Comprehensive CBSE Question Bank in Physics XII (Term-II)

CBSE II Question Bank in Physics CLASS 12 Features Short Answer Type Questions Long Answer Type Questions Strictly Based on the Latest CBSE Term-wise Syllabus Case Study Based MCQs Chapter Summary Very Short Answer Type Questions



Comprehensive CBSE Question Bank in Physics Term–II (For Class-XII)



Comprehensive CBSE Question Bank in Physics Term–II (For Class-XII) (According to the Latest CBSE Examination Pattern) By NARINDER KUMAR M.Sc. PES(I) Formerly, Senior Lecturer Department of Physics S.D. Govt. College, Ludhiana Punjab   laxmi Publications (P) Ltd (An iso 9001:2015 company) bengaluru • chennai • guwahati • hyderabad • jalandhar Kochi • kolkata • lucknow • mumbai • ranchi new delhi

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Contents Unit V: Electromagnetic Waves 1. Electromagnetic Waves (NCERT Textbook Chapter 8) ...................................... 1–35 Unit VI: Optics 2. Ray Optics and Optical Instruments (NCERT Textbook Chapter 9) ............. 36–118 3. Wave Optics (NCERT Textbook Chapter 10) ................................................ 119–163 Unit VII: Dual Nature of Radiation and Matter 4. Dual Nature of Radiation and Matter (NCERT Textbook Chapter 11) ....... 164–227 Unit VIII: Atoms and Nuclei 5. Atoms (NCERT Textbook Chapter 12) ........................................................... 228–264 6. Nuclei (NCERT Textbook Chapter 13) ........................................................... 265–282 Unit IX: Electronic Devices 7. Semiconductor Electronics (NCERT Textbook Chapter 14) ......................... 283–308



Syllabus Class XII (Code N. 042) (2021-22) Term–II (Theory) Time: 2 Hours Max Marks: 35 No. of Periods Marks Unit–V Electromagnetic Waves 02 Unit-VI Chapter–8: Electromagnetic Waves Unit-VII Optics 17 Unit-VIII Chapter–9: Ray Optics and Optical Instruments 18 Unit-IX Chapter–10: Wave Optics Dual Nature of Radiation and Matter 07 Chapter–11: Dual Nature of Radiation and Matter Atoms and Nuclei 11 Chapter–12: Atoms 11 Chapter 13: Nuclei Electronic Devices 07 07 Chapter–14: Semiconductor-Electronics: Materials, Devices and Simple Circuits Total 45 35 Unit V: Electromagnetic waves 2 Periods Chapter–8: Electromagnetic Waves Electromagnetic waves, their characteristics, their Transverse nature (qualitative ideas only). Electromagnetic spectrum (radio waves, microwaves, infrared, visible, ultraviolet, X-rays, gamma rays) including elementary facts about their uses. Unit VI: Optics 18 Periods Chapter–9: Ray Optics and Optical Instruments Ray Optics: Refraction of light, total internal reflection and its applications, optical fibers, refraction at spherical surfaces, lenses, thin lens formula, lensmaker’s formula, magnification, power of a lens, combination of thin lenses in contact, refraction of light through a prism. Optical instruments: Microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers.

Chapter–10: Wave Optics Wave optics: Wave front and Huygen’s principle, reflection and refraction of plane wave at a plane surface using wave fronts. Proof of laws of reflection and refraction using Huygen’s principle. Interference, Young’s double slit experiment and expression for fringe width, coherent sources and sustained interference of light, diffraction due to a single slit, width of central maximum. Unit VII: Dual Nature of Radiation and Matter 7 Periods Chapter–11: Dual Nature of Radiation and Matter Dual nature of radiation, Photoelectric effect, Hertz and Lenard’s observations; Einstein’s photoelectric equation-particle nature of light. Experimental study of photoelectric effect Matter waves-wave nature of particles, de-Broglie relation Unit VIII: Atoms and Nuclei 11 Periods Chapter–12: Atoms Alpha-particle scattering experiment; Rutherford’s model of atom; Bohr model, energy levels, hydrogen spectrum. Chapter–13: Nuclei Composition and size of nucleus Nuclear force Mass-energy relation, mass defect, nuclear fission, nuclear fusion. Unit IX: Electronic Devices 7 Periods Chapter–14: Semiconductor Electronics Materials, Devices and Simple Circuits Energy bands in conductors, semiconductors and insulators (qualitative ideas only) Semiconductor diode-I-V characteristics in forward and reverse bias, diode as a rectifier; Special purpose p-n junction diodes: LED, photodiode, solar cell.

Unit VI: Optics Chapter 2: Ray Optics and Optical Instruments (NCERT Textbook Chapter 9) SUMMARY OF THE CHAPTER  The branch of Physics which deals with the study of nature, production and propagation of light is called optics.  The branch of Physics that deals with the measurement of light energy is called photometry. This energy is called radiant energy.  The luminous flux of a source of light may be defined as the luminous or visible energy emitted per second from the source. It is denoted by . It is measured in Lumen.  The luminous intensity (I) of a light source in any direction is the luminous flux emitted by that source in a unit solid angle in that direction. Its SI unit is candela (Cd). 1 Cd = 1 lumen/steradian = 1 lmSr–1 We can write,  = 4I  The illuminance of a surface is defined as the luminous flux incident normally on a unit area of the surface. E=  A Also, E=   4I I 4r2 4r2 or E = r2  When light travelling in one medium falls on the surface of a second medium, the following three effects may occur: (i) Some part of the incident light is turned or bounced back into the first medium. This is known as reflection of light. 36

Ray Optics and Optical Instruments 37 Normal Incident ray Reflected ray Medium I Reflecting ii surface r Medium II Refracted ray (ii) Some part of the incident light gets transmitted into the second medium along a changed direction. This is called refraction of light. (iii) The remaining third part of light energy gets absorbed by the second medium. This is known as absorption of light.  The phenomenon of change in the path of light, as it goes from one medium to another, is called refraction of light. The main cause of refraction is change in the velocity of light in going from one medium to the other.  Whenever light goes from one medium to another, the frequency of light, colour of light and phase of light never change, only the velocity of light and wavelength of light change.  Snell’s law: (i) The incident ray, the refracted ray and the normal, all lie in the same plane. (ii) The ratio of sine of angle of incidence to the sine of angle of refraction is always constant, and is equal to the refractive index of the medium. sin i = 2  12 sin r 1 where 12 represents refractive index of medium 2 wrt medium 1.  The refractive index of a medium for a light of given wavelength may be defined as the ratio of the speed of light in vacuum to its speed in that medium. Speed of light in vacuum or  = c Refractive index = v Speed of light in medium Refractive index of a medium with respect to vacuum is also called absolute refractive index. The refractive index of a medium may be defined as the ratio of wavelength of light in vacuum to its wavelength in that medium.  The lens formula is a mathematical relation between the object distance u, image distance v and focal length f of a spherical lens. 11 = 1 vu f In words, we can say that: 1  1 =1 Image distance Object distance Focal length

38 Physics—XII This formula is valid for both convex and concave lenses for both real and virtual images.  According to the principle of reversibility of light, as the final path of a ray of light after any number of reflections and refractions is reversed, the ray of light retraces it entire path. ag × ga = 1 or ag = 1 g a  Consider a ray of light suffering refraction. (i) When 1 = 2, the ray of light does not bend at all, as shown in figure (a). (ii) When 1 < 2, the ray of light bends towards the normal, as shown in figure (b). (iii) When 1 > 2, the ray of light bends away from the normal, as shown in figure (c). m1 m1 m1 m2 m2 m2 (a) (b) (c)  The ratio of real depth to the apparent depth is called refractive index () of the medium.  = Real depth Apparent depth  The phenomenon of reflection of light into a denser medium from an interface of the denser medium and a rarer medium, is called total internal reflection. There are two essential conditions for total internal reflection: (i) light travels from a denser medium to a rarer medium (ii) angle of incidence in denser medium should be greater than the critical angle for the pair of media in contact.  The critical angle is defined as that angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°. The critical angle is denoted by C and its value depends upon the nature of the media in contact. ab = 1 or  = 1 sin C sin C  When a ray of light undergoes refraction, while travelling from a rarer to a denser medium at a spherical surface, it is governed by the relation, 1  2 = 2  1 uv R However, when refraction occurs from denser to rarer medium, we interchange 1 to 2 in the above relation, to get 2  1 = 2  1 uv R

Ray Optics and Optical Instruments 39  The lens marker’s formula is given by, 1 = (  1)  1  1  f  R1 R2    For convex lens, R1 is positive and R2 is negative. For concave lens, R1 is negative and R2 is positive.  The linear magnification produced by a lens is defined as the ratio of the size of the image (h2) to the size of the object. It is denoted by m. m = h2  v h1 u In case of a concave lens, m is positive, when image formed is virtual and negative, as image formed is real.  The power is taken as positive for a convex lens and for a diverging lens, power is taken as negative. The SI units of power is dioptre (D). 1 = (  1)  1  1  f  R1 R2     P= 1  (  1)  1  1 f  R1   R2  The reciprocal of the focal length expressed in metre is power in dioptre. One dioptre is the power of a lens of focal length one metre.  The Optical instruments are the devices which make use of mirrors, lenses and prisms and are primarily used to extend the range of vision of human eye. The human eye is the most remarkable instrument.  The distance between infinity and 25 cm point is called the range of normal vision.  The minimum distance from the eye, at which the eye can see the object clearly without getting any strain is called the least distance of distinct vision.  The farthest point from the eye at which an object can be seen clearly by the eye is known as far point of the eye.  The maximum variation of its power for focussing on near and distant objects is called power of accommodation.  There are mainly four common defects of vision which can be corrected by the use of suitable eye glasses. These defects are: 1. Myopia or near-sightedness. 2. Hypermetropia or far-sightedness. 3. Presbyopia. 4. Astigmatism.  A myopic eye is corrected by using a concave lens of suitable focal length. A hypermetropic eye is corrected by using a convex lens of suitable focal length.  Presbyopia is similar to hypermetropia and can be corrected by using a convex lens of suitable focal length. Astigmatism can be corrected by using a cylindrical lens.  A simple microscope or a magnifying glass is just a convex lens of short focal length, which is held close to the eye. When the final image is formed at the least distance of distinct vision, the magnifying power is given by, D m=1+ f

40 Physics—XII When the final image is formed at infinity, the magnifying power is given by, m= D f Here, D = 25 cm for a normal eye.  A compound microscope is an optical instrument, which is used for viewing highly magnified images of tiny objects. Magnifying power of a compound microscope is given by, m= v0 1  D  u0  fc   where 0 and v0 are distances of object and image respectively from the objective lens.  An astronomical telescope is an optical instrument that is used for observing distinct images of heavenly bodies like stars, planets etc. (i) Astronomical Telescope is used for observing far off objects in the sky. In normal adjustment (final image at ), angular magnification magnifying power, m = f0  fe (ii) When final image is seen at least distance of distinct vision (D), then m= f0 1  fe   fe D   The magnifying power of an astronomical telescope may be defined as the ratio of the angle subtended at the eye by the final image at the least distance of distinct vision to the angle substended at the eye by the object at infinity, when seen directly. Important Results 1. According to Snell’s law, 2 = sin i  2 1 sin r 2. The relative refractive index of medium 2 with respect to medium 1 is defined as the ratio of speed of light (v1) in medium 1 to the speed of light (v2) in medium 2 and is denoted by 12. So, 12 = v1 v2 As refractive index is the ratio of two similar physical quantities, so it has no units. 3. ag = 1 Also, aw = 1 g a w a 4. = Real depth Apparent depth

Ray Optics and Optical Instruments 41 5.  = 1 , where C is critical angle and  is refractive index of denser medium sin C w. r. t. rarer medium. 6. ab × bc × ca = 1 7. When refraction occurs from rarer to denser medium,  1  2 = 2  1 uv R 8. When refraction takes place from denser to rarer medium, we interchange 1 and 2 and get  2  1 = 1  2 uv R 9. According to lens maker’s formula 1 = ( – 1) 1  1 f    R1 R2  10. Magnification of lens, m = h2 v h1 u 11. The total magnification m equal to the product of the magnifications m1, m2 and m3 ........, produced by the individual lenses. m = m1 × m2 × m3 × ... 12. Power of a lens, P = 1  100 f (m) f (m) Also, P = 1  (  1)  1  1 f  R1   R2  13. For n thin lenses in contact, we have 1 = 1  1  1  ...  1 f f1 f2 f3 fn Also, Equivalent power, P = P1 + P2 + P3 + ... + Pn 14. In case of a simple microscope: (i) When the final image is formed at the least distance of distinct vision, the magnifying power is m=1+ D f D (ii) When the final image is formed at infinity, the magnifying power is m = . f 15. In a compound microscope, the objective is a convex lens of short focal length and small aperture, while the eyepiece is a convex lens of short focal length and large aperture. 16. When the final image is formed at the least distance D of distinct vision, the length of the compound microscope, L = v0 + ue

42 Physics—XII 17. When the final image is formed at infinity, the length of the compound microscope, L = v0 + fe. 18. In case of a compound microscope: (i) Magnifying power, m = m0 × me (ii) When the final image is formed at the least distance of distinct vision, m= v0 1  D    L 1  D u0  fe  f0    fe  (iii) When the final image is formed at infinity, m = v0 . D   L . D u0 fe f0 fe 19. In case of an Astronomical telescope: (i) In normal adjustment, m =  f0 fe Distance between objective and eyepiece = f0 + fe (ii) When the final image is formed at the least distance of distinct vision, m=  f0 1  fe  . fe D  (iii) Distance between objective and eyepiece  f0  ue  f0  fe D fe  D 20. In case of a Terrestrial telescope: (i) In normal adjustment, m = f0 fe (ii) Distance between objective and eyepiece = f0 + 4f + fe where f = focal length of the erecting lens. 21. In case of a Galileo’s telescope: (i) In normal adjustment, m = f0 fe (ii) Distance between objective and eyepiece = f0 – fe 22. In case of a Reflecting telescope: m = f0 . fe

Ray Optics and Optical Instruments 43 REFRACTION OF LIGHT VERY SHORT ANSWER TYPE QUESTIONS (1 Mark) 1. One surface of an equiconvex lens is made plane by grinding. What would be the effect on focal length and power? Ans. The focal length is doubled and the power becomes half. 2. A convex lens forms a virtual image of an object. What is the position of the object? Ans. The object lies between optical centre and principal focus of the lens. 3. What is the maximum angle of refraction when a ray of light is refracted from glass into air? Ans. 90°. 4. A ray of light is incident normally on a glass slab. What is the angle of refraction? Ans. Zero degree. 5. What is the power of the combination of a convex lens and a concave lens of the same focal length? Ans. Zero. 6. How is power of a lens related to its focal length? Ans. Power in dioptre is equal to the reciprocal of the focal length in metre. 7. Can relative refractive index of a medium w.r.t. another be less than unity? Ans. Yes. As an example, the refractive index of water w.r.t. glass is less than unity. 8. The lateral shift produced by a parallel-sided glass slab is zero. What is the angle of incidence? Ans. The angle of incidence is zero degree. 9. Name two factors on which the normal shift depends. Ans. The refractive index and the thickness of the refracting medium. 10. For which angle of incidence, a parallel sided glass slab produces maximum lateral shift. Ans. 90°. 11. Can total internal reflection take place when light travels from rarer to denser medium? Ans. No. 12. Is it possible to use a lens of refractive index n in a medium of refractive index n? Ans. No. 13. Can we get a complete image of an object if a portion of the lens is broken? Ans. Yes. The intensity of the image will be reduced. 14. A glass lens of refractive index 1.45 disappears when immersed in a liquid. What is the value of the refractive index of the liquid? Ans. 1.45 15. Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid? [CBSE 2012] Ans. The lens will act as a plane sheet when the refractive index of glass of biconvex lens is equal to the refractive index of the liquid in which lens is immersed.

44 Physics—XII 16. Do the frequency and wavelength change when light passes from a rarer to a denser medium? [CBSE 2012 Comptt.] Ans. The frequency remains unchanged but wavelength decreases. 17. If the wavelength of light incident on a convex lens is increased, how will its focal length change? [CBSE 2013 Comptt.] Ans.  increases, n decreases, f increases. 18. A ray of light of frequency of 5 × 1014 Hz is passed through a liquid. The wavelength of light measured inside the liquid is found to be 450 × 10–9 m. Calculate the refractive index of the liquid. Ans. n= c  c  5 3  108  109 = 1.33 v   1014  450 Refractive index is a number. 19. The image of an object formed by a lens on the screen is not in sharp focus. Suggest a method to get a clear focussing of the image on the screen without disturbing the positions of the object, the lens or the screen. Ans. For getting a sharp image, we have to use light of suitable wavelength. This is because focal length of a lens depends upon wavelength. 20. A glass slab is placed over a page in which letters are printed in different colours. Will the image of all the letters lie in the same plane? Ans. Since the refractive index of glass is different for different colours therefore the images of different colours will not be raised up equally. 21. To a fish under water viewing obliquely a fisherman standing on the bank of a lake, does the man look taller or shorter than what actually he is? Ans. Taller. 22. A lens of large aperture suffers from spherical aberration. Why? Ans. When the lens has a large aperture, the paraxial and marginal rays come to focus at different points on the principal axis of the lens. SHORT ANSWER TYPE–I QUESTIONS (2 Marks) 1. Does refraction of light mean a change in the direction of light propagation? If not, then what does ‘ refraction’ exactly mean? Ans. The refraction of light does not mean a change in the direction of light propagation. The direction of propagation of light can change even on reflection from a mirror. The word ‘refraction’ actually means a change in the speed of light as light travels from one medium to another. 2. What do you understand by the term ‘superior mirage’? Ans. The optical illusion of objects floating in air is called superior mirage. It is also known as looming. This occurs in very cold regions. This is due to total internal reflection. 3. What do you understand by the term ‘inferior mirage’? Ans. The optical illusion that water is present at some distant place is called inferior mirage. This generally occurs on very hot summer days. This is due to total internal reflection.

Ray Optics and Optical Instruments 45 4. Show analytically from the lens equation that when the object is at the principal focus, the image is formed at infinity. Ans. When object is at principal focus, u = – f. Now, 1 = 11 11 1 f vuv f or = 0  v =  v So, when the object is at the principal focus, the image is formed at infinity. 5. A 35 mm slide with a 24 mm × 36 mm picture is projected on a screen placed 12 m from the slide. The image of the slide picture on the screen measures 1.0 m × 1.5 m. Determine the location of the projection lens, and its focal length. Ans. Linear magnification  1000  125 24 3   u  125 u  = 12, i.e., u =  9 m, v = 375 m  3  32 32 Using 11 = 1 , f = 27.5 cm vu f The projection lens should be 28.1 cm from the slide and has focal length = 27.5 cm. 6. A man, wanting to obtain the picture of a zebra, photographs a white donkey after fitting a glass with black streaks on the lens of the camera. Can the man succeed in his purpose? Ans. No. This is because the black streaks would merely decrease the intensity of light. Thus, the brightness of the image will be reduced. 7. The refractive index of diamond has a high value (2.42). Is this fact of some use to a diamond cutter? Ans. Yes. The large value of refractive index makes critical angle small (24°). A skilled diamond cutter exploits the large range of angles of incidence (24° to 90°) to ensure that the light entering the diamond is totally reflected from many faces before getting out. This produces sparkling effect. 8. The critical angle for glass-air pair is ic. Comment on the critical angle for glass- water pair. Ans. For glass-air pair, sin ic = 1 nga If ic is the critical angle for glass-water pair, then sin ic = 1  nwa = nwa sin ic ngw nga nwa > 1  sin ic > sin ic or ic > ic. So, the critical angle for glass-water pair is greater than the critical angle for glass- air pair. 9. Mention two situations in which Snell’s law of refraction fails. Ans. (i) When light is incident normally on the surface of a refracting medium, both i and r become zero. sin i  sin r is indeterminant.

46 Physics—XII (ii) When angle of incidence in the denser medium is greater than critical angle, Snell’s law cannot be satisfied. 10. A ray of light incident on a concave lens becomes parallel to the principal axis after refraction. Show this situation with the help of a ray diagram. Ans. In the ray diagram, the incident ray is directed towards F1 the first principal focus of the concave lens. After refraction, this ray becomes parallel to the principal axis. 11. An object is placed at the principal focus of a convex lens. F C Determine the position of the image making use of lens f equation. Ans. u = – f From the lens equation, 1 1 1 or 1 =0 = v f f v  v= So, the image is formed at infinity. 12. A lens whose radii of curvature are different is forming the image of an object placed on its axis. If the surfaces of the lens facing the object and image are interchanged, will the position of the image change? Ans. No. We know that 1 = (n21  1)  1  1  . f  R1 R2    When we interchange R1 and R2, the value of f does not change except for the sign. So, the image will be formed at the same position. 13. Give three advantages of a totally reflecting prism over a plane mirror. Ans. (i) The reflection is 100%. (ii) Only one bright image is formed. This is due to the absence of multiple reflections. (iii) Silvering is not required. 14. Consider a small portion of the image on the screen, say the lowermost portion. Does this portion get illumination only from the lowermost part of the source? Or only from the uppermost part of the source? Or from all points of the source? Explain. Ans. Because the image of the slide on the screen is real and inverted, the ‘lower’ portion of the image corresponds to ‘upper’ portion of the slide. But each point of the slide is illuminated by all points of the source. Thus every portion of the screen image gets its illumination from all parts of the source. 15. A magician during a show makes a glass lens n = 1.47 disappear in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water? Ans. The refractive index of the liquid must be equal to 1.47 in order to make the lens disappear. 1 = (n21  1)  1 – 1   n21  n2  f  R1 R2   n1    

Ray Optics and Optical Instruments 47 If n1 = n2, the refractive index of the lens and the surrounding medium is the same. 1 = 0 or f = . The lens in the liquid will act like a plane sheet of glass. No, the f liquid is not water. It could be glycerine. 16. The lens shown in figure is made of two different materials. A point object is placed on the principal axis of this lens. How many images will be obtained? Ans. Since, the lens is made of two different materials therefore the lens has two different refractive indices. So, the lens would have two different focal lengths. Thus, two images will be obtained. 17. A double convex lens, made from a material of refractive index n1 , is immersed in a liquid of refractive index n2 where n2 > n1. What change, if any, would occur in the nature of the lens? Ans. 1   n1  1   1  1  , n1 1, f –ve. The lens would behave like a diverging (concave) f  n2   r1 r2  n2     lens. 18. A converging lens is kept coaxially in contact with a diverging lens—both the lenses being of equal focal lengths. What is the focal length and power of the combination? Ans. 1  1  1 = 0 or F = . Again, P = 1  1 = 0 Ff f F 19. When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a decrease in the energy carried by the light wave? Justify your answer. Ans. According to the relation E = h, the energy of light wave depends upon frequency . The frequency does not change during refraction. So, energy also remains the same. Thus, the decrease in speed does not imply a decrease in energy. 20. A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens? [CBSE 2014] O Ans. Rays coming out of lens are incident normally on plane 20 cm mirror. So, reflected rays will trace the path of incident rays. Thus image is formed on the object itself. Clearly, focal length of lens is 20 cm. 21. Using lens maker’s formula, show how the focal length of a given lens depends on the colour of light incident on it. [CBSE 2015 Comptt.] Ans. 1 = (n21 – 1) 1  1  f  R2   R1  Since the refractive index of the medium (n2) (glass) with respect to air (n1) depends on the wavelength or colour of light, therefore focal length of the lens would change with colour.

48 Physics—XII 22. The following data was recorded for values of object distance and the corresponding values of image distance in the experiment on study of real image formation by a convex lens of power +5 D. One of these observations is incorrect. Identify this observation and give reason for your choice. S. No. 1 2 34 56 Object distance (cm) 25 30 35 45 50 55 Image distance (cm) 97 61 37 35 32 30 [CBSE Sample Paper 2017] Ans. Given: A convex lens of power = +5 D, f = 100 cm = 20 cm. 5 For 3rd observation given in the data table, we find that for object distance = 35 cm (which is less than 2f), the image distance (v) must be greater than 2f(40 cm), while reading given in table is 37 cm. Hence 3rd observation is wrong. 23. For same value of angle of incidence, the angles of refraction in three media are 15°, 20°, and 25° respectively. In which medium, the velocity of light will be minimum? [CBSE Sample Paper 2017, 2012] Ans. From Snell’s law, n = sin i  c sin r v For given (i), v  sin r The minimum value of the given angle of refraction is 15°. So, the velocity of light will be minimum for the medium with 15° as the angle of refraction. 24. How long will the light take in travelling a distance of 500 m in water ? Also calculate the equivalent path. Given: Refractive index of water = 4 . 3 Ans. n = Speed of light in vacuum Speed of light in water 4 = 3  108 or v = 2.25 × 108 m s–1 3v t= 500 = 2.22 × 10–6 s 2.25  108 Equivalent path = n × distance travelled in water = 4  500 m = 666.67 m 3 25. Velocity of light in glass is 2 × 108 m s–1 and that in air is 3 × 108 m s–1. By how much would an ink dot appear to be raised when covered by a glass plate 6 cm thick? Ans. n= c  3  108 = 1.5 v 2  108 Apparent depth = Real depth = 6 cm = 4 cm n 1.5 Shift = Real depth – Apparent depth = 6 cm – 4 cm = 2 cm

Ray Optics and Optical Instruments 49 26. Calculate the value of  for which light incident normally on face AB grazes along the face BC. ng = 3 , nw = 4 . [CBSE Sample Paper 2016] 2 3 A B AB ng q q q nw C C Ans. If ic is the critical angle for face BC, then sin ic = nw = 4/3 = 8 ng 3/2 9 But  = ic  sin  = 8 or  = sin–1  8  or  = 62.7° 9  9  27. A ray PQ incident normally on the refracting face BA is refracted in the prism BAC made of material of refractive index 1.5. Complete the path of ray through the prism. From which face will the ray emerge? Justify your answer. [CBSE 2016] A A P P Q Q 30° 60° C 60° C B B Ans. sin ic = 1 1 2 = 0.6667 or ic = 41.8° n 1.5 3 For face AC, i = 30° i.e., i < ic. The ray will not suffer total internal reflection. The ray will emerge from face AC bending away from the normal. 28. Calculate the speed of light in a medium whose critical angle is 30°. Ans. n= 1  1 =2 sin ic sin 30 Also, n = c or v = c or v = 3  108 m s–1 = 1.5 × 108 m s–1 vn 2

50 Physics—XII 29. Refractive index of glass is 1.5. Calculate velocity of light in glass if velocity of light in vacuum is 3 × 108 m s–1. Also calculate critical angle for glass-air interface. Ans. n = c or v = c  3  108 m s–1 = 2 × 108 m s–1 v n 1.5 Again, sin ic = 1 1 2 = 0.6667 or ic = sin–1 (0.6667) = 41°49 n 1.5 3 30. Figure shows a rectangular glass slab of 3 cm thickness. A Rays of light from a point O on the lower surface strike the upper surface and, after reflection, outline a ic 3 cm circle of diameter 4.8 cm on the lower face. Find the refractive index of the glass slab. ic Ans. tan ic = ON  2.4 = 0.8 O NB AN 3 ic = tan–1 (0.8) = 38° 40 n = 1  1  1 = 1.6 sin ic sin 38 40 0.6248 31. Determine the critical angle for a glass-air surface, if a ray of light which is incident in air on the surface is deviated through 15°, when its angle of incidence is 40°. Ans. Clearly, r + 15° = 40° or r = 25° 40° n = sin 40  0.6428 = 1.52 r sin 25 0.4226 Again, sin ic = 1 1 = 0.6579 or ic = 41.14° 15° n 1.52 32. An optical fibre (n = 1.72) is surrounded by a glass coating (n = 1.50). Find the critical angle for total internal reflection at the fibre-glass interface. Ans. Using Snell’s law, 1.72 sin c = 1.50 sin 90° 90° or sin c = 1.50  75 1.72 86 qc or c = sin1  75  = sin–1 (0.872) = 60.69°  86  33. Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed? [CBSE 2016] Ans. R = 20 cm, n2 = 1.5, n1 = 1, u = –100 cm, v = ? n2 = n2  n1  n1 or 1.5 = 0.5  1 = 1.5 or v = 100 cm v Ru v 20 100 100 So, the real image is formed at 100 cm in the denser medium.

Ray Optics and Optical Instruments 51 34. Locate the image formed by refraction in the n1 = 1.0 n2 = 1.5 situation shown in figure. C S Ans. n2  n1 = n2  n1 25 cm 20 cm vu R 1.5  1 = 1.5  1 v  25 20 On simplification, v = – 100 cm = – 1 m So, the image will be formed, from the surface, on the side of S at a distance of 1 m. 35. What curvature must be given to the n1 = 1 n2 = 1.5 bounding surface of a refracting medium (n = 1.5) for the virtual image of an object I OP in the adjacent medium (n = 1) at 10 cm to be formed at a distance of 40 cm? Ans. n1 = 1, n2 = 1.5, u = – 10 cm, 10 cm v = – 40 cm, R = ? n2  n1 = n2  n1 40 cm vu R 1.5  1 = 1.5  1  40 10 R On simplification, R = 8 cm 36. Figure shows a solid glass sphere of radius 5 cm that has a small air bubble O trapped at a distance 2 cm from the centre C. The refractive index of the material of glass is 1.5. Find the apparent position of the bubble when viewed through the surface P I OC of the sphere from an outside point. Ans. R = – 5 cm, u = – 3 cm, n2 = 1.5, n1 = 1 n1  n2 = n1  n2 vu R 1  1.5 = 1  1.5 P IO v 3 5 On simplification, v = – 2.5 cm 37. A beam of light travelling in air strikes a glass sphere of 20 cm diameter converging towards a point 40 cm behind the pole of the spherical surface. Find the position of the image, if the refractive index of glass is 1.5. Ans. n2  n1 = n2  n1 or vu R 20 cm 40 cm 1.5  1 = 1.5  1 v 40 10 v = 20 cm 3v = 60 or

52 Physics—XII 38. An empty spherical flask of diameter 30 cm is placed in water of refractive index 4 . A parallel beam of light strikes the flask. 3 Where does it get focussed, when observed from within the flask? Ans. n1  n2 = n1  n2 vu R 1  4/3 = 1 4  1 or v = – 45 cm v  3 45 15 39. The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. If focal length of the lens is 12 cm, find the refractive index of the material of the lens. Ans. R1 = 10 cm, R2 = – 15 cm, f = 12 cm Using 1 = (n21  1)  1  1  f  R1 R2    1 = (n21  1)  1  1  or n21 = 1.5 12  10 15  40. Find the radius of curvature of the convex surface of a plano-convex lens, whose focal length is 0.3 m and the refractive index of the material of the lens is 1.5. Ans. f = 0.3 m, n21 = 1.5, R1 = , R2 = R = ? 1 = (n21  1)  1  1  f  R1 R2    1 = (1.5  1)   1 or R = – 15 cm 0.3  R  41. A convex lens is used to obtain a magnified image of an object on a screen 10 m from the lens. If the magnification is 19, find the focal length of the lens. Ans. Since the convex lens forms the image on the screen therefore the image formed is real and inverted.  m = – 19 Again, the image lies on the other side of the object.  v = + 10 m We know that f v  – 19 = f  10 or f = 0.5 m m= f f 42. The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, calculate the object and image distances. Ans. m = 4, f = 20 cm, u = ?, v = ? m= f or 4 = 20 or u = – 15 cm f u 20  u Again, m = 4 = v or v = 4u or v = – 60 cm u

Ray Optics and Optical Instruments 53 43. A convergent beam of light passes through a diverging P lens of focal length 0.2 m and comes to focus 0.3 m behind the lens. Find the position of the point at which the beam would converge in the absence of the lens. Ans. f = – 0.2 m, v = + 0.3 m, u = ? u 11 = 1 This problem is based on typical case of vu f positive value of object distance. or 1 = 1 1 or 1 1  1 u v f u 0.3  0.2 or 1 = 50 m or u = 6 m = 0.12 m u6 50 44. Two thin lenses of powers –4D and 2D are placed in contact coaxially. Find the focal length of the combination. [CBSE 2012 Comptt.] Ans. Net power, P = P1 + P2 or P = –4 D + 2 D = –2 D Focal length, f = 1  1 m = –50 cm P 2 45. A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal length 20 cm. Determine the power of combination. Will the system be converging or diverging in nature? [CBSE 2013] Ans. f1 = +25 cm, f2 = – 20 cm Peq. = P1 + P2 = 100  100 = 100  100 =–1D f1 f2 25 20 The system will be a diverging lens as it has negative power. 46. As shown in figure, a ray is incident on medium A. Find n for block C such that the ray through this medium will be parallel to the ray incident on A. A n = 1.45 B n = 1.55 Ans. Since the incident and final emergent rays are parallel therefore both must be in the same medium. So, the refractive index of block C is same as that of air. C 47. Figure shows object PQ in front of a right-angled Q A B (45° – 90° – 45°) glass prism. The critical angle of glass P C is 42°. Redraw this figure tracing the complete path of rays from P and Q into and out of the prism. P¢ Q¢ Ans. Figure shows the complete path of rays from P and Q into and out of the prism. The angles of incidence at faces AB and BC is 45° (> 42°). So, the rays suffer total internal reflection both at AB and BC. The final image PQ is an inverted image.

54 Physics—XII 48. In the previous question, which letter would appear to be maximum raised and which letter would appear to be minimum raised? Ans. While the blue letter image will be maximum raised, the red letter image will be least raised. The refractive index of glass is minimum for red light. So, the normal shift t 1  1 is minimum for red light. n  49. Within a glass slab, a double convex air bubble is formed. How would the air bubble behave? Ans. Although, the geometrical shape of the air bubble shall resemble a double convex lens, the bubble would behave as a divergent lens. This is because the refractive index of the material of the bubble is less than the refractive index of the surrounding medium. 50. Images formed by totally reflected light are brighter than the images formed by ordinary reflected light. Why? Ans. This is because, in total internal reflection, 100% of incident light gets reflected. On the other hand, in ordinary reflection, 20%—30% of incident light is lost. So, images formed in the first case are brighter than the images formed in the second case. 51. Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of reflected light? For what angle of incidence is the reflected ray normal to the incident ray? Ans. Reflection does not change wavelength. Frequency remains unchanged both in reflection and refraction.   = 5000 Å. Again,  = c  3  108 Hz = 6 × 1014 Hz  5000  1010 Again, i = r and i + r = 90°.  i = 45° 52. Why goggles (Sun glasses) have zero power even though their surfaces are curved? Ans. Both the surfaces of the Sun glasses are curved in the same direction. Moreover, the curvature is the same for both the surfaces.  P = (n21 – 1) 1  1 = 0.  R R  53. Explain why a convex lens behaves as a diverging lens when immersed in carbon disulphide? While the refractive index of glass is 1.5, the refractive index of carbon disulphide is 1.65. Ans. The refractive index of glass with respect to carbon disulphide is less than 1. So, according to the relation, 1 = (n21 – 1) 1  1  , f  R2   R1  f comes out to be negative. This explains the diverging action of convex lens in carbon disulphide.

Ray Optics and Optical Instruments i > ic 55 54. Explain the shining of an air bubble in water. Water Ans. For water-air interface, the critical angle is 48.8°. When Air bubble light travelling in water is incident at an angle greater than 48.8°, the light suffers total internal reflection. So, the air bubble in water shines brilliantly. SHORT ANSWER TYPE–II QUESTIONS (3 Marks) 1. Show that the optical centre of a plano-convex P1 C P2 P1 C P2 or a plano-concave lens is situated at the pole of the curved surface of the lens. Ans. In plano-convex and plano-concave lenses, one surface is plane whose radius of curvature is infinite. If we assume that in a plano-convex or a plano-concave lens, the first surface is plane, then R1 = .  CP1 = R1   or CP2 = 0. CP2 R2 R2 So, the optical centre C of these lenses is situated at the pole of the curved surface of the lens (Figures (a) and (b)). 2. An equiconvex lens of focal length f is cut into two equal halves in thickness. What is the focal length of each half? Ans. 1 = (n21  1)  2  f  R  For each half, 1 = n 1 or f = 2 or f  = 2f f R f 3. Use the lens equation to deduce algebraically the following: “A concave lens produces a virtual and diminished image independent of the location of the object.” Ans. 11 1 uf vu =f or v = uf Both u and f are negative. So, v is negative. Clearly, the image is formed in front of the lens. Thus, the image is virtual, whatever may be the location of the object. Again, m= f f u Since both u and f are negative therefore m is positive. Moreover, m < 1. So, the image is diminished, whatever may be the location of the object.

56 Physics—XII 4. One end of a cylindrical glass rod (n = 1.5) of radius 1.0 cm is rounded in the shape of a hemisphere. The rod is immersed in water  n  4  and an object is placed in the water 8 cm n2 = 1.5  3  along the axis of the rod at a distance of n1 = 4 8.0 cm from the rounded edge. Determine the 3 location of the image of the object. Ans. n2  n1 = n2  n1 vu R 4 1.5  4 On simplification, 1.5 = 0 or v = . 1.5  3 =3 v v 8 1 5. Refer to Figs. (a), (b) and (c). Give relationship between n1 and n2 in each case. n1 n1 n1 n2 n2 n2 Ans. (a) n1 = n2, because the ray of light does not bend at all. (b) n2 > n1, because the ray of light bends towards the normal. (c) n2 < n1, because the ray of light bends away from the normal. 6. Light falls from glass (n = 1.5) to air. Find the angle of incidence for which the angle of deviation is 90°. Ans. sin ic = 1  1 0.6667 or ic = 41.8° n 1.5 Deviation = 90° – ic = 90° – 41.8° = 48.2° This is the maximum attainable deviation in refraction. So, the given data favours total internal reflection. In reflection, deviation = 180° – 2i or 90° = 180° – 2i or 2i = 90° or i = 45°. 7. A right-angled isosceles glass prism is made from glass of refractive index 1.5. Show that a ray of light incident normally on (i) one of the equal sides of this prism is deviated through 90°. (ii) the hypotenuse of this prism is deviated through 180°. Ans. For glass of refractive index 1.5 for glass-air interface, the critical angle has an approximate value of 42°.   sin1 1   sin1 1   4148 ic n  1.5  

Ray Optics and Optical Instruments 57 A B 45° 45° 90° BC 45° 45° AC (a) (b) (i) When light ray is incident normally on one equal side of a right-angled isosceles prism, the ray is deviated through 90° as shown in Fig. (a). (ii) When a ray of light is incident normally on the hypotenuse face of the prism, the ray of light is deviated through 180° (Fig. (b)). 8. In figure, the direct image formed by the lens O¢ 15 cm O (f = 10 cm) of an object placed at O and that 50 cm formed after reflection from the spherical mirror are formed at the same point O. What is the radius of curvature of the mirror? Ans. v = 15 cm, f = 10 cm 1  1 = 1 or u = – 30 cm 15 u 10 Distance of O from the mirror = 20 cm For the rays to reverse, O must be at the centre of curvature of the mirror.  R = 20 cm. 9. Calculate the distance d so that a real image of an O 15 cm d object at O, 15 cm in front of a convex lens of focal length 10 cm, be formed at the same point O. The radius of curvature of the mirror is 20 cm. Will the image be inverted or erect? Ans. u = – 15 cm, f = 10 cm 1  1 = 1 or v = 30 cm v 15 10 Distance of this image from the mirror must be 20 cm for the mirror to reverse the light.  d = (30 + 20) cm = 50 cm The final image is inverted. 10. A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason. [CBSE 2014] Ans. 1 = (n21 – 1) 1  1  f  R2   R1  1 =  n2  1   1  1 f  n1   R1     R2 

58 Physics—XII n2 = 1.25 = 0.94 n1 1.33 The value of (n21 – 1) is negative. So, ‘f’ will be negative. Thus the lens will behave as a diverging lens. 11. Two monochromatic rays of light are incident A 45° normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism '1' '2' for the two rays ‘1’ and ‘2’ are respectively 1.35 and 1.45. Trace the path of these rays entering through the prism. [CBSE 2014] Ans. For ray ‘1’, sin C1 = 1 =1 = 0.74 B 45° C n1 1.35 C1 = sin–1 (0.74) = 47.8° > 45° Since the angle of incidence (45°) is less than the critical angle (47.8°) therefore the ray will 45° be refracted. 1 45° Refraction For ray ‘2’, sin C2 = 1 =1 = 0.69 2 Total internal n2 1.45 45° reflection C2 = sin–1 (0.69) = 43.6° 45° Since the angle of incidence (45°) is more than the critical angle (43.6°) therefore the ray will suffer total internal reflection. 12. A biconvex lens made of a transparent material of refractive index 1.5 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason. [CBSE 2014] Ans. 1 = (n21 – 1) 1  1  f  R2   R1  1 =  n2  1   1  1  f  n1   R1 R2      n2 = 1.5 = 1.13 > 1 n1 1.33 The value of (n21 – 1) is positive. So, f will be positive. Thus the lens will behave as a converging lens. 13. A ray PQ is incident normally on the face AB of a P A triangular prism of refracting angle of 60°, made of a 60° transparent material of refractive index 2 , as shown in Q 3 figure. Trace the path of the ray as it passes through the BC prism. Also calculate the angle of emergence and angle of deviation. [CBSE 2014 Comptt.]

Ray Optics and Optical Instruments 59 Ans. sin ic = 1 = 3 or ic = 60° A n 2 P 60° Angle of incidence at face AC of the prism = 60° 30° Q 60° Since ic = i therefore the refracted ray grazes the surface B C AC. Angle of emergence = 90° Angle of deviation = 30° 14. A double convex lens having both faces of the same radius of curvature has refractive index 1.55. Find out the radius of curvature of the lens required to get the focal length of 20 cm. [CBSE 2014 Comptt.] Ans. 1 = (n21 – 1) 1  1  f  R2   R1  1 = (1.55 – 1) 1  1 20  R  R  1 = 0.55 × 2 or R = 0.55 × 2 × 20 = 22 cm 20 R 15. Draw a ray diagram to show how a right-angled isosceles prism may be used to bend the path of light 45° 45° rays by 90°. Write the necessary condition in terms of B A 90° the refractive index of the material of this prism for the ray to bend to 90°. [CBSE 2016] Ans. For total internal reflection, 45° > ic 45°  sin 45° > sin ic A¢ 1 B¢ d 2 > sin ic 45° 1 > 1 or n > 2 2n 16. A light ray is incident at an angle of 45° with the normal to a 2 cm thick plate (n = 2.0). Find the shift in the path of the light as it emerges out from the plate. Given: 7 = 2.646. Ans. Applying Snell’s law, n2 sin r = n1 sin i, r 2 sin r = 1 sin 45° i = 45° or sin r = 1 22 Now, cos r = 1  sin2 r  1  1  7 8 22

60 Physics—XII Lateral displacement, d = t sin (i  r) = t [sin i cos r – cos i sin r] cos r cos r 22 21  7 1  1  4  7  1 =  = 7  4  7  2 2 2 2 2 2   4  = 7  1  2.646  1 = 1.646 cm = 0.62 cm 7 2.646 2.646 17. A ray of light enters the side of a water-filled glass aquarium making an angle of 30° with a normal to the aquarium face. Calculate the angle made by the ray within the glass and within the water. Given: na = 1, nga = 1.5, nwa = 1.33. Ans. Applying Snell’s law at the first interface, 1.5 sin  = 1 × sin 30° 30° Normal 1 Air or sin  = 3 = 0.3333 or  = sin–1 (0.3333) = 19.469°  19.5° Glass Applying Snell’s law at the second interface, we get nga q q nwa sin  = nga sin  Water nga sin  = 1.5  0.3333 = 0.3759 nwa q¢ nwa 1.33 or sin  = or  = sin–1 (0.3759) = 22.079  22.1° 18. The bottom of a container is a 4.0 cm thick glass (n = 1.5) slab. The container contains two immiscible liquids A and B of depths 6.0 cm and 8.0 cm respectively. What is the apparent position of a scratch on the outer surface of the bottom of the glass slab when viewed through the container ? Refractive indices of A and B are 1.4 and 1.3 respectively. Ans. In this problem, we shall consider three media-glass, liquid A and liquid B. The ‘real thickness’ of glass, liquid A and liquid B is 4.0 cm, 6.0 cm and 8.0 cm respectively. The ‘refractive index’ of glass, liquid A and liquid B is 1.5, 1.4 and 1.3 respectively. We know that Apparent thickness = Real thickness n  Apparent thickness of glass slab = 4.0 cm = 2.67 cm 1.5 Apparent thickness of liquid A = 6.0 cm = 4.29 cm 1.4 8.0 Apparent thickness of liquid B = 1.3 cm = 6.15 cm So, apparent position of scratch below upper liquid level is (2.67 + 4.29 + 6.15) cm, i.e., 13.11 cm.

Ray Optics and Optical Instruments 61  Total upward shift = (4 + 6 + 8) cm – 13.11 cm = 4.89 cm Refractive index has neither units nor dimensions. It is a scalar. 19. A small object is placed at the centre of the bottom of a cylindrical vessel of radius 3 cm and height 4 cm filled completely with water. Consider the ray leaving the vessel through a corner. Suppose this ray and the ray along the axis of the vessel are used to trace the image. Find the apparent depth of the image and the ratio of real depth to the apparent depth under the assumptions taken. Refractive index of water = 4 . 3 Ans. sin r 4 or sin r  4 sin i = 3 3 3 or or 5 r Now, A 3 cm P or sin r = 4  3  4  0.8 35 5 AD r r RD i r = sin–1 (0.8) = 53.13° I 4 cm tan r = 3 5 cm AD i AD = 3 3 cm tan 53.13 1.33 O 3 cm = 2.2556 cm Ratio of real depth and apparent depth = 4  1.33 = 1.77 3 20. A fish in a water tank sees the outside world as if it (the fish) is at the vertex of a cone such that the circular base of the cone coincides with the surface of water. Given the depth of water, where fish is located, being h and the critical angle for water-air interface being ic, find out by drawing a suitable ray diagram the relationship between the radius of the cone and the height ‘h’. [CBSE 2012 Comptt.] Ans. As is clear from Fig. (a), r h ic tan ic = r h ic or r = h tan ic ...(1) (a) We know that sin ic = 1 From Fig. (b), n n tan ic = 1 ic 1 n2  1 n2 – 1 From equation (1), r = h (b) n2  1

62 Physics—XII 21. A light ray is incident normally on the face AB of a right-angled prism ABC (n = 1.5) as shown in Fig. (a). What is the largest angle  for which the light ray is totally reflected at the surface AC? A A f f 90°– f f C BC q q (a) B (b) Ans. sin c  1 or sin c  1  2 n 1.5 3 Now,  > c , sin  > sin c , sin  > 2 But 3  +  = 90° or  = 90° –  sin (90° – ) > 2 or cos  > 2 or  < cos–1  2  3 3  3  So, largest value of  is cos–1  2  .  3  22. Figure shows a triangular prism of glass. A ray incident normally on one face is totally reflected. What can you A C 90° air conclude about the minimum value of index of refraction of glass? [CBSE 2016] Ans. Figure shows a right-angled prism. A ray PQ is incident PQ q normally on the face AB of the prism. It passes on undeviated and is incident on the face BC at R. For R total reflection to occur,  > ic But  = 45° 45° B  45° > ic or ic < 45° or sin ic < sin 45° or 1 1 or n> 2 n2 nmin. = 2 = 1.414 C 23. Figure shows a transparent hemisphere of radius A 3.0 cm made of a material of refractive index 2.0. 45° (i) A narrow beam of parallel rays is incident on the n = 2.0 hemisphere as shown in figure. Are the rays totally B 3.0 cm reflected at the plane surface? (ii) Find the image formed by the refraction at the first surface. Assume that the sphere is completed. Ans. (i) sin ic = 1 1 or ic = 30° n2 Now, i, which is 45°, is greater than ic. So, the rays suffer total internal reflection.

Ray Optics and Optical Instruments 63 (ii) n2  n1 = n2  n1 vu R 2 1 = 2 1  1 or v = 6 cm v  33 So, the image is formed at a point which is diametrically opposite to A. Here it is assumed that the sphere is completed. 24. Write the Lens Maker’s formula and use it to obtain the range of values of n (the refractive index of the material of the lens) for which the focal length of an equiconvex lens, kept in air, would have a greater magnitude than that of the radius of curvature of its two surfaces. [CBSE 2016 Comptt.] Ans. Lens Maker’s formula, 1 = (n21 – 1) 1  1  f  R2   R1  For equiconvex lens, R1= R and R2 = – R  1 = (n21 – 1) 2 R f  R  or f = 2(n21  1) For f to be greater than R, 2(n21 – 1) < 1 or 2 n21 < 3 or n21 < 1.5 So, the required range is 1.0 < n21 < 1.5 25. A convex lens of refractive index 1.5 has a focal length of 20 cm in air. Calculate the change in its focal length when it is immersed in water of refractive index 4 . 3 [CBSE Sample Paper 2017 (similar)] Ans. 1 = (nga – 1) 1  1  fa  R2   R1  1 = (1.5 – 1) 1  1  or 11 =1 ...(1) 20  R2  R1 R2 10  R1  Now, 1 =  ng  1   1  1  fw  nw   R1 R2      1 = 3  3  1 1 or 1=1 or fw = 80 cm fw  2 4 10  fw 80 Change in focal length = (80 – 20) cm = 60 cm = 0.60 m 26. A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in: (i) medium A of refractive index 1.65 (ii) medium B of refractive index 1.33. Explain, giving reasons, whether it will behave as a converging lens or a diverging lens in each of these two media. [CBSE 2011]

64 Physics—XII Ans. (i) Focal length fA in medium A of refractive index 1.65 is given by fA = (ngA – 1) 1 – 1    1.5  1   1 – 1 =  0.15  1 – 1  R2   1.65  R1  1.65  R1   R1   R2   R2  As fA is negative, the lens is diverging. (ii) Focal length fB in medium B of refractive index 1.33 is given by fB = (ngB – 1) 1 – 1  =  1.5  1   1 – 1  = 0.17  1 – 1  R2   1.33   R1 R2     R1    1.33  R1 R2  As fB is positive, the lens is converging. 27. A concave lens has the same radii of curvature for both sides and has a refractive index 1.6 in air. In the second case, it is immersed in a liquid of refractive index 1.3. Calculate the ratio of the focal lengths of the lens in the two cases. Ans. 1 =  ng  1   1 – 1 =  1.6  1   1 – 1  ...(1) fa  na   R1   1   R1 R2     R2    1 =  ng  1   1 – 1  =  1.6  1   1 – 1  ...(2) fl  nl   R1 R2   1.3   R1 R2        Dividing (2) by (1), we get fa = 1.6  1.3  1 = 0.3 = 0.3846 fl 1.3 1.6  1 1.3  0.6 28. An illuminated object and a screen are placed 90 cm apart. Determine the focal length and nature of the lens required to produce a clear image on the screen, twice the size of the object. Ans. Let x be the distance of the object from the lens. As the distance between the object and the screen is 90 cm therefore the distance of the image from the lens is (90 – x) cm. Now, u = – x cm, v = (90 – x) cm. Since the image is formed on the screen therefore the image is real.  m=–2 Using m = v , 90  x u – 2 =  x or x = 30 cm Now, v = (90 – x) cm = 60 cm u = – 30 cm, v = + 60 cm Using 1  1  1 , f = 20 cm vu f So, a convex lens of focal length 20 cm is required. 29. A double convex lens of glass of refractive index 1.6 has its both surfaces of equal radii of curvature of 30 cm each. An object of height 5 cm is placed at a distance of 12.5 cm from the lens. Calculate the size of the image formed. Ans. n21 = 1.6, R1 = + 30 cm, R2 = – 30 cm F I1 GH KJf 1 1  = (n21 – 1)   R2  = (1.6 – 1) 1 1 21 or f = 25 cm  R1  30 30 = 0.6 × 30 = 25

Ray Optics and Optical Instruments 65 Now, 11 = 1 or 1 1 1 =– 1  1 =– 1 or v = – 25 cm vu f v u f 12.5 25 25 The negative sign indicates that the image is formed on the same side of the object. Again, m = I  v ; I   25 = 2 or I = +10 cm O u 5  12.5 The positive sign indicates that the image is virtual and erect. The image is magnified two times. 30. A particle executes a simple harmonic motion of L MR amplitude 1.0 cm along the principal axis of a convex 19 cm lens of focal length 12 cm. The mean position of oscillation is at 20 cm from the lens. Find the 20 cm amplitude of oscillation of the image of the particle. 21 cm Ans. When the particle is at the right extreme position, u = – 19 cm, v1 = ?, f = 12 cm 1 1 = 1 or 1 = 11 7 or v1  12  19 cm v1 u f v1 12 19 12  19 7 When the particle is at the left extreme position, u = – 21 cm, v2 = ?, f = 12 cm 1 1 = 1 or 1 11  1  1  9 or v2  12  21 cm v2 u f v2 f u 12 21 12  21 9 Amplitude of oscillation of image = v1  v2 = 1 12  19  12  21  cm = 2.2857 cm 2 2  7 9  31. A beam of light converges to a point P. A lens is P placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm? Ans. In this problem, the object is virtual. 12 cm (a) u = +12 cm, f = +20 cm, v = ? 1 = 11 or 111 1  1 or 1  32 or v = 7.5 cm f vu v u f 12 20 v 240 The image is formed on the right of the lens (7.5 cm from the lens) and is real. (b) u = +12 cm, f = – 16 cm, v = ? 1 = 1 1 1 1 = 1 P v u f 12 16 48 or v = 48 cm 12 cm The image is formed on the right of the lens (48 cm from the lens) and is real.

66 Physics—XII 32. A convex lens of focal length 10 cm is placed coaxially 5 cm away from a concave lens of focal length 10 cm. If an object is placed 30 cm in front of the convex lens, find the position of the final image formed by the combined system. Ans. For first lens, u1 = – 30 cm, f1 = 10 cm 1 = 11 1 1 1 or v1 = 15 cm v1 f1 u1 10 30 15 Image formed by first lens serves as a virtual object for second lens at a distance (15 – 5) cm to the right of the second lens. For second lens, u2 = 10 cm, f2 = – 10 cm 1 = 1  1  1  1 =0 or v2 =  v2 f2 u2 10 10  Image is formed at infinity. The rays will emerge parallel to the axis. 33. Two lenses of powers 10 D and –5 D are placed in contact. (i) Calculate the power of the new lens. (ii) Where should an object be held from the new lens, so as to obtain a virtual image of magnification 2? Ans. (i) P = P1 + P2 = (10 – 5)D = 5 D (ii) m = +2, f = 1 = 1 = 20 cm P 2 m= f or u = f – f or u = 20 – 20 = – 10 cm uf m 2 34. Two lenses of powers +15 D and –5 D are in contact with each other forming a combination lens. (i) What is the focal length of this combination? (ii) An object of size 3 cm is placed at 30 cm from this combination of lenses. Calculate the position and size of the image formed. Ans. (i) P = P1 + P2 = (15 – 5) D = 10 D F= 11 m = 10 cm P 10 (ii) 11 = 1 or 11 1 or 1 1  1 or v = 15 cm vu F vFu v 10  30 Again, I = v  15 1 or I= 1 × 3 cm = – 1.5 cm O u  30 2 2 The negative sign indicates that the image is real and inverted. 35. The image of an object, formed by a combination of a convex lens (of focal length f) and a convex mirror (of Image radius of curvature R), set up, as Object O shown is observed to coincide with the object. Redraw this diagram to mark on it the position of the centre of ad curvature of the mirror. Obtain the expression for R in terms of the (a) distances, marked as a and d, and the focal length f of the convex lens. [CBSE 2016 Comptt.]

Ray Optics and Optical Instruments 67 Ans. The centre of curvature C is shown in the redrawn diagram. [Fig. (b)] For convex lens: u = – a, v = R + d 1 = 11 Image C f vu Object R 1 1 1 a = f Rd a = 1 1 d Rd a (b) 1 = 11 = af Rd fa af R = af – d af 36. A convex lens, of focal length 25 cm, and a convex mirror, of radius of curvature 20 cm, are placed co-axially 40 cm apart from each other. An incident beam, parallel to the principal axis, is incident on the convex lens. Find the position and nature of the image formed by this combination. [CBSE 2016 Comptt.] Ans. The object, being at infinity, the image formed by the convex lens, is at its focus, i.e., 25 cm from the lens. This image becomes the ‘object’ for convex mirror. Now, for the convex mirror, object distance = (40 – 25) cm = 15 cm u = – 15 cm, R = +20 cm Now, 11 = 1 = 2 vu f R 1  1 = 2 or 1 = 1  1 = 1 or v = +6 cm v 15 20 v 10 15 6 The final image is, therefore, a virtual image that appears to be formed (behind the convex mirror) at a distance of 6 cm from it. 37. A convex lens, of focal length 20 cm, has a point object placed on its principal axis at distance of 40 cm from it. A plane mirror is placed 30 cm behind the convex lens. Locate the position of image formed by this combination. Ans. We first consider the effect of the P L 30 cm M lens. For the lens, we have u = –40 cm and f = +20 cm 40 cm Q Q1 Using the lens formula, we get 10 cm 10 cm 1 1 = 1  v1 = +40 cm v1 (40) 20 Had there been the lens only the image would have been formed at Q1. The plane mirror M is at a distance of 30 cm from the lens L. We can, therefore, think of Q1 as a virtual object, located at a distance of 10 cm, behind the plane mirror M. The plane mirror therefore forms a real image (of this virtual object Q1) at Q, 10 cm in front of it.

68 Physics—XII 38. A convex lens is placed in contact with a plane mirror. An axial point object, at a distance of 20 cm from this combination, has its image coinciding with itself. What is the focal length of the convex lens? Ans. Figure (a) shows a convex lens L L LM in contact with a plane mirror M. P is the point object, kept in front P P of this combination at a distance of 20 cm, from it. Since the image (a) M (b) M of the object is coinciding with the object itself, the rays from the object, after refraction from the lens, should fall normally on the mirror M, so that they retrace their path and form an image coinciding with the object itself. This will be so, if the incident rays from P form a parallel beam perpendicular to M, after refraction from the lens. For clarity, M has been shown at a finite distance from L, in figure (b). For lens L, since the rays from P, form a parallel beam after refraction, P must be at the focus of the lens. Hence focal length of the lens is 20 cm. 39. Use the lens equation to deduce algebraically the following: ‘‘An object placed within the focus of a convex lens produces a virtual and enlarged image’’. Ans. 1  1 = 1 or 1  1  1 vu f vfu Convex lens: f > 0, u < 0 (object on the left) For 0 < | u | < f 1 = 1  1  0 i.e., v < 0 (image on left; virtual) v f |u| Also for this case, 1 < 1 i.e., | v | > | u | (image enlarged) |v| |u| 40. The plane side of a plano-convex lens is silvered. The lens acts like a concave mirror of 30 cm focal length. The refractive index of the lens is 1.5. What is the radius of curvature of this lens? Ans. Radius of curvature of concave mirror = 60 cm Let C represent the centre of curvature of the concave mirror. When the object is at C, the image will also be formed at C. This would be possible if the rays C strike the silvered plane surface normally. This would mean that C also represents the principal focus of the lens. So, f = 60 cm 1 = (n21  1)  1  1  f  R1 R2    1 = (1.5  1)  1  1 or R = 30 cm. 60  R  

Ray Optics and Optical Instruments 69 41. A point object is placed at a distance of 15 cm on the axis of a convex lens of focal length 12 cm. On the other side of the lens, a convex mirror is placed at a distance of 20 cm from the lens such that the image formed by the combination coincides with the object itself. Calculate the focal length of the convex mirror. Ans. Using lens equation, 1  1 = 1 OI v  15 12 or 1 = 1 1 = 1 or v = 60 cm 15 cm 20 cm R v 12 15 60 If the point I where the image is supposed to be formed by the lens coincides with the centre of curvature of the convex mirror, then the light rays would retrace their path after reflection at the convex mirror and the image formed by the combination would coincide with the object.  R = (60 – 20) cm = 40 cm f = R = 40 cm = 20 cm 22 42. A weight in the form of a glass hemisphere is placed, flat surface down, on the page of a book. Calculate the position and magnification of the image of the type at the centre of the flat surface of the hemisphere. The index of refraction of glass is 1.5. Ans. n1  n2 = n1  n2 vu R 1  1.5 = 1  1.5 = 0.5 v R –R R 1 = 0.5  1.5 or v = – R R vR m = n2v = 1.5( R) = 1.5 n1u 1( R) 43. Two thin lenses, when in contact, produce a combination of power + 10 D. When they are 0.25 m apart, the power reduces to + 6 D. Calculate the focal lengths of the two lenses. Ans. P1 + P2 = 10 ...(i) Using P = P1 + P2 – d P1P2, 6 = 10 – 0.25 P1P2 or P1P2 = 16 Now, P1 – P2 = (P1  P2 )2 – 4P1P2 = 100  64 = 6 ...(ii) From (i) and (ii), P1 = 8 D and P2 = 2 D f1 = 1m = 0.125 m, f2 = 1 m = 0.5 m 8 2

70 Physics—XII 44. An object and its real image are located at distances 25 cm and 40 cm respectively from the two principal foci of a lens. Calculate the focal length of the lens. Ans. 1 = f 1 1 O F1 F2 I f  40  (f  25) = 11 25 cm 40 cm f  40 f  25 (f + 40) (f + 25) = f(f + 25) + f(f + 40) or f = 40  25 cm = 31.6 cm 45. An object is placed 21 cm in front of a concave mirror of radius of curvature 10 cm. A glass slab of thickness 3 cm and refractive index 1.5 is then placed close to the mirror in the space between the object and the mirror. Find the position of the final image formed. Ans. Apparent shift = 3 1  1  = 1 cm 1.5  u = – 20 cm, f = – 5 cm O O¢ P Using 1  1 = 1 , v = – 6.67 cm u v f The reflected ray will pass through the slab again suffering a further shift of 1 cm. So, the final image will be at a distance of 7.67 cm from the pole. LONG ANSWER TYPE QUESTIONS (5 Marks) 1. State the laws of refraction and cause of refraction. Ans. The refraction of light occurs in accordance with the following two laws of refraction. (i) The incident ray, the normal to the refracting surface at the point of incidence and the refracted ray all lie in the same plane. (ii) For a given pair of media and for light of a given wavelength, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. A This law is sometimes called Snell’s law of refraction. sin i ...(1) i Denser According to this law, sin r = n21. O r Rarer Here n21 is a constant called the refractive index of the second A B medium with respect to the first medium. The refractive index (a) n21 is a characteristic of the pair of media. It also depends upon the wavelength of light. However, it is independent of i Rarer the angle of incidence. O Denser From Eq. 1, if n21 > 1, r < i, i.e., the refracted ray bends r towards the normal. In such a case, medium 2 is said to be optically denser (or denser, in short) than medium 1. On the B other hand, if n21 < 1, r > i, the refracted ray bends away (b) from the normal. This is the case when incident ray in a denser medium refracts into a rarer medium.

Ray Optics and Optical Instruments 71 Cause of Refraction The bending of light or refraction is due to the change in the speed of light as it passes from one medium to another. Larger the change in the speed of light, more is the bending due to refraction. According to Snell’s law of refraction, n21 = sin i = v1 sin r v2 (i) If v1 > v2, then n21 > 1 and sin i > sin r or i > r. So, the refracted ray bends towards the normal. The medium 2 is said to be optically denser than medium 1. Thus, a ray of light bends towards the normal as it refracts from a rarer medium into a denser medium. (ii) If v1 < v2, then n21 < 1 and sin i < sin r or i < . So, the refracted ray bends away from the normal. The medium 2 is said to be optically rarer than medium 1. Thus, a ray of light bends away from the normal as it refracts from a denser medium into a rarer medium. 1 2. State the principle of reversibility of light and use it to arive at result nba = nab . Ans. If a ray of light, after suffering any number of reflections and/or refractions has its path reversed at any stage, it travels back to the source along the same path in the opposite direction. This is called principle of reversibility of light. Consider an interface XY which separates a rarer A medium ‘a’ from a denser medium ‘b’. Let a ray of light AO be incident on the interface as shown in X i figure at an angle of incidence i. After refraction, the ray follows the path OB such that r is the angle O aY of refraction. If a plane mirror is placed normal to b the path of the refracted ray, then the ray is r reversed and retraces the whole path in the B opposite direction. Let us now make use of the principle of reversibility of light to arrive at an important result. Principle of reversibility of light. When ray of light travels from rarer to denser medium, the refractive index of denser medium with respect to rarer medium is given by sin i ...(1) nba = sin r When the path of light is reversed, then r will be treated as angle of incidence and i will be treated as angle of refraction because the ray of light would now travel from denser to rarer medium. So, refractive index of rarer medium with respect to denser medium is given by sin r ...(2) nab = sin i Multiplying (1) by (2), we get nba × nab = sin i  sin r = 1 or nba = 1 sin r sin i nab Conclusion. The refractive index of denser medium w.r.t. rarer medium is the reciprocal of the refractive index of rarer medium w.r.t. denser medium.

72 Physics—XII 3. Prove that the ray of light will leave the parallel glass slab at the same angle at which it entered the glass slab on the opposite side with a lateral displacement. Ans. Consider a ray of light AB passing from air (medium a) through a parallel sided glass slab (medium b) into air (medium a). The ray of light will clearly suffer two refractions. Since the medium on both sides of glass is the same therefore the ray of light will get laterally shifted without any deviation. This is proved below. When ray of light is refracted from air to A i1 glass, then the refractive index of glass B (medium b) w.r.t. air (medium a) is given by Medium a (Air) nba = sin i1 ...(1) r1 d sin r1 i2 N When ray of light is refracted from glass to t Medium b y (Glass) air, then the refractive index of air w.r.t. glass B¢ C Medium a r2 is given by (Air) nab = sin i2 ...(2) sin r2 Multiplying (1) and (2), we get nba × nab = sin i1  sin i2 D sin r1 sin r2 Refraction through parallel-sided slab. 1 But we know that nba = nab  1 × nab = sin i1  sin i2 nab sin r1 sin r2 Since the slab is parallel sided,  i2 = r1  sin i1 = 1 or i1 = r2 sin r2 Conclusion. The ray of light will leave the parallel glass slab at the same angle at which i it entered the glass slab on the opposite side. r a r b However, it gets laterally displaced. 4. Figure shows two parallel slabs of different c materials placed one above the other. Two materials of upper and lower slabs have been r¢ designated as b and c respectively. The medium r¢ above and below the compound slab is the same and has been designated as ‘a’. Prove that a nba × ncb = nca. i Ans. If we consider refraction of light at the interface of ‘a’ and ‘b’, then sin i ...(1) nba = sin r Refraction through compound slab.

Ray Optics and Optical Instruments 73 If we consider refraction of light at the interface of ‘b’ and ‘c’, then sin r ...(2) ncb = sin r Considering refraction of light at the interface of c and a, we get nac = sin r ...(3) sin i Multiplying equations (1), (2) and (3), we get 1 or nba × ncb = nca nba × ncb × nac = 1 or nba × ncb = nac 5. Experimentally demonstrate the phenomenon of total internal reflection. Ans. All optical phenomena can be demonstrated very easily with the use of a laser torch or pointer, which is easily available nowadays. Take a glass beaker with clear water in it. Stir the water a few times with a piece of soap, so that it becomes a little turbid. Take a laser pointer and shine its beam through the turbid water. You will find that the path of the beam inside the water shines brightly. Shine the beam from below the beaker [Fig. (a)] such that it strikes at the upper water surface at the other end. The beam undergoes partial reflection (which is seen as a spot on the table (a) (b) below) and partial refraction [which comes out in the air and is seen as a spot on the roof. Now direct the laser beam from one side of the beaker such that it strikes the upper surface of water (c) more obliquely [Fig. (b)]. Adjust the direction of laser beam until you find the angle for which the Observing total internal reflection refraction above the water surface is totally in water with a laser beam absent and the beam is totally reflected back to water. This is total internal reflection at its (refraction due to glass of beaker neglected being very thin). simplest. Pour this water in a long test tube and shine the laser light from top, as shown in Fig. (c). Adjust the direction of the laser beam such that it is totally internally reflected every time it strikes the walls of the tube. This is similar to what happens in optical fibres. 6. Lenses are constructed by a material of refractive index 1.50. The magnitude of the radii of curvature are 20 cm and 30 cm. Find the focal lengths of the possible lenses with the above specifications. Ans. (i) R1 = 20 cm, R2 = – 30 cm 1 = (n21  1)  1  1  f  R1 R2    1 = (1.5  1) 1  1 or f = 24 cm f  20 30 

74 Physics—XII (ii) R1 = – 20 cm, R2 = – 30 cm 1 = (1.5  1)   1  1 f  20 30  or f = – 120 cm (iii) R1 = – 20 cm, R2 = + 30 cm 1 = (1.5  1)   1  1  f  20 30  or 1 = 1 or f = – 24 cm f 24 (iv) 1 = (1.5  1)  1  1 f  20 30  or 1 = 1 32 or f = 120 cm f 2  60  7. Derive thin lens formula for convex and concave lenses. Ans. Lens equation or lens formula is a mathematical relation between distance ‘u’ of object from the optical centre of the lens, distance v of image from the optical centre and focal length f of lens. This relation is sometimes called Gaussian form of lens equation. (i) For Convex Lens AD (a) Real Image. Consider an object AB held F 2F B¢ perpendicular to the principal axis of the 2F B F convex lens and placed between F and 2F. A C real, inverted and magnified image AB is A¢ formed as shown in Fig. (a). (a) Triangles ABC and ABC are similar.  AB = CB ...(i) AB CB Again, triangles DCF and ABF are similar.  AB = FB or A B  FB ...(ii) CD CF AB CF From equations (i) and (ii), CB = FB or CB  CB – CF CB CF CB CF According to the Cartesian sign convention, CB = – u, CB = + v, CF = + f  v vf u = f or – uv + uf = vf or uf – vf = uv Dividing by uvf, we get 11 1 vu f

Ray Optics and Optical Instruments 75 (b) Virtual Image. Proceeding as in the case of A¢ real image, AD CB = FB or CB  CF + CB B¢ F B C CB CF CB CF (b) F According to the Cartesian sign convention, CB = – u, CB = – v and CF = + f v f v  u = f or v = f  v or uf – uv = vf uf Dividing by uvf, we get 1  1  1 or 11 1 vf u vu f (ii) For Concave Lens Proceeding as in the case of convex lens, CB = FB or CB  CF  CB AD CB CF CB CF A¢ Applying Cartesian sign convention, 2F B F B¢ C CB = – u, CB = – v and CF = – f   v  f  ( v) or v  v f (c) u = f u f or uv – uf = – vf or – uf + vf = – uv or uf – vf = uv Dividing by uvf, we get 1  1  1 vu f 8. Define linear magnification of a lens and derive the expression of magnification for convex and concave lenses. Ans. Linear magnification produced by a lens is defined as the ratio of the size of the image to the size of the object. It is generally denoted by m. If h represents the height of the object and h represents the height of the image, then linear magnification is given by m = h A C B¢ h h¢ h Magnification in terms of v and u B (i) Convex lens. Fig. (a) shows the formation of image by a convex lens. s CBA and CBA are similar. AB CB A¢  AB = CB uv (a)