a telescope whose objective and eyepiece have focal lengths 4m and 10cm respectively. What will the angular diameter of the image of the moon. Solution MP   f0 400  40 . Angle subtended by the moon at the objective 3.5  103 = 0.009 radian. fe  = 3.8  105 10 Thus angular diameter of the image = MP × visual angle = 40 × 0.009 = 0.36 radian = 0.36  180  21 3.14 E 41

JEE-Physics 1 11 111  ue  fe     (ii) When final image is formed at infinity ve ue fe  ue fe MP  v0 D   f0 D   f0  v0 D   h2 D  u0    u0   f0   h1    fe  f0  fe   fe   fe  . Length of the tube L = v + f 0 e Sign convention for solving numerical u = –ve, v = +ve, f = +ve, 0 00 u = –ve, v = –ve, f = +ve, m = –ve, m = +ve, M = –ve e ee 0 e Example A thin convex lens of focal length 5 cm is used as a simple microscope by a person with normal near point (25 cm). What is the magnifying power of the microscope ? Solution D 25 Here, f = 5 cm; D = 25 cm, M = ? MP  1   1   6 f5 Example A compound microscope consists of an objective lens of focal length 2.0 cm and an eye piece of focal length 6.25 cm, separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm) (b) infinity ? Solution Here, f = 2.0 cm; f = 6.25 cm, u = ? 0e 0 1 1 1 1 1 1 1 1 1  4 5          (a) v = – 25 cm ve ue fe  ue ve fe 25 6.25 25 25  u = – 5 cm e e As distance between objective and eye piece = 15 cm; v = 15 – 5 = 10 cm 0 1 1  1 1  1 1 1 1  1 5  u0  10  2.5 cm v0  f0  v0  10 2 10 4 u0 u0 f0 v 0   D  10 1  25   20 u 0 1  2.5 6.25  Magnifying power = | |  fe  (b)  v = , u = f = 6.25 cm  v = 15 – 6.25 = 8.75 cm. e ee 0 1 1  1  1  1 1  1 1  2  8.75  u0  17.5  2.59cm v0  f0 u0 v0  8.75  17.5 6.75 u0 f0 20 v0   D  v0  D 8.75 25  13.51 | u0| 1   u0 ue  Magnifying power =  fe  2.59 6.25 ASTRONOMICAL TELESCOPE f0 NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 fe  Fe F0 R A\"  A'   O B' ue B\" ve A telescope is used to see distant object, objective lens forms the image A'B' at its focus. This image A'B' acts as a object for eyepiece and it forms final image A\"B\". 40 E

JEE-Physics Example A man with normal near point 25 cm reads a book with small print using a magnifying glass, a thin convex lens of focal length 5 cm. (a) What is the closest and farthest distance at which he can read the book when viewing through the magni- fying glass ? (b) What is the maximum and minimum MP possible using the above simple microscope ? Solution (a) As for normal eye far and near point are  and 25 cm respectively, so for magnifier vmax= – and 11 1 f v  u  f  u  / v 1 v = –25 cm. However, for a lens as min So u will be minimum when v = minimum = –25 cm i.e. u min  5   25  4.17cm 6 5 / 25 1 Ans u will be maximum when v = maximum =  i.e., u = 5 = – 5 cm max  5 1    So the closest and farthest distance of the book from the magnifier (or eye) for clear viewing are 4.17 cm and 5 cm respectively. (b) As in case of simple magnifier MP = (D/u). So MP will be minimum when u = max = 5 cm  MP min  2 5  5  D and MP will be maximum when u = min = (25/6) cm 5    MP max  25  6  1  D   25 / 6 COMPOUND MICROSCOPE Compound microscope is used to get more magnified image. Object is placed infront of objective lens and image is seen through eye piece. The aperture of objective lens is less as compare to eye piece because object is very near so collection of more light is not required. Generally object is placed between F – 2F due to this a real inverted and magnified image is formed between 2F – . It is known as intermediate image A'B'. The intermediate image act as a object for eye piece. Now the distance between both the lens are adjusted in such a way that intermediate image falls between the optical centre of eye piece and its focus. In this condition, the final image is virtual, inverted and magnified. eye lens object lens object F0 A' C h C A\" A F0 NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 B final image B\" Total magnifying power = Linear magnification × angular magnification MP = m m = v0 D 0e u0 ue (i) When final image is formed at minimum, distance of distinct vision. MP  v0   D   f0   D   f0  v0   D   h2   D u0 1 fe   u0 1 fe  f0 1 fe  h1 1 fe    f0     Length of the tube = v + |u | 0e E 39

JEE-Physics LATER AL CHROMATIC ABERR ATION As the focal–length of the lens varies from red BR BV f  lateral chromatic color to color, the magnification m =  u  f  violet aberration produced by the lens also varies from color to A hV hR color. AV Therefore, for a finite–size white object AB, the FV AR images of different colors formed by fR the lens are of different sizes. B The formation of images of different colors in different sizes is called lateral chromatic aberration. The difference in the height of the red image B A and the violet image B A  is known RR VV as lateral chromatic aberration. LCA = h – h RV A C H RO M AT I S M If two or more lens combined together in such a way that this combination produce image at a same point then this combination is known as achromatic combination of lenses.  + ' = 0  1  2  0  1   f1 fy f'y f1 f2 2 f2 111  For combination of lens. F f1 f2 (Apply sign convention in numerical) OPTICAL INSTRUMENTS Simple microscope When object is placed between focus and optical centre a virtual, magnified and erect image is formed h' h Ih  O F FC visual angle with instrument () h maximum visual angle for unaided eye ()  MP  Magnifying power (MP) = u D NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 h u D (i) When the image is formed at infinity : by lens equation 11 1 1 1 1uf So DD   MP   v u f  u f uf (ii) If the image is at minimum distance of clear vision D : 1 1 1 1 11      [v = –D and u = –ve] D u f u D f DD DD Multiplying by D both the sides  1   MP   1  uf uf 38 E

JEE-Physics Example The refractive indices of flint glass for red and violet colours are 1.644 and 1.664. Calculate its dispersive power. Solution Here, r = 1.644, v = 1.664,  = ? Now y  v  r  1.664  1.644  1.654   v  r  1.664  1.644  0.0305 2 2 y 1 1.654  1 Example In a certain spectrum produced by a glass prism of dispersive power 0.031, it was found that r = 1.645 and v = 1.665. What is the refractive index for yellow colour ? Solution Here,  = 0.031, r = 1.645 v = 1.665, y = ?   v  r  y 1  v  r  1.665 1.645  0.020  0.645  y = 0.645 + 1 = 1.645 y 1  0.031 0.31 Example A combination of two prisms, one of flint and other of crown glass produces dispersion without deviation. The angle of flint glass prism is 15°. Calculate the angle of crown glass prism and angular dispersion of red and violet. ( for crown glass = 1.52,  for flint glass = 1.65,  for crown glass 0.20,  for flint glass = 0.03). Solution Here, A = 15°, A' = ?,  = 0.03, ' = 0.02,  = 1.65, ' = 1.52, For no deviation,  + ' = 0 0.65  15 ( – 1)A + (' – 1)A' = 0  (1.65 – 1)15° + (1.52 –1)A' = 0  A' = 0.52 = –18.75° Negative sign indicates that two prisms must be joined in opposition. Net angular dispersion (v – r)A + ('v – 'r)A' = ( – 1)A + ' (' –1)A' = 0.03 (1.65 – 1)15° + 0.02 (1.52 – 1) (–18.75°) = 0.2925 – 0.195 = 0.0975° CHROMATIC ABERR ATION The image of a object in white light formed by a lens is usually white light red FV FR colored and blurred. This defect of image is called chromatic violet aberration and arises due to the fact that focal length of a O red lens is different for different colors. For a single lens fR– fY= fY violet 1     1   1  1  as  of lens is maximum for violet f  R1 R2  and NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65   while minimum for red, violet is focused nearest to the lens while red farthest from it. It is defect of lens. Longitudinal or Axial Chromatic Aberration When a white object O is situated on the axis of a lens, then images of different colors are formed at different points along the axis. The formation of images of different colors at different positions is called 'axial' or lon- gitudinal chromatic aberration. The axial distance between the red and the violet images I – I is known as RV longitudinal aberration. When white light is incident on lens, image is obtained at different point on the axis because focal length of lens depend on wavelength. f    f > f R V f – f = f y  Axial or longitudinal chromatic aberration R V If the object is at infinity, then the longitudinal chromatic aberration is equal to the difference in focal–lengths (f f ) for the red and the violet rays. RV E 37

JEE-Physics DISPERSIVE POWER () It is ratio of angular dispersion () to mean colour deviation (y ) Dispersive power       (V  R )A  V  R    V  R y (y  1)A y  1 y 1 Refractive index of mean colour y  V  R . Dispersive power depends only on the material of the prism. 2 COMBINATION OF PRISM Deviation without dispersion ( = 0°) Two or more than two thin prism are combined in such a way that deviation occurs i.e. emergent light ray makes angle with incident light ray but dispersion does not occur i.e., light is not splitted into seven colours.  Total dispersion =  =  = (V – R)A + ('V –  ' )A' white R For no dispersion = 0 ; (V – R)A + ('V – 'R)A' = 0 white light A R W V A' Therefore, A '   (V  R )A  '  R' V –ve sign indicates that prism angles are in opposite direction. Dispersion without deviation    Two or more than two prisms combine in such a way that dispersion white light A R L occurs i.e., light is splitted into seven colours but deviation do not W A' occur i.e., emergent light ray becomes parallel to incident light ray. R V Total deviation  ; –A + ('–1)A' = 0  A '   ( 1)A  ' 1 –ve sign indicates that prism angles are in opposite direction. GOLDEN KEY POINTS • Dispersive power like refractive index has no units and dimensions and depends on the material of the prism and is always positive. • As for a given prism dispersive power is constant, i.e., dispersion of different wavelengths will be different and will be maximum for violet and minimum for red (as deviation is maximum for violet and minimum for red). • As for a given prism    a single prism produces both deviation and dispersion simultaneously, i.e., a single prism cannot give deviation without dispersion or dispersion without deviation. Example White light is passed through a prism of angle 5°. If the refractive indices for red and blue colours are 1.641 and 1.659 respectively, calculate the angle of dispersion between them. NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 Solution As for small angle of prism  = ( – 1)A, B = ( 1.659 – 1) × 5° = 3.295° and R = (1.641 – 1) × 5° =3.205° so  =B – R = 3.295° – 3.205° = 0.090° Example Prism angle of a prism is 10o. Their refractive index for red and violet color is 1.51 and 1.52 respectively. Then find the dispersive power. Solution Dispersive power of prism    v  r  but y  v  r  1.52  1.51  1.515  y 1  2 2   Therefore   1.52  1.51  0.01  0.019 1.515 1 1.515 36 E

JEE-Physics Example A ray of light passes through an equilateral prism such that angle of incidence is equal of emergence and the later is equal to 3/4th of the angle of prism. Calculate the angle of deviation. Refractive index of prism is 1.5. Solution 3 A = 45°,  = ? A = 60°,  = 1.5 ; i1 = i2 = 4 A +  = i + i 60° +  = 45° + 45°  = 90° – 60° = 30° 1 2 Example A prism of refractive index 1.53 is placed in water of refractive index 1.33. If the angle of prism is 60°, calculate the angle of minimum deviation in water. (sin 35.1° = 0.575) Solution Here, a = 1.33, a = 1.53, A = 60°,  = ? w g  a g  1.53  1.15 w g  sin A  m gw m a w 1.33 2 sin A 2  sin(A  m )  w g  sin A 60  0.575  A  m = sin–1 (0.575) = 35.1° 2 2  1.15 sin 2 2  m = 35.1 × 2 – 60 = 10.2° GOLDEN KEY POINTS • Angle of prism or refracting angle of prism means the angle between the faces on which light is incident and from which it emerges. • If the faces of a prism on which light is incident and from which it emerges are parallel then the angle of prism will be zero and as incident ray will emerge parallel to itself, deviation will also be zero, i.e., the prism will act as a transparent plate. • If  of the material of the prism is equal to that of surroundings, no refraction at its faces will take place and light will pass through it undeviated, i.e.,  = 0. DISPERSION OF LIGHT When white light is incident on a prism then it is splitted into seven colours. This phenomenon is known as dispersion. Prism introduces different refractive index with different wavelength As  = (–1) A   >  So   >    >  m(red) min R V VR m(violet) ANGULAR DISPERSION NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 It is the difference of angle of deviation for violet colour and red colour Angular dispersion  = V –  R = (V – 1)A – (R – 1)A = (V – R) A  It depends on prism material and on the angle of prism  = (V – R )A A angular A dispersion  white light ray red white light ray R Y V B orange B R yellow Y green  V blue C indigo violet C E 35

JEE-Physics CONDITION OF MINIMUM DEVIATION For minimum deviation In this condition i = i = i  r = r = r and since r + r = A r + r = A  2r = A   r  A 1 2 12 1 2 2 Minimum deviation  = 2i  – A; i  A  min , A r min 2 2 if prism is placed in air  1; 1 × sin i =  sin r angle of deviation  max   A  min  A sin A  min   2  2  2  sin   sin    sin A  min 2 if angle of prism is small A < 10° then sin    i=ig i=e e=ig e=90° i=90° A  min  2  A  min  A   min = A  min = ( –1)A angle of incidence A A 2 CONDITION FOR MA XIMUM DEVIATION/GR AZING EMERGENCE  Angle of incidence (i )for grazing emergence A g For i , e = 90° g Applying Snell's law at face AC µsinr =1 × 1  sinr 1 r = sin 1  1  =c ig r1 r2 2 2 =µ ; 2  µ    r + r = A  r= A – c 1 2 1 Again, Applying Snell's law at face AB B C 1 × sin i = µsinr ; 1 × sin i = µsin(A – c) 1 g g sini = µ[sinAcosc – cosAsinc] g sin 1  µ2   sin c  1, cos c  µ2  1      as µ µ   ig  1 sin A  cos A    If i increases beyond i, r increases thus r decreases and becomes less than c and ray emerges. 1 2 g Thus i  i  ray emerges, otherwise TIR. max = i + 90° – A g g NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 NO EMERGENCE CONDITION A Let maximum incident angle on the face AB i = 90° A max sin r1 1  sin C ; 1 × sin 90° =  sin r;  r = ...(i) 1  1 C N if TIR occur at face AC then r > C ...(ii) r1 r2 2 i1 r + r = A ...(iii) r2 > C 12 A>2C from (i) and (ii) r + r > C + C  r + r > 2C ...(iv) 12 1 2 B C AA A1 1 from (iii) and (iv) A  2C  2  C  sin 2  sin C  sin 2    sin A   2 E 34

JEE-Physics PRISM A prism is a homogeneous, transparent medium (such as glass) enclosed by two plane surfaces inclined at an angle. These surfaces are called the 'refracting surfaces' and the angle between them is called the 'refracting angle' or the 'angle of prism'. The section cut by a plane perpendicular to the refracting surfaces is called the 'principal section' of the prism.  prism prism prism prism It is not a prism principal section of prism i1 for this inclination i1 equilateral prism A=0   A B A C 45° A   90° 45° B 90° B C right angled isosceles prism right angled prism DEVIATION PQ = incident ray A QR = Refracted ray RS = emergent ray A = Prism angle T K N' i = incident angle on face AB N  R 1 Q i2 S   i = emergent angle on face AC i1 r1  r2 2 P NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\02. Ray theory-Part2.p65 r = refracted angle on face AB O 1 r = incident angle on face AC 2 Angle of deviation on face AB.  i – r 1 1 Angle of deviation on face AC   i – r BC  2 2 Total angle of deviation         i1 – r) + (i – r) i1 + i – (r + r) .....(i) 1 2 2 1 2 2 In QOR r + r +  = 180° ...(ii) 1 2 In AQOR A +  = 180° ...(iii) from (ii) and (iii) r + r = A ...(iv) 12 from (i) and (iv) Total angle of deviation  = i + i –A 1 2 from Snell's law at surface AB  sin i =   sin r 1  1 and at surface AC  sin r =  sin i 2 2 E 33

JEE-Physics ROTATIONAL MOTION Till  so  far  we  have  learnt  kinematics  and  kinetics  of  translation  motion  in  which  all  the  particles  of  a  body undergo  identical  motions  i.e.  at  any  instant  of  time  all  of  them  have  equal  velocities  and  equal  accelerations and  in  any  interval  of  time  they  all  follow  identical  trajectories.  Therefore  kinematics  of  any  particle  of  a  body or  of  its  mass  center  in  translation  motion  is  representative  of  kinematics  of  the  whole  body.  But  when  a  body is  in  rotation  motion,  all  of  its  particles  and  the  mass  center  do  not  undergo  identical  motions.  Newton’s  laws  of motion,  which  are  the  main  guiding  laws  of  mechanics,  are  applicable  to  a  point  particle  and  if  applied  to  a rigid  body  or  system  of  particles,  they  predict  motion  of  the  mass  center.  Therefore,  it  becomes  necessary  to investigate  how  mass  center  and  different  particles  of  a  rigid  body  move  when  the  body  rotates.  In  kinematics of  rotation  motion  we  investigate  relations  existing  between  time,  positions,  velocities  and  accelerations  of different  particles  and  mass  center  of  a  rigid  body  in  rotation  motion. Rigid  Body A  rigid  body  is  an  assemblage  of  a  large  number  of  material  particles,  which  do  not  change  their  mutual distances  under  any  circumstance  or  in  other  words,  they  are  not  deformed  under  any  circumstance. Actual  material  bodies  are  never  perfectly  rigid  and  are  deformed  under  action  of  external  forces.  When  these deformations  are  small  enough  to  be  considered  during  their  course  of  motion,  the  body  is  assumed  a  rigid body.  Hence,  all  solid  objects  such  as  stone,  ball,  vehicles  etc  are  considered  as  rigid  bodies  while  analyzing their  translation  as  well  as  rotation  motion. To  analyze  rotation  of  a  body  relative  motion  between  its  particles  cannot  be  neglected  and  size  of  the  body becomes  a  considerable  factor.  This  is  why  study  of  rotation  motion  is  also  known  as  mechanics  of  rigid  bodies. Rotation  Motion  of  a  Rigid  Body Any  kind  of  motion  of  a  body  is  identified  by  change  in  position  or  change  in  orientation  or  change  in  both.  If  a body  changes  its  orientation  during  its  motion  it  said  to  be  in  rotation  motion. In  the  following  figures,  a  rectangular  plate  is  shown  moving  in  the  x-y  plane.  The  point  C  is  its  mass  center.  In the  first  case  it  does  not  changes  orientation,  therefore  is  in  pure  translation  motion.  In  the  second  case  it changes  its  orientation  by  during  its  motion.  It  is  a  combination  of  translation  and  rotation  motion. y  y   New  A   A  t  orientation   C  t+ t  C  B  C  C  B  t+ t  Original  t  orientation   NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 O  x  O  x    P u re  T r a n s l a t io n   Com bination of trans lation and rotation         Rotation  i.e.  change  in  orientation  is  identified  by  the  angle  through  which  a  linear  dimension  or  a  straight  line drawn  on  the  body  turns.  In  the  figure  this  angle  is  shown  by  . Example Identif y  Translation  and  rotation  motion A  rectangular  plate  is  suspended  from  the  ceiling  by  two  parallel  rods  each  pivoted at  one  end  on  the  plate  and  at  the  other  end  on  the  ceiling.  The  plate  is  given  a  side- push  to  oscillate  in  the  vertical  plane  containing  the  plate.  Identify  motion  of  the plate  and  the  rods. E1

JEE-Physics Solution. Neither  of  the  linear  dimensions  of  the  plate  turns  during  the  motion.  Therefore,  the  plate  does  not  change  its orientation.  Here  edges  of  the  body  easily  fulfill  our  purpose  to  measure  orientation;  therefore,  no  line  is  drawn on  it. The  plate  is  in  curvilinear  translation  motion  and  the  rods  are  in  rotation  motion. Types  of  Motions  involving  Rotation Motion  of  body  involving  rotation  can  be  classified  into  following  three  categories. I Rotation  about  a  fixed  axis. I I Rotation  about  an  axis  in  translation. III Rotation  about  an  axis  in  rotation Rotation  about  a  fixed  axis Rotation  of  ceiling  fan,  potter’s  wheel,  opening  and  closing  of  doors  and  needles  of  a  wall  clock  etc.  come  into this  category. When  a  ceiling  fan  it  rotates,  the  vertical  rod  supporting  it  remains  stationary  and  all  the  particles  on  the  fan move  on  circular  paths.  Circular  path  of  a  particle  P  on  one  of  its  blades  is  shown  by  dotted  circle.  Centers  of circular  paths  followed  by  every  particle  are  on  the  central  line  through  the  rod.  This  central  line  is  known  as axis  of  rotation  and  is  shown  by  a  dashed  line.  All  the  particles  on  the  axis  of  rotation  are  at  rest,  therefore  the axis  is  stationary  and  the  fan  is  in  rotation  about  this  fixed  axis. C e il in g  F a n    D o o r    P              Axis of rotation  Axis of rotation  A  door  rotates  about  a  vertical  line  that  passes  through  its  hinges.  This  vertical  line  is  the  axis  of  rotation.  In  the figure,  the  axis  of  rotation  is  shown  by  dashed  line. Axis  of  rotation Axis of rotation  r  An  imaginary  line  perpendicular  to  plane  of  circular  paths  of  particles          of  a  rigid  body  in  rotation  and  containing  the  centers  of  all  these  circular P  NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 paths  is  known  as  axis  of  rotation. A  It  is  not  necessary  that  the  axis  of  rotation  pass  through  the  body. B  Consider system  shown in the figure, where  a block is fixed on a rotating disk.  The  axis  of  rotation  passes  through  the  center  of  the  disk  but  not Q  Axis of rotation   E through  the  block. Important  observations Let  us  consider  a  rigid  body  of arbitrary shape  rotating about  a  fixed                                   axis  PQ  passing  through  the  body.  Two  of  its  particles  A  and  B  shown are  moving  on  their  circular  paths.  All  of  its  particles,  not  on  the  axis  of  rotation,  move  on  circular paths  with  centers  on  the  axis  or  rotation.  All  these  circular  paths are  in  parallel  planes  that  are  perpendicular  to  the  axis  of  rotation. 2

JEE-Physics  All  the  particles  of  the  body  cover  same  angular  displacement  in  the  same  time  interval,  therefore  all  of them  move  with  the  same  angular  velocity  and  angular  acceleration.  Particles  moving  on  circular  paths  of  different  radii  move  with  different  speeds  and  different  magnitudes of  linear  acceleration.  Furthermore,  no  two  particles  in  the  same  plane  perpendicular  to  the  axis  of rotation  have  same  velocity  and  acceleration  vectors.  All  the  particles  on  a  line  parallel  to  the  axis  of  rotation  move  circular  paths  of  the  same  radius  therefore have  same  velocity  and  acceleration  vectors.  Consider  two  particles  in  a  plane  perpendicular  to  the  rotational  axis.  Every  such  particle  on  a  rigid  body in  rotation  motion  moves  on  circular  path  relative  to  another  one.  Radius  of  the  circular  path  equals  to the  distance  between  the  particles.  In  addition,  angular  velocity  and  angular  acceleration  equals  to  that  of rotation  motion  of  the  body. Rotation  about  an  axis  in  translation Rotation  about  an  axis  in  translation  includes  a  broad  category  of  motions.  Rolling  is  an  example  of  this  kind  of motion.  A  rod  lying  on  table  when  pushed  from  its  one  end  in  its  perpendicular  direction  also  executes  this  kind of  motion.  To  understand  more  let  us  discuss  few  examples. Consider  rolling  of  wheels  of  a  vehicle,  moving  on  straight  level  road.  Relative C  to  a  reference  frame,  moving  with  the  vehicle  wheel  appears  rotating  about  its stationary  axel.  The  rotation  of  the  wheel  from  this  frame  is  rotation  about fixed  axis.  Relative  to  a  reference  frame  fixed  with  the  ground,  the  wheel appears  rotating  about  the  moving  axel,  therefore,  rolling  of  a  wheel  is superposition  of  two  simultaneous  but  distinct  motions  –  rotation  about  the  axel fixed  with  the  vehicle  and  translation  of  the  axel  together  with  the  vehicle. Important  observations  Every  particle  of  the  body  always  remains  in  a  plane  perpendicular  to  the  rotational  axis.  Therefore,  this kind  of  motion  is  also  known  as  general  plane  motion.  Relative  to  every  particle  another  particle  in  a  plane  perpendicular  to  axis  of  rotation  moves  on  circular path.  Radius  of  the  circular  path  equals  to  the  distance  between  the  particles  and  angular  velocity  and angular  acceleration  equals  to  that  of  rotation  motion  of  the  body.  Rotation  about  axis  in  translation  is  superposition  of  pure  rotation  about  the  axis  and  simultaneous translation  motion  of  the  axis. Rotation  about  an  axis  in  rotation. NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 In  this  kind  of  motion,  the  body  rotates  about  an  axis  that  also  rotates  about  some  other  axis.  Analysis  of rotation  about  rotating  axes  is  not  in  the  scope  of  JEE,  therefore  we  will  discus  it  to  have  an  elementary  idea only. As  an  example  consider  a  rotating  top.  The  top  rotates  about                    Rotation about  Precession of the  its  central  axis  of  symmetry  and  this  axis  sweeps  a  cone  about  a central axis  central axis  vertical  axis.  The  central  axis  continuously  changes  its  orientation, therefore  is  in  rotation  motion.  This  type  of  rotation  in  which  the axis  of  rotation  also  rotates  and  sweeps  out  a  cone  is  known  as precession. Another  example  of  rotation  about  axis  in  rotation  is  a  table-fan  swinging  while  rotating.  Table-fan  rotates about  its  horizontal  shaft  along  which  axis  of  rotation  passes.  When  the  rotating  table-fan  swings,  its  shaft rotates  about  a  vertical  axis. E3

JEE-Physics Angular  displacement,  angular  velocity  and  angular  acceleration Rotation  motion  is  the  change  in  orientation  of  a  rigid  body  with        t =  0  N e w  o rie n ta tio n    time.It  is  measured  by  turning  of  a  linear  dimension  or  a  straight A  line  drawn  on  the  body. t   A  In  the  figure  is  shown  at  two  different  instants  t  0   and  t  a B  B  rectangular  plate  moving  in  its  own  plane.  Change  in  orientation Original orientation  during  time  t  equals  to  the  angle    through  which  all  the  linear dimensions  of  the  plate  or  a  line  AB  turns. If  the  angle    continuously  changes  with  time  t,  instantaneous  angular  velocity    and  angular  acceleration    for rotation  of  the  body  are  defined  by  the  following  equations.   d [1] dt   d 2  d   d [2] dt 2 dt d Direction  of  angular  motion  quantities Angular  displacement,  angular  velocity  and  angular  acceleration  are  known  as  angular  motion  quantities. Infinitesimally  small  angular  displacement,  instantaneous  angular  velocity  and  angular  acceleration  are  vector quantities.  Direction  of  infinitesimally  small  angular  displacement  and  instantaneous  angular  velocity  is  given  by the  right  hand  rule.  For  a  disk  rotating  as  shown  in  the  figure,  the  angular  velocity  points  upwards  along  the axis  of  rotation.   d   Axis of rotation  Axis of rotation           The  direction  of  angular  acceleration  depends  on  whether  angular  velocity  increases  or  decreases  with  time. For  increasing  angular  velocity,  the  angular  acceleration  vector  points  in  the  direction  of  angular  velocity  vector and  for  decreasing  angular  velocity,  the  angular  acceleration  vector  points  opposite  to  the  angular  velocity vector.  NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65                      An g u l a r  a c ce le r a t io n : In c r e a s in g  a n g u l a r  s p e e d   An g u l a r  a c ce le r a t io n : D e c r e a s in g  a n g u l a r  s p e e d   In  rotation  about  fixed  axis  and  rotation  about  axis  in  translation,  the  axis  of  rotation  does  not  rotate  and angular  velocity  and  acceleration  always  point  along  the  axis  of  rotation.  Therefore,  in  dealing  these  kinds  of motions,  the  angular  motion  quantities  can  used  in  scalar  notations  by  assigning  them  positive  sign  for  one direction  and  negative  sign  for  the  opposite  direction. 4E

JEE-Physics These  quantities  have  similar  mathematical  relations  as  position  coordinate,  velocity,  acceleration  and  time have  in  rectilinear  motion.  A  body  rotating  with  constant  angular  velocity   and  hence  zero  angular  acceleration  is  said  to  be  uniform rotation.  Angular  position    is  given  by  equation   o  t [3]  Thus  for  a  body  rotating  with  uniform  angular  acceleration  ,  the  angular  position    and  angular  velocity   can  be  expressed  by  the  following  equation.   o  t [4]   1 t 2 1 t  o  ot  2  o  2 o  [5]  2 2   o  2   o [6] Angular  motion  quantities  in  rotation  and  assumption  of  axis  of  rotation Rotation  is  identified  by  change  in  orientation,  which  is  measured  by  turning  of  a  linear  dimension  of  the  body or  a  line  drawn  on  the  body.  It  remains  unchanged  relative  to  all  inertial  frames. Therefore,  if  we  assume  axis  of  rotation  anywhere  but  parallel  to  the  original  one,  angular  displacement, angular  velocity  and  angular  acceleration  of  rotation  motion  remain  the  same. Example A  wheel  is  rotating  with  angular  velocity  2  rad/s.  It  is  subjected  to  uniform  angular  acceleration  2.0  rad/s2. (a)  How  much  angular  velocity  does  the  wheel  acquire  after  10  s? (b)  How  many  complete  revolution  it  makes  in  this  time  interval? Solution. The  wheel  is  in  uniform  angular  acceleration,  therefore  from  eq.  [4]   o  t  Substituting  values  of  o,    and  t,  we  have   2  2  10  22   rad/s From  eq.[5],  we  have   o 1 o  t  Substituting  o  0   for  initial  position,  and  o from  above  equation,  we  have 2   0  1 2  1010  60   rad. 2 In  one  revolution,  the  wheel  rotates  through  2  radians.  Therefore,  number  of  complete  revolutions  n  is NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 n    60  2 2 Example A  disk  rotates  about  a  fixed  axis.  Its  angular  velocity    varies  with  time  according  to  equation    at  b .  At the  instant  t  =  0  its  angular  velocity  is  1.0  rad/s  and  at  angular  position  is  2  rad  and  at  the  instant  t  =  2  s, angular  velocity  is  5.0  rad/s.  Determine  angular  position    and  angular  acceleration    when  t  =  4  s. Solution. The  given  equation    at  b has  form  similar  to  eq.[4],  therefore  motion  is  rotation  with  uniform  angular acceleration. Initial  angular  velocity  =  o  b  1.0 rad/s Angular  acceleration   a Substituting  these  values  in  eq.[5],  we  get   1 at2  0 t  c 2 Since  at  t  =  0,    =  1.0  rad/s,  we  obtain  the  constant  c. Initial  angular  position  =  o  c  2.0 rad E5

JEE-Physics Since  at  t  =  2.0  s  angular  velocity  is  5.0  rad/s,  from  given  expression  of  angular  velocity,  we  have   at  b  Substituting  b  =  1.0  rad/s,  t  =  2.0  s  and    =  5.0  rad/s,  we  have a  2.0 rad/s2 Now  we  can  write  expressions  for  angular  position,  angular  velocity  and  angular  acceleration.   t 2  t  2.0 (1)   2.0t  1.0 (2) From  the  above  equations,  we  can  calculate  angular  position,  angular  velocity  and  angular  acceleration  at t  =4.0  s 4  22 rad,  4  9.0 rad/s,    2.0 rad/s2 Example   An  early  method  of  measuring  the  speed  of  light  makes  use  of  a  rotating  slotted  wheel.  A  beam  of  light  passes through  slot  at  the  outside  edge  of  the  wheel,  as  shown  in  figure  below,  travels  to  a  distant  mirror,  and  returns to  the  wheel  just  in  time  to  pass  through  the  next  slot  in  the  wheel.  One  such  slotted  wheel  has  a  radius  of 5.0  cm  and  500  slots  at  its  edge.  Measurements  taken  when  the  mirror  is  L  =  500  m  from  the  wheel  indicate a  speed  of  light  of  3.0  ×  105  km/s. (a) What  is  the  (constant)  angular  speed  of  the  wheel  ? (b) What  is  the  linear  speed  of  a  point  on  the  edge  of  the  wheel? L Light Light beam source Mirrtoor l ipgehrtp beenadmicular Rotating slotted wheel Solution During  the  time  light  goes  from  the  wheel  to  the  mirror  and  comes  back  again,  the  wheel  turns  through (a) an  angle  of      2 2 2 500m (b) =  1.26  ×  10–2 rad.  That  time  is  t  =  = 2.998 x108 m / s =  3.34  ×  10–4  s 500 c So  the  angular  speed  of  the  wheel  is     1.26 x102 rad   =  3.8  ×  103  rad/s NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 t = 3.34 x106 s Linear  speed  of  a  point  on  the  edge  of  a  wheel  v=r  =  3.8  ×  103  ×  0.05  =  1.9  ×  102  m/s Example A  pulsar  is  rapidly  rotating  neutron  star  that  emits  a  radio  beam  like  a  lighthouse  emits  a  light  beam.  We  receive a  radio  pulse  for  each  rotation  of  the  star.  The  period  T  of  rotation  is  found  by  measuring  the  time  between pulses.  The  pulsar  in  the  Crab  nebula  has  a  period  of  rotation  of  T  =  0.033  s  that  is  increasing  at  the  rate of  1.26  x  10–5  second/year. (a) What  is  the  pulsar's  angular  acceleration? (b) If  its  angular  acceleration  is  constant,  how  many  years  from  now  will  the  pulsar  stop  rotating? (c) The  pulsar  originated  in  a  supernova  explosion  seen  in  the  year  1054.  What  was  the  initial  T  for  the  pulsar? (Assume  constant  angular  acceleration  since  the  pulsar  originated.). 6E

JEE-Physics Solution (a) The  angular  velocity  in  rad/s     =  2 / T . d 2 dT The  angular  acceleration  =    dt   =  – T2   dt dT 1.26  105 s / y For  the  pulsar  described  dt   =  3.16  107 s / y   =  4.00  ×  10–13     2   So  0 .0 3 3 s 2  4.00  10 13   =  –  2.3  ×  10–3  rad/s2   The  negative  sign  indicates  that  the  angular  acceleration  is  opposite  the  angular  velocity  and  the  pulsar  is slowing  down. (b)    0   + t   for  the  time  t  when  =0.  2 2  t  =  –     =– aT =  –  2.3  109 red / s2 0.033s   =  8.3  ×1010  s  .    This  is  about  2600  years. (c) The  pulsar  was  born  1992–  1054  =  938  years  ago. This  is  equivalent  to  (938  y)  (3.16  ×  107  s/y)  =  2.96  x  1010  s.  Its  angular  velocity  was  then 2 2   0  t   =   t   =     (–2.3  x  10–9  rad/s2)  (–2.96  x  1010s)  =  258  rad/s. T 0.033s 2 Its  period  was     T  =    =    2.4  ×  10–2  s.  Example A  turn  table  is  rotating  in  a  horizontal  plane  about  the  vertical  axis  passing  through  its  centre  with  an  angular velocity  20  rad/s.  It  carries  upon  it  a  flywheel  rotating  with  an  angular  velocity  40  rad/s  about  a  horizontal axle  mounted  in  bearings.  Find  the  angular  velocity  of  the  wheel  as  seen  by  an  observer  in  the  room. Solution As the  axis  of  the  turn  table  is  vertical its  angular  velocity     is  directed vertical. The  axis  of  flywheel is  horizontal r   therefore  its  angular  velocity  F  is  directed  horizontal,  hence  the  resultant  angular  velocity  is  R  F  T  2F  2T  402  202   =  20 5   rad/s |R|   T R R   lies  in  a  plane  which  makes  an  angle    with  the  horizontal plane,  given  by    =  tan–1   T   tan 1 1   F   2  F z  Kinematics  of  rotation  about  fixed  axis NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 In  figure  is  shown  a  rigid  body  of  arbitrary  shape  rotating  about  the  z- axis.  In  the  selected  frame  (here  the  coordinate  system)  all  the  three  axes  are  at  rest,  therefore  the  z-axis  that  is  the  axis  of  rotation  is  at  rest  and  the  body  is  in  fixed  axis  rotation.  All  of  its  particles  other  than    r C    those  on  the  z-axis  move  on  circular  paths  with  their  centers  on  the  z-   v axis.  All  these  circular  paths  are  parallel  to  the  x-y  plane.  In  the  figure, s  P   x  one  of  its  particles  P  is  shown  moving  with  velocity  v   on  a  circular  path   of  radius  r  and  center  C.  Its  position  vector  is  R .  It  were  at  the  line  Cx R  at  t  =  0  and  at  the  position  shown  at  the  instant  t.  During  time  interval t,  it covers the  circular arc  of length s and its  radius  vector  turns through O  x  angle   y  In  an  infinitesimally  small  time  interval  dt  let,  the  particle  covers E7

JEE-Physics infinitesimally  small  distance  ds  along  its  circular  path.                         [7] ds  d  r  d  R                               [8] v ds d r d R dt dt dt From  eq.  [7]  and  [8]  we  have v                             [9]  r  R The  above  equation  tells  us  the  relation  between  the  linear  and  angular  velocity.  Now  we  explore  relation between  the  linear  and  angular  accelerations.  For  the  purpose,  differentiate  the  above  equation  with  respect to  time.                   [10] a dv d r  dr  r  v dt dt dt The  first  term  on  the  RHS  points  along  the  tangent  in  the  direction  of  the  velocity  vector  and  it  is  known  as  tangential  acceleration  aT   same  as  we  have  in  circular  motion.  In  addition,  the  second  term  point  towards  the  center  C.  It  is  known  as  centripetal  acceleration  or  normal  component  an   of  acceleration  same  as  in  circular motion.  Now  we  have   [11] Tangential  acceleration aT    r Normal  acceleration     2  [12] an  v r How  to  Locate  Axis  of  Rotation B  vA    OA   vD    OD   Every  particle  in  a  plane  perpendicular  to  the  axis  of  rotation    move  with  different  velocities  and  accelerations,  moreover,  they  D  all  have  the  same  angular  velocity  and  angular  acceleration. A  Such  a  section  of  a  body  in  rotation  is  shown  here.  The  particles C  A,  B  and  C  at  equal  distance  from  the  axis  of  rotation  move  with O  equal  speeds  vA  and  the  particle  D  moves  with  speed  vD  on concentric  circular  paths.  The  location  of  rotational  axis  can  be determined  by  any  of  the  two  graphical  techniques.  Lines  perpendicular  to  velocity  vectors  and  passing  through the  particles,  whose  velocity  vectors  are  neither  parallel nor  antiparallel  intersect  at  the  axis  of  rotation.  See  pairs of  particles  A  and  B,  B  and  C  and  B  and  D.  Lines  perpendicular  to  velocity  vectors  and  passing  through  the  particles,  whose  velocity  vectors  are either  parallel  or  antiparallel,  coincide  and  intersect  the  line  joining  tips  of  their  velocity  vectors  at  the axis  of  rotation.  Refer  pairs  of  particles  A  and  C,  A  and  D  and  C  and  D. Example A  belt  moves  over  two  pulleys  A  and  B  as  shown  in  the  figure.  The  pulleys   are  mounted  on  two  fixed NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 horizontal  axels.  Radii  of  the  pulleys  A  and  B  are  50  cm  and  80  cm  respectively.  Pulley  A  is  driven  at  constant angular  acceleration  0.8  rad/s2  until  the  pulley  B  acquires  an  angular  velocity  of  10  rad/s.  The  belt  does  not slide  on  either  of  the  pulleys. A  C  B  (a)  Find  acceleration  of  a  point  C  on  the  belt  and  angular  acceleration  of  the  pulley  B. E (b)  How  long  after  the  pulley  B  achieve  angular  velocity  of  10  rad/s. 8

JEE-Physics Solution. Since  the  belt  does  not  slide  on  the  pulleys,  magnitude  of  velocity  and  acceleration  of  any  point  on  the  belt  are same  as  velocity  and  tangential  acceleration  of  any  point  on  periphery  of  either  of  the  pulleys. Using  the  above  fact  with  eq.[11],  we  have   aC  A rA  B rB aT    r  Substituting  rA  =  0.5  m,  rB  0.8 m  and    =  0.8  rad/s2,  we  have aC  =  0.4  m/s2  and  B  aC  ArA  0.5 rad/s2 rB rB From  eq.  [4],  we  have   o  t  t  B  Bo B Substituting  Bo  0 ,  B  10 rad/s  and  B  0.5 rad/s2, we  have  t  =  20  s Kinematics  of  rotation  about  axis  in  translation y  C B In  this  kind  of  motion,  the  body  rotates  about  an  axis  and  the  axis v  x  moves  without  rotation.  Rolling  is  a  very  common  example  of  this  kind O  A of  motion. A  As  an  example  consider  a  rod  whose  ends  A  and  B  are  sliding  on  the x  and  y-axis  as  shown  in  the  figure.  Change  in  its  orientation  measured by change in angle     indicates that the  rod is in  rotation.  Perpendiculars drawn  to  velocity  vector  of  its end points intersect  at  the axis of  rotation, which  is  continuously  changing  it  position. Instantaneous  Axis  of  Rotation  (IAR) It  is  a  mathematical  line  about  that  a  body  in  combined  translation  and  rotation  can  be  conceived  in  pure rotation  at  an  instant.  It  continuously  changes  its  location. Now  we  explore  how  the  combined  translation  and  rotational  motion  of  the  rod  is  supperposition  of  translation motion  of  any  of  its  particle  and  pure  rotation  about  an  axis  through  that  particle. Consider  motion  of  the  rod  from  beginning  when  it  was  parallel  to  the  y-axis.  In  the  following  figure  translation motion  of  point  A  is  superimposed  with  pure  rotation  about  A. NODE6 E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-3\\Rotation\\Eng\\Theory.p65 P u r e   t r a n s l at i o n   of   A   P u re r o ta tio n  ab ou t A   Com bined translation and rotation   B vA y  y  y    B     B  v BA    AB      v BA    AB         vA vB     v      vA A vA O  A  x  O  A  x  O  A  x  E9