76 Physics—XII According to Cartesian sign convention, AB = – h, AB = h CB = + v, CB = – u h = v or h v h u hu v A Magnification = u (ii) Concave lens. Fig. (b) shows the formation of image by a concave lens. s CBA and CBA are similar. h A¢ h¢ AB CB B F B¢ v C AB CB u f According to Cartesian sign convention, AB = h, AB = h, CB = – v, CB = – u (b) h v h = u v Magnification = u Magnification in terms of v and f We know that 11 = 1 or v v = v vu f v u f vv v v vf or 1– u = f or u = f –1= f or v = f v or f v u f m= f Magnification in terms of u and f We know that 11 = 1 or u u = u or uu +1 vu f v u f vf or u = uf or vf f v f u = u f or m= f u 9. Define power of a lens. Write its SI unit. Find the relation between focal length and power of the lens. Ans. Power of a lens is the ability of the lens to bend a ray of light falling on it. Clearly, a lens of shorter focal length bends the incident light more, while converging it in case of a convex lens and diverging it in case of a concave lens. The power P of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light falling at unit distance from the optical centre (See figure).
Ray Optics and Optical Instruments 77 h If h = 1, then P = tan = 1 tan = f : f 1 d or P = = f for small value of . h 1 d F O P= f f The power of a lens is the reciprocal of its focal Power of a lens. length. So, shorter the focal length of a lens, the greater is its power. Since focal length of a converging lens is positive therefore its power is positive. Similarly, the power of a diverging lens is negative. Opticians express the power of a lens in terms of a unit called the dioptre. It is regarded as the SI unit of optical power. The power of a lens is said to be one dioptre if the focal length of the lens is 1 metre. 1 D = 1 m–1 When focal length is in cm, P = 100 dioptre. f 10. You are given three lenses L1, L2 and L3 each of focal length 20 cm. An object is kept at 40 cm in front of L1, as shown. The final real image is formed at the focus ‘I’ of L3. Find the separations between L1, L2 and L3. [CBSE 2012] L1 L2 L3 40 cm I 20 cm x1 x2 Ans. Applying lens formula for the case of L1, 1 = 11 1 1 1 or v1 = 40 cm v1 f1 u1 20 40 40 Image by L3 is formed at the focus of L3. So, the object should lie at infinity for L3. So, L2 will produce image at infinity. Clearly, for L2, the object should be at the focus of L2. The image by L1, is formed at 40 cm right of L1, which is 20 cm left of L2. Distance x1 between L1 and L2 = (40 + 20) cm = 60 cm Again, distance between L2 and L3 does not matter as the image by L2 is formed at infinity. So, x2 can take any value. f = + 10 cm f = + 30 cm 11. Find the position of the image formed by the lens combination in the given figure. Ans. Image formed by the first lens 11 =1 30 cm 5 v1 u1 f1 cm or 1 1 1 10 cm v1 30 10 f = – 10 cm (a)
78 Physics—XII or 11 = 1 or v1 = 15 cm v1 30 10 The image formed by the first lens serves as the object for the second. This is at a distance of (15 – 5) cm i.e., 10 cm to the right of the second lens. It is a virtual object. Now, 11 = 1 or 1 1 1 0 or v2 = v2 10 10 v2 10 10 The virtual image is formed at an infinite distance to the left of the second lens [See Fig. (b)]. This acts as an object for the third lens. Again, 1 1 = 1 v3 30 11 (b) or v3 = 30 or v3 = 30 cm The final image is formed 30 cm to the right of the third lens. 12. Prove that a convex lens produces an n times magnified image when the object distances from the lens have magnitude f f . Here f is the magnitude of the focal n length of the lens. Hence find the two values of object distance for which a convex lens of power 2.5 D will produce an image that is four times as large as the object. Ans. m= f uf When image is real, m = – n. f or u + f = f or u = f f ...(1) –n= uf n n When image is virtual, m = n. f f or u = f f ...(2) n= uf or u + f = n n It follows from (1) and (2) that the magnitude of the object distance is f f . n Focal length of lens = 1 m = 1 100 cm = 40 cm 2.5 2.5 Now, 40 ± 4 = u 40 or u = – 30 cm or – 50 cm
Ray Optics and Optical Instruments 79 13. A convex lens and a convex mirror (of radius of curvature 20 cm) are placed co-axially with the convex mirror placed at a distance of 30 cm from the lens. For a point object at a distance of 25 cm from the lens, the final image due to this combination, coincides with the object itself. What is the focal length of the convex lens? Ans. The final image, formed by the L M combination, is coinciding with the A object itself. This implies that the B rays, from the object, are retracing O1 O2 C their path, after refraction from the lens and reflection from the 25 cm A¢ 30 cm B¢ mirror. The (refracted) rays are, therefore, falling normally on the 20 cm mirror. It follows that the rays AB and AB when produced, are meeting at the centre of curvature, C of the mirror. Hence O2C = 20 cm, the radius of curvature of the mirror. From the figure, we then see that for the convex lens u = –25 cm and v = +(30 + 20) cm = +50 cm. If f is the focal length of the lens, we have 1 1 = 1 or f = 50 cm = 16.67 cm 50 (25) f 3 14. A convex lens, of focal length 20 cm, is placed co-axially with a convex mirror of radius of curvature 20 cm. The two are kept 15 cm apart from each other. A point object is placed 60 cm in front of the convex lens. Find the position of the image formed by this combination. [CBSE 2014] Ans. The ray diagram, for the image P L Q formed by the combination, is A Q1 shown in figure. 60 cm O O¢ 30 cm For the convex lens, we have A¢ u1 = – 60 cm and f = +20 cm 15 cm Using the lens formula, we get 1 1 = 1 or v1 = 30 cm v1 ( 60) 20 Had there been only the lens L, the image of P would have been formed at Q1 which acts as a virtual object for the convex mirror. OQ1 = distance of this virtual object (Q1) from convex mirror = OQ1 – OO = (30 – 15) cm = 15 cm So, for the convex mirror, u2 = +15 cm and R = +20 cm, Using the mirror formula, we get 1 1 = 2 i.e., 11 2 or v2 = +30 cm v2 u2 , v2 15 20 R Thus the final image (Q) formed is a virtual image formed at a distance of 30 cm behind the convex mirror.
80 Physics—XII 15. A convex lens, of focal length 20 cm, and a concave mirror, of focal length 10 cm, are placed co-axially 50 cm apart from each other. An incident beam parallel to the principal axis, is incident on the convex lens. Locate the position of the (final) image formed due to this combination. Ans. The incident beam, on lens L, is 50 cm parallel to its principal axis. Hence the lens forms an image Q1 at its focal M point, i.e., at a distance OQ1 (= 20 cm) from the lens. This image, Q1, now O Q1 Q acts as a real object for the concave mirror. For the mirror, we then have: L 20 cm 15 cm u = –30 cm, and f = –10 cm, 30 cm Using mirror formula, we get 1 1 = 1 or v = –15 cm v (30) (10) The lens – mirror combination, therefore, forms a real image Q at a distance of 15 cm from M. The ray diagram is as shown in figure. 16. An equiconvex lens with radii of curvature of magnitude r each, is put over a liquid layer poured on top of a plane mirror. A small needle, with its tip on the principal axis of the lens, is moved along the axis until its inverted real image coincides with the needle itself. The distance of the needle from the lens is measured to be ‘a’. On removing the liquid layer and repeating the experiment Liquid the distance is found to be ‘b’. Given that two values of distances measured represents (a) the focal length values in the two cases, obtain a formula for the refractive index of the liquid. Ans. Fig. (b) shows the formation of image. The liquid layer Q P P¢ Q¢ can be regarded as forming a plano-concave lens. The first value (= a) of the measured distance is, therefore, the focal length of the combination of the given lens and the liquid lens. The second value (= b) represents the focal length of the lens itself. If f represents the focal length of the liquid lens, then using 1 1 1 , we get F f1 f2 1 = 11 or 1 1 1 = ba ...(1) (b) a bf f a b ab ...(2) Using Lens maker’s formula for the liquid lens, we get 1 = (nL 1) 1 1 = nL 1 f r r
86 Physics—XII 1. A ray of light travels in optical fibre due to (a) refraction (b) total internal reflection (c) reflection (d) polarization Ans. (b) 2. An optical fibre (n = 1.72) is surrounded by a glass coating (n = 1.50). The critical angle for total internal reflection at the fibre-glass interface will be: (a) 59.69° (b) 62.61° (c) 60.69° (d) 58.76° Ans. (c) Hint and Solution: 90° Using Snell’s law, 1.72 sin c = 1.50 sin 90° qc or sin c = 1.50 75 1.72 86 or c = sin–1 75 = sin–1 (0.872) = 60.69° 86 3. Which of the following is incorrect statement? (a) Light undergoes successive total internal reflections as it moves through an optical fibre. (b) Optical fibres consist of thousands of extremely thin and long strands of high quality glass/quartz. (c) Optical fibres are used for optical signal transmission. (d) None of these Ans. (d) 4. Figure shows a cross section of a ‘light- i¢ i¢ Cladding pipe’ made of a glass fibre of refractive r Core index 1.68. The outer covering of the pipe is made up of material refractive t index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which the total internal reflection inside the pipe takes place as shown in the figure? (a) 59° (b) 31° (c) 60° (d) 53.5° Ans. (c) Hint and Solution: sin iC = outer = 1.44 inner 1.68 Hence, iC = 59° Hence, r = 90° – iC = 31° rmax = 31° sin imax = 1.68 sin rmax Therefore, imax = 60° Thus, total internal reflection will take place for all angles for which i < 60°.
Ray Optics and Optical Instruments 87 OPTICAL INSTRUMENTS VERY SHORT ANSWER TYPE QUESTIONS (1 Mark) 1. The difference in the focal lengths of the two lenses is larger in which case–microscope or telescope? Ans. Telescope. 2. What is the formula for angular magnification for a telescope in normal adjustment? Ans. fo . fe 3. Who designed the compound microscope? Ans. Galileo. 4. Name an optical instrument in which we make use of the phenomenon of total internal reflection. Ans. Prism binocular. 5. What type of lens is used as eye-lens in Galileo telescope? Ans. Concave lens. 6. Give two uses of optical instruments. Ans. (i) magnification (ii) resolution. 7. What is the size and nature of the image of an object placed in contact with a convex lens? Ans. Same size, virtual and erect. 8. What is the ideal position of the eye for observing the image in a microscope or telescope? Ans. The position of the eye-ring. 9. What happens when the focal length of the objective of a telescope is increased? Ans. The angular magnification increases. 10. Who designed the first telescope? Ans. Galileo designed the first telescope in 1609. 11. What do you understand by the term eye-ring? Ans. It is the image of the objective formed by the eyepiece. 12. What is visual angle? Ans. Visual angle is the angle subtended by an object at the eye. 13. How does the visual angle vary with the distance of the object from the eye? Ans. The visual angle decreases with the increase in distance of the object from the eye. 14. Why an object seen from a distance appears to be of a small size? Ans. The visual angle subtended, by a distant object, at the eye of the observer is small. 15. What is meant by range of an astronomical telescope? Ans. The range of an astronomical telescope is the distance up to which a star can be viewed with sufficient clarity.
88 Physics—XII 16. Why the objective and eyepiece of a compound microscope have short focal lengths? Ans. This increases the magnifying power because the magnifying power is inversely proportional to the focal lengths of the objective and eyepiece. 17. What would be the effect on the length of telescope if the focal length of the objective is increased? Ans. Since the length of the telescope in normal adjustment is the sum of the focal lengths of the objective and the eyepiece therefore the length of the telescope will be increased. 18. The objective of telescope A has a diameter 3 times that of the objective of telescope B. How much greater amount of light is gathered by A compared to B? Ans. Light-gathering capacity of A is 9 times that of B. 19. Why no cross-wires can be used in Galileo telescope? Ans. This is because the intermediate image in this telescope is virtual. 20. In a simple microscope, why the focal length of the lens should be small? Ans. This is because the angular magnification is inversely proportional to the focal length. 21. Name one essential difference between a terrestrial telescope and an astronomical telescope. Ans. While a terrestrial telescope forms an erect image, the astronimical telescope forms an inverted image. 22. How can you increase the magnifying power of a telescope? Ans. This can be increased by increasing the focal length of the objective and/or decreasing the focal length of the eyepiece. 23. Why should the objective lens of a compound microscope have small focal length? Ans. m= L D . Clearly, small fo increases m. fo 1 fe 24. How will you distinguish between a compound microscope and a telescope simply by seeing it? Ans. In the case of a compound microscope, the length of the tube is fixed. In the case of a telescope, the length is adjustable. 25. A reflecting type telescope has a concave reflector of radius of curvature 120 cm. Calculate the focal length of eyepiece to secure a magnification of 20. Ans. fo = R 120 cm = – 60 cm; fe = fo 60 cm = 3 cm. 22 m 20 26. Given: Galilean telescope with fo = 150 cm, fe = – 7.5 cm. What is the separation between the objective and the eyepiece? Ans. fo = 150 cm, fe = – 7.5 cm Length L = | fo | – | fe | = 150 – 7.5 = 142.5 cm. 27. What do you understand by the term ‘focal plane’ of a lens? Ans. Focal plane is a plane perpendicular to the principal axis and at a distance equal to the focal length of the lens. 28. “Prism Binocular is nothing but a pair of astronomical telescopes.” Comment on this statement. Ans. The given statement is not exactly true. The length of a prism binocular is smaller than the length of the telescope. The binocular produces an erect image without any lateral inversion. The image is highly magnified.
Ray Optics and Optical Instruments 89 SHORT ANSWER TYPE–I QUESTIONS (2 Marks) 1. Name the type of lenses used for correcting a (i) myopic, (ii) hypermetropic and (iii) an astigmatic eye? Ans. (i) Concave lens (ii) Convex lens (iii) Cylindrical lens. 2. List the advantages of a reflecting telescope, especially for high resolution astronomy. Ans. No chromatic aberration due to the objective because no refraction is involved; spherical aberration reduced by using a mirror of the shape of a paraboloid; brighter image than in a refracting telescope of equivalent size because in the latter, intensity of light is partially lost due to reflection and absorption by the objective lens glass; mirror entails grinding and polishing of only one side; high resolution (as well as brightness of a point object) achieved by using a mirror of large aperture which is easier to support (its back side being available) than a lens of the same aperture. 3. How does the magnification of a magnifying glass differ from its magnifying power? Ans. The magnification of a magnifying glass is v . The magnifying power is the ratio of u the angles subtended by the image and the object. In practice, the angles are small. So, magnifying power becomes equal to magnification. 4. The objective of telescope A has a diameter 3 times that of the objective of telescope B. Show that the range of A is three times the range of B. (Range tells you how far away a star of some standard absolute brightness can be spotted by the telescope). Ans. Consider a star at a distance l that is barely visible in a telescope using an objective of diameter d. The intensity of light from a star (of the same absolute brightness) at a distance 2l is reduced by a factor of 4. To receive the same total amount of light, the diameter of the objective should be 2d. Thus if the distance of the star is doubled, the diameter of the objective should be doubled for the star to be again barely visible. Thus the distance a telescope can penetrate through the sky (range) is proportional to the diameter of its objective. 5. Why must both the objective and the eyepiece of a compound microscope have short focal lengths? Ans. Both the objective and the eyepiece of a compound microscope should have short focal lengths to have greater magnifying power as the magnifying power of a compound microscope is given by m = L D . fo 1 fe 6. Which two main considerations are kept in mind while designing the ‘objective’ of an astronomical telescope? Ans. The two main considerations: (i) Large light-gathering power (ii) Higher resolution (resolving power) [Both these requirements are met better when an objective of large focal length as well as large aperture is used.] 7. Write two important limitations of a refracting telescope over a reflecting type telescope. [CBSE 2013] Ans. (i) It suffers from chromatic aberration due to refraction and hence the image obtained is multicoloured and blurred.
90 Physics—XII (ii) As a lens of large aperture cannot be manufactured easily, its light gathering power is low and hence cannot be used to see faint stars. 8. A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. Find the focal length of the eyepiece. Ans. m = vo D or m = fo vo D or m = vo D ...(1) uo fe fo fe 1 fo fe Since image is formed at infinity therefore ue = | fe |. vo = 6.5 – fe Substituting values in (1), – 100 = 1 6.5 fe 25 0.5 fe or – 4fe = 1 – (13 – 2fe) or – 6fe = – 12 or fe = 2 cm 9. The sum of focal lengths of two lenses of a refracting telescope is 105 cm. The focal length of one lens is 20 times that of the other. Determine the total magnification of the telescope when the final image is formed at infinity. [CBSE 2014 Comptt.] Ans. fo + fe = 105, fo = 20 fe, 20 fe + fe = 105 or fe = 105 = 5 cm 21 fo = 20 × 5 cm = 100 cm Now, M = fo 100 20 fe 5 10. An astronomical telescope uses two lenses of powers 10 D and 1 D. What is its magnifying power in normal adjustment? Ans. Power of eyepiece, Pe = 10 D Power of objective, Po = 1 D Magnifying power, M = fo Pe or M = 10 = – 10 fe Po 1 11. The magnifying power of an astronomical telescope in the normal adjustment position is 100. The distance between the objective and the eyepiece is 101 cm. Calculate the focal lengths of the objective and of the eyepiece. Ans. m = – fo ; – 100 = – fo fe fe or fo = 100 fo ...(i) Now, fo + fe = 101 cm ...(ii) From (i) and (ii), 101 fe = 101 or fe = 1 cm; fo = 100 × 1 cm = 100 cm 12. Why do we prefer an objective of large aperture in a telescope? Ans. The objective of large aperture permits a wider beam of light to be incident on it. So, the refracted light produces large illumination in the eye. Thus, even distant stars which are too faint to be seen with naked eye become clearly visible. However, the objective of very large aperture will create the problems of aberrations.
Chapter 3: Wave Optics (NCERT Textbook Chapter 10) SUMMARY OF THE CHAPTER Huygen’s principle: (i) Every point on a given wave front (known as primary wave front) acts as a source of secondary wavelet, which travel in all directions with the velocity of light in the medium. (ii) Any surface touching these secondary wavelets tangentially in the forward direction at any instant yields the new wave front at that instant. It is known as secondary wave front. The wavelengths of light changes when the speed of light changes—which happens when light crosses an interface from one medium to another. Consider some monochromatic light having wavelength and speed c in vacuum. Suppose it has wavelength and speed v in a medium of refractive index . = c = v = Thus, greater the refractive index of a medium, smaller the wavelength of light in that medium. According to the superposition principle, when two or more waves travelling through a medium superpose one another, the resultant displacement ( y ) of the newly formed wave is equal to the vector sum of the displacements (y1, y2, y3, ...) due to individual waves at that instant, i.e. y = y1 y2 y3 . . . In case of a single source of light, the distribution of light energy in the surrounding medium is uniform in all the directions. However, in case of two coherent sources of light emitting continuous waves of same amplitude, same wavelength and in the same phase (or some constant difference in phase), the distribution of light energy no longer remains uniform in all directions. There occurs some kind of modification in energy distribution, which is called interference. Thus, interference of light is the phenomenon of redistribution of light energy in the medium due to superposition of light waves from two coherent sources. The interference is constructive at the points, where the resultant intensity of light is maximum and the interference is destructive at the points, where the resultant intensity of light is minimum. Thomas Young in 1802, demonstrated the phenomenon of interference of light by a simple experiment, known as Young’s double slit experiment. 119
120 Physics—XII The alternate bright and dark bands, running Screen parallel to the lengths of slits appear on the screen. These are the interference bands or interference fringes. The bright and dark Mono- S A bands are placed alternately and are equally chromatic B spaced. In case S is the source of white light, the interference fringes are coloured and their source of Dark widths are unequal. band light The dots () represent the positions of Bright constructive interference, where crest of one Superposition of wave fronts band wave falls on the crest of the other and the trough of one wave falls on the trough of the other. The resultant amplitude and hence intensity of light is maximum at these positions. Also, the crosses (×) represent the positions of destructive interference, where crest of one wave falls on the trough of the other and vice-versa. The resultant amplitude and hence intensity of light is minimum at these positions. Suppose the waves from two coherent sources of light are given by, y1 = a sin t ...(i) y2 = b sin (t + ) ...(ii) where a and b are the respective amplitudes of the waves and is phase difference. Then, the resultant wave has an amplitude R given by, y = R sin (t + ) ...(iii) and R = a2 b2 2ab cos ...(iv) Since the resultant intensity I is directly proportional to the square of amplitude of the resultant wave, I R2 (a2 + b2 + 2ab cos ) (i) For constructive interference, = 0, 2, 4, . . . i.e. = 2n, where n = 0, 1, 2, . . . If x is the path difference between the two waves reaching any arbitrary point (say P), corresponding to phase difference , then x = n, where n = 0, 1, 2, . . . (ii) For destructive interference, = , 3, 5, . . . or = (2n – 1), where n = 1, 2, . . . The corresponding path difference will be given by x = (2n – 1) , where n = 1, 2, . . . 2 For constructive interferences, cos = 1 Rmax = (a + b) and Imax (a + b)2 or Imax = k(a + b)2 For destructive interference, cos = –1 Rmin = (a – b) and Imin (a – b)2 or Imin = k(a – b)2
Wave Optics 121 Imax = (a b)2 Imin (a b)2 Also, IR = I1 + I2 + 2 I1I2 cos If w1 and w2 are widths of two slits from which intensities of light I1 and I2 emanate, we can write I1 = w1 a2 I2 w2 b2 During interference, energy is being transferred from the regions of destructive interference to the regions of constructive interference. The energy is neither created nor destroyed during interference. Thus, the principle of energy conservation is being obeyed in the process of interference. In figure A and B are fine slits (distance d apart) that X are illuminated by a strong source of monochromatic light P of wavelength . XY is a screen at a distance D from the A x slits. In case of nth bright fringes. dO M xn = nD B d D and for dark fringes, xn = (2n – 1) D Y 2d The separation between the centres of two consecutive bright or consecutive dark fringes is width of the fringe, called fringe width. The fringe width is denoted by , where = D d The coherent sources of light, which emit continuous light waves of the same wavelength, same frequency and in same phase or have a constant difference in phase are called coherent sources. The phenomenon of bending of light around corners of an obstacle or aperture in the path of light is called diffraction. Due to bending, the light penetrates into the geometrical shadow of the obstacle and deviates from its linear path. The phenomenon of diffraction is more prominent when the slit width is of the order of wavelength of light. The narrower the slits, the greater the spreading. As seen in figure (c), when slit width is of the order of wavelength of light, diffraction is more pronounced. Diffraction is common to all types of waves. It is readily observed in case of sound and radio waves but not so in case of visible light, because for visible light is of the order of 10–6 m and obstacles/apertures of this size are hardly available. In case of diffraction at a single slit, the condition for nth secondary minima is, path difference = a sin n = n or sin n = n a where n gives the direction of the nth secondary maximum and n = 1, 2, 3, . . .
122 Physics—XII Also, for nth secondary maximum, a sin n = (2n + 1) or sin n = (2n + 1) where n = 1, 2, 3, . . . 2 2a Incident wave Diffracted wave Diffracted wave Diffracted wave l l l a a a Screen a = 7l a = 3l a = 1.5l (a) (b) (c) The point O corresponds to the position of central bright maximum and the points with path difference, a sin = , 2, ... are secondary minima. If f is focal length of the lens, held very close to the slit, then f = D = distance of the slit from the screen, sin ~ x x , q X fD A P where is half the angular width of central maximum of diffraction pattern of single slit. As sin = Bq q x a O x = or x = D f C Y Da aa D Also, width of central maximum = 2x = 2D = 2f aa The width of central maximum is the distance between first secondary minimum on the either side. Y Intensity X¢ – l O l 2l 3l X – 3l – 2l Path difference
Wave Optics 123 Important Results 1. Velocity of light (c) is given by, c = 2. Refractive index () of the medium is given by = c(air or vacuum) air v(medium) mcd 3. If y1, y2, y3, ..., yn are the displacements due to the different waves acting separately, then as per the principle of superposition, the resultant displacement when all the waves act together is given by their vector sum: y y1 y2 y3 ... yn 4. For constructive interference, path difference is given by, xnd = n, D where n = 1, 2, 3, . . . for 1st, 2nd, 3rd... bright fringes respectively. 5. For destructive interference, path difference is given by, xnd = (2n – 1) , D2 where n = 1, 2, 3, . . . for 1st, 2nd, 3rd . . . dark fringes respectively. 6. Fringe width, = D d 7. Wavelength of light used, = d D 8. Imax = k(a + b)2 9. Imin = k(a – b)2, where k is a constant and a, b are the amplitudes of two interfering waves. 10. If w1 and w2 are slit widths from which intensities of light I1 and I2 emanate, we have w1 = I1 a2 w2 I2 b2 11. The amplitude (R) of the resultant wave is given by, R = a2 b2 2ab cos is phase angle 12. IR = I1 + I2 + 2 I1I1 cos , where IR is resultant intensity. 13. If a transparent sheet (refractive index ) of thickness t is kept in one of the paths of interfering waves, the lateral shift of fringes (y0) is given by y0 = ( 1)t 14. The condition for diffraction minima is, a sin n = n, where n = 1, 2, 3, ... for 1st, 2nd, 3rd dark bands respectively.
124 Physics—XII 15. The condition for diffraction maxima is a sin n = (2n + 1) , where n = 1, 2, 3, . . . 2 for 1st, 2nd, 3rd, . . . bright bands respectively. 16. The width of the central maximum is given by, 2x = 2D 2f aa HUYGENS’ PRINCIPLE AND INTERFERENCE VERY SHORT ANSWER TYPE QUESTIONS (1 Mark) 1. How is a wavefront related to the direction of corresponding rays? Ans. The wavefront is perpendicular to the direction of rays. 2. Can the phenomenon of rectilinear propagation, reflection and refraction be explained on the basis of wave nature of light? Ans. Yes. 3. Can we perform double slit experiment with ultraviolet light? Ans. No. 4. If no particular colour of light or wavelength is specified, then the refractive index of the medium refers to which colour? Ans. Yellow colour. 5. Name one effect which could not be explained by Huygens’ wave theory. Ans. Photoelectric effect. 6. What is the effect on fringe width if the distance between the coherent sources is decreased? Ans. Fringe width increases. 7. Why a thin film of oil on the surface of water appears coloured? Ans. It is a case of interference in thin films. 8. Can we produce interference with white light? Ans. Yes. Coloured fringes will be formed. 9. In Young’s double slit experiment, a thin plate of some transparent material is introduced in the path of one of the interfering rays. What would happen? Ans. The fringes would suffer a displacement. 10. What will be the effect on the interference pattern if the phase difference between the two interfering waves changes continuously? Ans. There will be no sustained interference. 11. Can you produce interference with sound waves in air? It should be noted here that sound waves are longitudinal. Ans. Yes. Longitudinal waves do show interference.
Wave Optics 125 12. Can we produce interference with two electric bulbs placed side by side? Ans. No. Interference cannot be produced by independent sources of light. 13. Name any five factors on which the speed of light in vacuum depends. Ans. The speed of light in vacuum is a universal constant. It does not depend on any factor. 14. Does the speed of light in vacuum depend upon (i) direction of propagation (ii) wavelength (iii) intensity of wave? Ans. (i) No, (ii) No, (iii) No. 15. A region is illuminated by two sources of light. The intensity I at each point is found to be equal to I1 + I2 where I1 is the intensity of light at the point when source 2 is absent. I2 is similarly defined. Are the sources coherent or incoherent? Ans. Incoherent. 16. Does the speed of light in vacuum depend upon the nature of source? Ans. The speed of light in vacuum is independent of the nature of source. 17. Is the speed of light in glass independent of the colour of light? Ans. No. The refractive index and hence the speed of light in a medium depends on the wavelength. 18. In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light? Ans. For a given frequency, the intensity of light in the photon picture is determined by the number of photons per unit area. 19. What is the effect on the interference pattern in Young’s double slit experiment when: screen is moved closer to the plane of slits? Explain your answer. Ans. Fringe width, = D d When screen is moved closer to the plane of the slits, the distance D decreases, the fringe width also decreases. 20. In Young’s double slit experiment, three lights, blue, yellow and red are used successively. For which colour, will the fringe width be maximum? Ans. As r > y > b and = D , therefore fringe width will be maximum for red colour. d 21. What is the relation of a wavefront with a ray of light? Ans. The rays of light are always normal to the wavefront. 22. What happens when a thin transparent film is placed just in front of one of the two slits in the Young’s double slit experiment? Ans. The interference pattern shall get displaced. 23. Define the term coherence for light waves. Ans. When phase difference between interfering light waves remains constant, the light waves are called coherent. 24. What is the shape of the wavefront of a beam of parallel rays? Ans. The wavefront is plane in shape.
126 Physics—XII 25. What is the geometrical shape of the wavefront of the light diverging from a point source? Ans. Spherical. 26. Name the shape of a wavefront originating from a point source. Ans. Spherical wavefront. 27. The monochromatic source of light in Young’s double slit experiment is replaced by another monochromatic source of shorter wavelength. What will be the effect? Ans. Both the fringe width and the angular separation decrease. 28. What is the shape of the wavefront emitted by a light source in the form of a narrow slit? Ans. Cylindrical. 29. State the path difference between two waves for destructive interference. Ans. Path difference, = (2n + 1) , where n = 0, 1, 2, ...... 2 30. State the path difference between two waves for constructive interference in terms of . Ans. Path difference, = (2n) , where n = 0, 1, 2, ...... 2 31. What is the effect on the interference fringes in Young’s double slit experiment if the separation between the two slits is increased? Ans. Both the fringe width and the angular separation decrease. 32. What is the phase difference corresponding to a path difference of ? Ans. 2. 33. What is the main condition to produce interference of light? Ans. The two sources of light must be coherent sources. 34. Why bubbles of colourless soap solution appear coloured in sunlight? Ans. This is due to interference of white light from the thin film of soap bubbles. 35. Give an example of interference of light in everyday life. Ans. When thin soapy water films and other thin films are exposed to white light, they exhibit colours due to interference. 36. When a wave undergoes reflection from a denser medium, what happens to its phase? Ans. A phase change of takes place. 37. If a wave undergoes refraction, what happens to its phase? Ans. No change in its phase. 38. What is the phase difference between any two points of a wavefront? Ans. Zero. 39. What is the geometrical shape of the wavefront when a plane wave passes through a convex lens? Ans. The wavefront will be converging in nature. 40. How does the angular separation of interference fringes change in Young's experiment, if the distance between the slits is increased? Ans. Angular separation = 1 d d When separation between two slits is increased, angular separation decreases.
Wave Optics 127 41. If s is the size of the source and its distance is a from the plane of the two slits, what should be the criterion for the interference fringes to be seen? Ans. For interference fringes to be seen, the condition s should be satisfied. ad 42. Draw a diagram to show refraction of a plane wavefront incident in a convex lens and hence draw the refracted wavefront. Ans. Incident plane Refracted wavefronts wavefront 43. Differentiate between a ray and a wavefront. Ans. Ray defines the path of light. Wavefront is the locus of points in the light wave having the same phase. 44. What are coherent sources of light? Ans. Coherent sources of light are those sources which emit light waves of same wavelength, same frequency and in same phase or having a constant phase difference. 45. State two conditions for two light sources to be coherent. Ans. 1. The phase difference between light waves emitted by the two sources should not change with the passage of time. 2. The light waves emitted by the two sources should be of the same wavelength. 46. What is a ray of light? Ans. The straight line perpendicular to the wavefront is called ray of light. 47. How would the angular separation of interference fringes in Young’s double slit experiment change when the distance between the slits and screen is doubled? [CBSE 2012] Ans. Angular separation = and is independent of slit-screen separation. So, angular d separation will not change. 48. How does the fringe width, in Young's double slit experiment, change when the distance of separation between the slits and screen is doubled? [CBSE 2012] Ans. = D , D . When D is doubled, the fringe width will become double. d 49. Define wavefront. [CBSE 2014 Comptt.] Ans. The locus of all particles of a medium which are vibrating in the same phase is called the wavefront. 50. When light travels from a rarer to a denser medium, it loses some speed. Does the reduction in speed imply a reduction in the energy carried by the light wave? [CBSE 2016, 2013]
128 Physics—XII Ans. No, the reduction in the speed of light does not imply the reduction in the energy of the light wave. This is because the energy carried by a wave depends on the amplitude of the wave and not on the speed of wave propagation. 51. In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light? [CBSE 2016] Ans. In the photon picture of light, intensity of light at a point is determined by the number of photons incident per unit area around that point. 52. What kind of fringes do you expect to observe if white light is used instead of monochromatic light? [CBSE 2018] Ans. The central fringe remains white. No clear fringe pattern is seen after a few (coloured) fringes on either side of the central fringe. 53. Bring out the essential difference between ‘source of light’ and ‘source of radio waves’. Ans. In radio wave sources, the charges oscillate in a particular orientation. In a visible light source, the atoms act as independent oscillators. So, the oscillations are randomly oriented. SHORT ANSWER TYPE–I QUESTIONS (2 Marks) 1. In a certain region in a thin film, we get 9 fringes with light of wavelength 4358 Å. How many fringes will be observed in the same region with light of wavelength 5893 Å? Ans. Clearly, n22 = n11 or n2 = n11 10 4358 = 7.4 2 5893 2. Name six properties of light which could be explained by Huygens’ wave theory. Ans. (i) Rectilinear propagation of light (ii) Interference (iii) Diffraction (iv) Polarisation (v) Reflection (vi) Refraction. 3. How can you justify the rejection of backward wavefront? Ans. The intensity at any point of a secondary wave is proportional to (1 + cos ) where is the angle between the central line and the normal [Stokes’ law]. The value of for backward wavefront is 180°. So, intensity is zero. 4. Does the speed of light in a medium depend upon the nature of the source? Ans. The speed of light in a medium is independent of the nature of the source. It is determined by the properties of the medium of propagation. This fact is also true for other waves e.g., sound waves, water waves, etc. 5. Mercury green light has a wavelength 5.5 × 10–5 cm. Deduce (i) frequency in MHz (ii) period (in micro second). Ans. Frequency, = 3 1010 Hz = 5.45 × 1014 Hz = 5.45 × 108 MHz 5.5 105 Period, T = 1 1 s = 1.8 × 10–15 s = 1.8 × 10–9 s 5.45 1014
Wave Optics 129 6. “In Young’s double slit experiment performed with a source of white light, only black and white fringes are observed”. Is this statement true? Ans. The given statement is false. When a source of white light is used in Young’s double slit experiment, the central bright fringe is white. But all other bright fringes are coloured. 7. Why should we have narrow sources to produce good interference fringes? Ans. A broad source is equivalent to a large number of narrow sources lying close to each other. This would result in overlapping of interference patterns. So, the resultant interference pattern is not distinct. 8. What changes in interference pattern in Young’s double slit experiment will be observed when (i) distance between the slits is reduced, (ii) the apparatus is immersed in water? 1 Ans. (i) d . So, increases with decrease in distance. (ii) Since decreases by a factor of n, therefore, is also reduced by a factor of n. 9. Sketch the wavefront corresponding to divergent rays. Ans. Rays Spherical wavefronts 10. Draw the type of wavefront that corresponds to a beam of white light coming from a very far off source. Ans. Rays Plane wavefronts 11. What will be the effect on the fringes of Young’s double slit experimental set-up if immersed in water? Ans. The fringes will become narrower. This is because the wavelength of light waves in water is less than the wavelength in air. Note that D d 12. In Young’s experiment on interference, what shall happen if monochromatic source is replaced by a source of white light? Ans. The interference patterns due to different component colours of white light overlap incoherently. The central bright fringes for different colours are at the same position. So, the central fringe is white. Since blue colour (neglecting violet and orange) has the minimum wavelength therefore the fringe closest on either side of the central white fringe is blue; the farthest is red. After a few fringes, no clear fringe pattern is seen.
130 Physics—XII 13. No interference pattern is detected when two coherent sources are infinitely close to each other. Why? Ans. We know that 1 . When d is very small, then is very large. Even a single d fringe may occupy the whole field of view. So, fringe pattern will not be detected. 14. What is the ratio of intensities at two points x and y on a screen in Young’s double slit experiment, where waves from S1 and S2 have path difference of (i) 0 and (ii) ? 4 Assume that the amplitudes of the waves from the two sources are equal. Ans. The equivalent phase differences are 0 and . 2 Ix = a2 a2 2a2 cos 0 4a2 2 Iy a2 a2 2a2 cos 2a2 1 2 15. If the screen is moved away from the plane of the slits, then what is the effect on the interference fringes in Young’s double slit experiment? Ans. The angular separation (/d) of the fringes remains constant. However, the actual separation of the fringes increases in proportion to the distance of the screen from the plane of the two slits. 16. In Young’s double slit experiment, explain with reason how the interference pattern changes, when width of the slits is increased. Ans. The width of each slit should be considerably smaller than the separation between the slits. When the slits are so wide that the condition is not satisfied, fringes are not seen. However, increase in the width of the slits does improve the brightness of the fringes. 17. Why is the interference pattern not detected when the two coherent sources are far apart? Ans. We know that = D or 1 . If d is large, then is very small. So, interference dd pattern is not detected. 18. How does the width of interference fringes in Young’s double slit experiment change when (a) the distance between the slit and the screen is decreased? (b) frequency of the source is increased? Justify your answer in each case. Ans. Fringe width, = D Dc . Clearly, D and 1. d d (a) When distance D between the slits and the screen is decreased, the fringe width decreases. (b) When the frequency is increased, the fringe width decreases.
Wave Optics 131 19. How does the fringe width of interference fringes change, when the whole apparatus of Young's experiment is kept in a liquid of refractive index 1.3? Ans. air = D ; water = air d 1.3 water = water D = 1 airD = 1 air d 1.3 d 1.3 So, the fringe width becomes 1 times the initial value. 1.3 20. State the reason, why two independent sources of light cannot be considered as coherent sources. Ans. Two independent sources of light cannot be coherent. This is because light is emitted by individual atoms when they return to ground state. Even the smallest source of light contains billions of atoms which obviously cannot emit light waves in the same phase. 21. What are coherent sources of light? Two slits in Young's double slit experiment are illuminated by two different sodium lamps emitting light of the same wavelength. Why is no interference pattern observed? Ans. Two sources of light having same frequency and a constant or a zero phase difference are said to be coherent. Light wave emitted from an ordinary source (like a sodium lamp) undergoes abrupt phase changes in times of the order of 10–9s. Thus, two independent sources of light will not have a fixed phase relationship and would be incoherent. 22. Two slits in Young’s double slit experiment are illuminated by two different sodium lamps emitting light of the same wavelength. Will you observe the interference pattern? Justify your answer. Ans. No interference pattern will be observed. Two independent sources of light cannot be coherent and hence cannot produce interference. 23. In Young’s double slit experiment, the fringe width obtained is 3.0 mm in air. If the apparatus is immersed in water (Refractive Index = 4/3), what will be the new fringe width? Ans. = 3 mm = 2.25 mm 4/3 24. What is the shape of the wavefront originating from (i) a point source and (ii) a line source? Ans. (i) The wavefront originating from a point source is spherical in shape. This is because the locus of all points equidistant from a point source is a sphere. (ii) All the points which are equidistant from the line source lie on the surface of a cylinder. So, the wavefront is cylindrical in shape. 25. What is the geometrical shape of wavefront of light emerging out of a convex lens, when a point source is placed at its focus? Ans. When a point source of light is placed at the focus of a convex lens, the light comes out of the lens as a parallel beam. A parallel beam of light corresponds to a plane wavefront. Thus, we can say that the light emerging out of the convex lens is in the form of a plane wavefront.
132 Physics—XII 26. What type of wavefront will emerge from a (i) point source, and (ii) distant light source? Ans. (i) Point source – Spherical wavefront (ii) Distant light source – Plane wavefront 27. What will be the effect on interference fringes if red light is replaced by blue light? Ans. = D d Wavelength of blue light is smaller than that of red light. So, fringe width will decrease. 28. When monochromatic light travels from a rarer to a denser medium, explain the following, giving reasons: (i) Is the frequency of reflected and refracted light same as the frequency of incident light? (ii) Does the decrease in speed imply a reduction in the energy carried by light wave? [CBSE 2013] Ans. (i) Yes, the frequency is the property of source. So, frequency does not change when light is reflected or refracted. (ii) No, decrease in speed does not imply reduction in energy carried by light wave. This is because the frequency does not change and according to the formula E = h, energy will be independent of speed. Energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation. 29. Does the appearance of bright and dark fringes in the interference pattern violate, in any way, conservation of energy? Explain. [CBSE 2015 Comptt.] Ans. The appearance of bright and dark fringes in the interference pattern does not violate the principle of conservation of energy. The light energy is distributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. There is no gain or loss of energy. 30. When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why? [CBSE 2016, 2013, 2011] Ans. Both reflection and refraction occur due to interaction of light with the atoms at the surface of separation. These atoms may be regarded as oscillators. Light incident on such atoms forces them to vibrate with the frequency of light. As the light emitted by these charged oscillators is equal to their own frequency of oscillation, so both the reflected and refracted lights have the same frequency as the frequency of incident light. 31. The ratio of the widths of two slits in Young's double slit experiment is 4:1. Evaluate the ratio of intensities at maxima and minima in the interference pattern. [CBSE 2015 Comptt.] Ans. w1 = 4 w1 = a12 4 , a1 = 2 2 Now, , w2 a22 1 a2 1 w2 1 Imax. a1 a2 2 a1 1 2 2 1 2 9 Imin. a1 a2 2 a2 2 1 1 = = 2 a1 1 a2
Wave Optics 133 32. Two coherent sources of intensity ratio 90 : 1 interfere. Deduce the ratio of the intensity between the maxima and minima in the pattern. Ans. I1 = 100 . I2 1 Now, I1 = a12 or 100 a12 or a1 = 10 or a1 = 10a2 I2 a22 1 a22 a2 For ‘maximum’, amplitude = a1 + a2 = 10a2 + a2 = 11a2 For ‘minimum’, amplitude = a1 – a2 = 10a2 – a2 = 9a2 Ratio of intensity between maxima and minima = (11a2 )2 1.49 (9a2 )2 33. In a two slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 × 10–2 m towards the slits, the change in fringe width is 3 × 10–5 m. If the distance between the slits is 10–3 m, calculate the wavelength of light used. Ans. = D, = D dd or = d = 10–3 3 105 m = 6 × 10–7 m = 6 × 10–7 × 109 Å = 6000 Å D 5 102 34. What is the effect on the interference fringes in a Young’s double slit experiment when the source slit is moved closer to the double-slit plane? Ans. Let s be the size of the source and S its distance from the plane of the two slits. For interference fringes to be seen, the condition s/S < /d should be satisfied ; otherwise, interference patterns produced by different parts of the source overlap and no fringes are seen. Thus, as S decreases (i.e. the source slit is brought closer), the interference pattern gets less and less sharp, and when the source is brought too close for this condition to be valid, the fringes disappear. Till this happens, the fringe separation remains fixed. 35. What is the effect on the interference fringes in a Young’s double slit experiment when the width of the source slit is increased? Ans. Same as in previous question. As the source-slit width increases, fringe pattern gets less and less sharp. When the source slit is so wide that the condition s/S < /d is not satisfied, the interference pattern disappears. 36. Sodium light has two wavelengths 1 = 589 nm and 2 = 589.6 nm. As the path difference increases, when is the visibility of the fringes a minimum? Ans. The visibility of the fringes is poorest when the path difference P is an integral multiple of 1 and a half-integral multiple of 2 (for example). As P is increased, this happens first when PP = 1 : P 1 12 = 0.29 mm. 1 2 2 2 2 1
134 Physics—XII 37. When monochromatic light is incident on a surface separating two media, both the reflected and refracted light have the same frequency as the incident frequency. Why? Ans. Reflection, refraction and scattering (in general) arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators which take up the frequency of light causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus, the frequency of scattered light is equal to the frequency of incident light. 38. A narrow pulse of light is sent through a medium. Will you expect the pulse to retain its shape as it travels through the medium? Ans. A pulse can be viewed as being made of harmonic waves with a large range of wavelengths. Since, the speed of propagation in a medium depends on wavelength, therefore, different wavelength components of the pulse travel with different speeds. So, the pulse will not retain its shape as it travels through the medium. SHORT ANSWER TYPE–II QUESTIONS (3 Marks) 1. If a particle is thrown horizontally at a speed of 3 × 108 m s–1, deduce the vertical fall in travelling 1 km distance. Given: g = 10 m s–2. Does that result depend upon the mass of the particle? Comment on your result, considering that Newton thought light is made up of corpuscles shot at a very large speed by the source. Ans. Horizontal distance, S = 1 km = 103 m Horizontal speed, v = 3 × 108 m s–1 (assumed to be uniform) Time taken, t= S 3 103 m s1 3 1 second v 108 m 105 If y be the vertical fall in time t, then y= 1 gt2 1 10 3 1 2 m = 5.5 × 10–11 m 2 2 105 The result is independent of the mass of the particle. As it is clear from the above result, the downward distance covered by the corpuscle is only 5.5 × 10–11 m for 103 m of horizontal distance covered by the corpuscle. So, we can neglect the effect of earth’s gravitation and safely assume that the corpuscles travel along a straight line. 2. Explain how Newton’s corpuscular theory predicts the speed of light in a medium, say water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment? Ans. In Newton’s corpuscular (particle) picture of refraction, the particles of light incident from a rarer to a denser medium experience a force of attraction normal to the surface. This results in an increase in the normal component of velocity but the component along the surface is unchanged. This implies that Since c sin i = v sin r or v sin i = n n > 1, c sin r v > c.
Wave Optics 135 The prediction is opposite to the experimental result (v < c). The wave picture of light is consistent with the experiment. 3. What is the effect on the interference fringes in a Young’s double slit experiment when the widths of two slits are increased? Ans. The angular size of the central diffraction band due to each slit is about /S where S is the width of each of the two slits. S should be sufficiently small so that these bands are wide enough to overlap and thus produce interference. This means /S >> /d i.e., the width of the each slit should be considerably smaller than the separation between the slits. When the slits are so wide that this condition is not satisfied, fringes are not seen. However, increase in the width of the slits does improve the brightness of the fringes. Thus, in practice, the two slits should be wide enough to allow sufficient light to pass through but narrow enough to cause enough diffraction from each slit to enable wavefronts from the two slits to overlap and interfere. 4. In Young’s double slit experiment, what is the effect of the following operations on interference fringes? (i) The screen is moved away from the plane of the slits. (ii) The monochromatic source is replaced by another monochromatic source of shorter wavelength. (iii) The monochromatic source is replaced by a source of white light. (iv) The width of the source slit is made wider. (v) The separation between the two slits is increased. (vi) The distance between the source slit and the plane of the slits is increased. (vii) The width of each of the two slits is of the order of wavelength of light source. Ans. (i) Angular separation of the fringes remains constant. The fringe width d increases in proportion to the distance of the screen from the plane of the two slits. (ii) Fringe width decreases. The angular separation also decreases. (iii) Coloured fringes will be obtained. But the centre of the pattern will be white. (iv) Overlapping of fringes will occur. The interference pattern will not be distinct. (v) Fringe width will decrease. The angular separation also decreases. (vi) Overlapping of fringes will occur. The interference pattern will not be distinct. (vii) Diffraction effects will superimpose on the interference pattern. 5. Figure shows an experimental set-up similar to Young’s double slit experiment to observe interference of light. S S1 P Here, SS2 – SS1 = 4 S2 O Write the condition of (i) constructive (ii) destructive interference at any point P in terms of path difference = S2P – S1P. Does the central fringe observed in the above set-up lie above or below O? Give reason in support of your answer.
136 Physics—XII Ans. (i) Constructive interference total = initial + = n, n = 0, 1, 2, ...... Given: initial = n or = n 1 4 4 4 (ii) Destructive interference, total = + = (2n – 1) , n = 1, 2, 3, ...... 4 2 or = 2n 3 2 2 For central fringe, n = 0 = – 4 The negative sign indicates that the central fringe is below point O. 6. In a double slit interference experiment, the two coherent beams have slightly different intensities I and I + I (I << I). Show that the resultant intensity at the maxima is nearly 4I while that at the minima is nearly zero. Ans. Imax. = I + (I + I) + 2 I(I I) cos 0 4I Imin. = I + (I + I) + 2 I(I I) cos 180 0 7. In Young's double slit experiment using monochromatic light of wavelength , the intensity of light at a point on the screen where path difference is , is K units. Find out the intensity of light at a point where path difference is . [CBSE 2012] 3 Ans. I = I1 + I2 + 2 I1I2 cos Let I0 be the intensity of either source. I1 = I2 = I0, I = 2I0 (1 + cos ) = 4I0 cos2 2 When p = , = 2 I = 4I0cos2 = 4I0 = K When p = , = 2 I = 4I0cos2 = 4I0 × 1 = K 33 3 4 4 8. If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern. [CBSE 2018] Ans. I2 = 50% of I1 = 0.5 I1 a22 = 0.5 a12, a2 = a1 2 Imax. (a1 a2 )2 a1 a1 2 1 1 2 2 1 2 Imin. (a1 a2 )2 2 1 2 2 1 = = = 1 2 = 34. 2 2 a1 a1 2
Wave Optics 137 9. The ratio of the intensities at minima to maxima in the interference pattern is 9 : 25. What will be the ratio of the widths of the two slits in the Young’s double slit experiment? [CBSE 2014] Ans. Intensity is proportional to width of slit. So, amplitude is proportional to the square root of the width of the slit. a1 = w1 ...(1) a2 w2 ...(2) Here w1 and w2 represent the widths of the two slits. Now, Imin. = (a1 a2 )2 a2 2 Imax. (a1 a2 )2 1 a1 a2 2 1 a1 a2 2 1 a2 9 1 a1 3 a1 or 8 a2 2 or a1 4 or 25 = or 5 a2 a1 a2 1 a2 2 1 a1 1 a1 From (1) and (2), w1 = 4 or w1 16 w2 1 w2 1 10. A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young’s double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide. [CBSE 2012] Ans. 1 = 800 × 10–9 m, 2 = 600 × 10–9 m, D = 1.4 m, d = 0.28 × 10–3 m Let n1th maximum corresponding to 1 coincide with n2th maximum corresponding to 2. Then n1 D1 = n2 D2 or n1 2 600 3 d d n2 1 800 4 The minimum integral value of n1 is 3 and that of n2 is 4. ymin. = n1 D1 3 1.4 800 109 m = 12 mm d 0.28 103 11. In Young’s double slit experiment, the two slits 0.15 mm apart are illuminated by monochromatic light of wavelength 450 nm. The screen is 1.0 m away from the slits. (a) Find the distance of the second (i) bright fringe, (ii) dark fringe from the central maximum. (b) How will the fringe pattern change if the screen is moved away from the slits?
138 Physics—XII Ans. d = 0.15 × 10–3m, = 450 × 10–9 m, D = 1.0 m (a) (i) Distance of nth bright fringe from central maximum = nD 2 1.0 450 109 m = 6 × 10–3 m = 6 mm d 0.15 103 (ii) Distance of nth dark fringe from central maximum = (2n – 1) D = (2 × 2 – 1) × 1.0 450 109 m = 4.5 × 10–3 m = 4.5 mm 2d 2 0.15 103 (b) Fringe width, = D d When the screen is moved away from the slits, D and hence increases. The angular width will remain the same. 12. In Young’s double slit experiment, monochromatic light of wavelength 630 nm illuminates the pair of slits and produces an interference pattern in which two consecutive bright fringes are separated by 8.1 mm. Another source of monochromatic light produces the interference pattern in which the two consecutive bright fringes are separated by 7.2 mm. Find the wavelength of light from the second source. What is the effect on the interference fringes if the monochromatic source is replaced by a source of white light? [CBSE 2012 Comptt.] Ans. Position of the nth bright fringe is given by nD from the central bright fringe. The d separation between two consecutive bright fringes is D . d With 1 = 630 nm, we have 1D = 8.1 mm d With 2, 2D = 7.2 mm d Dividing, we get 1 = 8.1 2 = 7.2 1 8 × 630 = 560 nm. 2 7.2 8.1 9 When the monochromatic light is replaced by a white light, (i) the central bright fring is white and (ii) all the other colours will form individual maxima with the least wavelength violet forming its bright fringe close to the central bright fringe. 13. Show that, in YDSE, the first slit, the centre of the S S1 q P double slit and the central fringe lie on a straight y2 y1 line as the source is moved. f Ans. At the central fringe, the total path difference should be zero. S2 D1 D2 SS2 + S2P – SS1 – S1P = 0 or (S2P – S1P) + (SS2 – SS1) = 0 or y1 d y2 d = 0 or y1 y2 or = – D1 D2 D1 D2
Wave Optics 143 Fringe width: = xn+1 – xn = (n 1) D nD D d dd I I0 Ox 6. In a modified set-up of Young's double slit experiment, it is given that SS2 – SS1 = 4 i.e., the source ‘S’ is not equidistant from the slits S1 P and S2. y (a) Obtain the conditions for constructive and S1 O destructive interference at any point P on the screen S d in terms of the path difference (S2P – S1P). (b) Obtain an expression for the fringe width. S2 (c) Does the observed central bright fringe lie above or below ‘O’? D [CBSE 2013 Comptt., CBSE Sample Paper 2013] Ans. Initial path difference between S1 and S2, 0 = SS2 – SS1 = 4 Path difference between disturbances from S1 and S2, at point P, = yd D Total path difference between the two disturbances at P, T = 0 + = yd 4 D For constructive interference, T = yd = n, where n = 0, 1, 2, ..... 4 D ynd = n 1 or yn = n 1 D ...(i) D 4 4 d For destructive interference, T = yd (2n 1) or yn d = 2n 1 1 4 D 2 D 2 2 or yn d = 2n 3 or yn = 2n 3 D D 2 2 2 2d
Unit VIII: Atoms and Nuclei Chapter 5: Atoms (NCERT Textbook Chapter 12) SUMMARY OF THE CHAPTER According to Thomson‘s model, every atom consists of a positively charged sphere of radius of the order of 10–10 m, where entire mass and positive charge of the atom are uniformly distributed. Inside the sphere, the electrons are embedded like seeds in a watermelon. The number of electrons is such that their total negative charge is equal to the positive charge of the atom. Therefore, the atom is electrically neutral. Thomson‘s model of atom could not explain large angle scattering of -particles from thin metal foils, as was observed by Rutherford. An -particle is helium nucleus containing two protons and two neutrons. Therefore, an alpha particle has two units of positive charge and four units of mass. Rutherford‘s -particle scattering experiment led to the discovery of atomic nucleus. Following observations were made from the scattering experiment: (i) Majority of the -particles passed straight through the gold foil or underwent scattering through small angles, (ii) Very few -particles scattered through large angles (greater than 90°), (iii) Rarely an -particle got scattered through an angle of 180°. Atomic number (Z) of an element is the number of protons present inside the nucleus of an atom of the element. It is also equal to the number of electrons revolving in different orbits around the nucleus of the atom. Mass number (A) of an element is the total number of protons and neutrons present inside the nucleus of the atom of an element. In order to calculate nuclear dimensions from the scattering experiment, distance of closest approach or distance of minimum approach of an -particle was obtained by Rutherford. If we neglect the loss of energy due to a-particle ++ + Nucleus interaction of -particle with the electrons, then at the distance of closest approach, ++ + Kinetic energy (KE) = Potential energy (PE) or 1 mv2 = 1 Ze(2e) 2 40 r0 r0 or r0 = 2Ze2 40 1 mv2 2 228
Atoms 229 Since an -particle cannot touch the periphery of the nucleus due to strong repulsion, the radius of the nucleus must be smaller than the calculated value of r0. The impact parameter (b) is the perpendicular distance of the velocity vector of the alpha particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom. Rutherford calculated analytically the relation between the impact parameter b and scattering angle , which can be written as, b= 1 Ze2 cot q 40 Ek 2 b Nucleus where Ek is the kinetic energy of the -particle when it is far away from the atom. Bohr’s atomic model is based on the following postulates: (i) An atom has a small and massive central core, called nucleus around which electrons revolve. The centripetal force required for rotation of electrons is provided by the electrostatic force of attraction between the electrons and the nucleus. (ii) The electrons are permitted to circulate only in those orbits for which angular momentum of an electron is an integral multiple of h , where h is Planck’s 2 constant. L = mvr = nh , n = 1, 2, 3, ... 2 (iii) An electron does not radiate energy while revolving in permissible orbits, called non-radiating or stationary orbits. (iv) An atom can emit or absorb radiation in the form of discrete energy photons only when an electron jumps from a higher to a lower orbit or from a lower to a higher orbit respectively. E2 – E1 = h Radius of permitted orbit in hydrogen-like atoms is, n2h2 r = 42mk Ze2 So, the radii of the permitted orbits are proportional to n2 and increase in the ratio of 1 : 4 : 9 : 16.... The parameter n is called the principle quantum number. Speed of electron in case of hydrogen atom is given by, v= c, where = 2ke2 n ch is a dimensionless quantity, called fine structure constant. = 1 137 v = 1 c 137 n
230 Physics—XII Total energy of the electron in nth orbit is, En =– 22 mk2 Z 2e4 n2h2 The negative value of the total energy indicates that the electron is bound to the nucleus by means of electrostatic attraction and some work is required to be done to pull it away from the nucleus. Whenever an electron makes a transition from a higher energy level n2 to a lower energy level n1, the difference of energy appears in the form of a photon. The frequency of the emitted photon is given by h = En2 En1 h = 22 mk2e4 1 1 or = 22 mk2e4 1 1 h2 n22 h3 n22 n12 n12 As c = , therefore wave number , which is the reciprocal of wavelength , is given by = 1 = 22 mk2e4 1 1 or = R 1 1 c ch3 n12 n12 n22 n22 where R = 22mk2e4 , is Rydberg’s constant whose value is 1.0973 × 107 m–1. ch3 The origin of the various series in the hydrogen (Lulytrmaavinosleerti)es spectrum are shown in the figure. When an electron jumps from any higher energy level Balmer series n2 = 2, 3, 4, ...... to a lower level n1 = 1, we get a set of spectral lines called Lyman series which lie in the n=1 Pa(sicnhfreanresder)ies ultraviolet region of the electromagnetic spectrum. n=2 Similarly, other spectral lines can be explained. n=3 Bsrearcieksett n=4 Bohr’s theory could successfully explain the spectrum n=5 Psfeurnieds of hydrogen-like single electron atoms and fails in the case of atoms with two or more electrons. Important Results 1. Kinetic energy (Ek) of -particle is given by, Ek = 1 mv2 1 Ze(2e) 2 40 r0 where r0 is the distance of closest approach of -particle, Z is atomic number, e is charge of the electron and 1 = 9 × 109 N m2 C–2 40
Atoms 231 2. Impact parameter (b) is given by, Ze2 cot 2 b= , where is the scattering angle. 40 Ek 3. Distance of closest approach, 2kZe2 where k = 1 = 9 × 109 Nm2 C–2 r0 = Ek , 40 4. The radius of a nucleus is of the order of a fermi, where 1 fermi (fm) = 10–15 m. 5. At the distance of closest approach (r0), the alpha particle is scattered through an angle of 180°. At this distance r0, the entire initial kinetic energy of the -particle gets converted into electrostatic potential energy. 6. The Rutherford’s model cannot explain the stability of an atom. 7. An -particle is a helium ion, i.e., a helium atom from which both the electrons have been removed. It has charge equal to + 2e and its mass is nearly four times the mass of a proton. 8. L = mvr = nh 2 9. h = En2 En1 n2h2 10. rn = 42 mkZe2 11. vn = 2 ke2 =. c, where = 2k e2 , is fine structure constant 1 . nh n ch 137 12. K.E. = k Ze2 2r 13. P.E. = k Ze2 r 14. Total energy, En = – 22 mk2 Z 2 e4 =– Rhc 13.6 eV, where R = 1.097 × 107 m–1 n2h2 n2 n2 15. K.E. = – En P.E. = – 2K.E. = 2En 16. Tn = 2rn n3h3 or Tn n3 vn = 42 mZk2 e4 17. Wave number, = 1 = R 1 1 where R = 22mk2e4 , is Rydberg’s constant. n12 ch3 n22 18. Ionisation potential = – 13.6 Z 2 volt n2
232 Physics—XII VERY SHORT ANSWER TYPE QUESTIONS (1 Mark) 1. What is the angular momentum of electron in the second orbit of Bohr’s model of hydrogen atom? Ans. h . 2. What is the value of fine structure constant? Ans. 1 . 137 3. What is the order of velocity of electron in the ground state of hydrogen atom? Ans. 106 m s–1. 4. What are the units of fine structure constant? Ans. It has no units. 5. Name the series of hydrogen atom which does not lie in the visible region. Ans. Lyman series. 6. Name one source of hydrogen spectrum. Ans. Hydrogen lamp. 7. What are the dimensions of the fine structure constant? Ans. It is dimensionless. 8. What is the expression for fine structure constant? Ans. The term e2 is fine structure constant. h 40 2 c 9. What will happen if electrons revolving around a nucleus come to rest? Ans. The electrons would fall into the nucleus on account of electrostatic attraction. 10. What is the physical meaning of ‘negative energy of an electron’? Ans. The physical meaning of negative energy is that the electron is bound to the atom. 11. Sharp lines are present in the spectrum of a gas. What does this indicate? Ans. This indicates that only certain definite frequencies of radiation are emitted in transitions between different energy states. 12. Name a physical quantity whose dimensions are the same as those of angular momentum. MNLAns. Planck’s constant Refer to Bohr’s postulate: Angular momentum is an integral multiple of h . 2 13. What does Rutherford scattering experiment prove? Ans. The atom consists of a tiny central core in which the whole of the positive charge and nearly all the mass of the atom is concentrated. 14. The ionisation potential of helium atom is 24.6 V. How much energy will be required to ionise it? Ans. The energy required to ionise helium atom will be 24.6 eV.
Atoms 233 15. Why are Bohr’s orbits called “stationary orbits”? Ans. This is because the electron can neither radiate nor absorb energy while revolving in an orbit. 16. Can a hydrogen atom absorb a photon having energy more than 13.6 eV? Ans. Yes, it can absorb. But the atom would be ionised. 17. Write the conclusion of Rutherford’s -particle scattering experiment. Ans. The whole of the positive charge of the atom is concentrated in a tiny central core in the atom. 18. Write the order of magnitude of the radius of nucleus. Ans. 10–15 m. 19. What is the relation for Paschen series lines of hydrogen spectrum? Ans. R 1 1 , where n = 4, 5, 6, ...... 32 n2 20. The size of the nucleus can be estimated by the scattering of which particles? Ans. -particles. 21. Name the series of hydrogen spectrum which has lowest wavelength. Ans. Lyman series. 22. Name the series of the hydrogen spectrum which lies in the visible region of the electromagnetic spectrum. Ans. Balmer series lies in the visible region of the electromagnetic spectrum. 23. The wavelengths of some of the spectral lines obtained in hydrogen spectrum are 1216 Å, 6463 Å and 9546 Å. Which one of these wavelengths belongs to the Paschen series? Ans. 9546 Å. 24. What is the difference between -particle and helium atom? Ans. -particle is only the nucleus of helium atom. 25. Which atomic part was discovered by Rutherford? Ans. Nucleus. 26. What is the ratio of volume of atom to the volume of nucleus? Ans. 1015. 27. How is impact parameter related to angle of scattering? Ans. Impact parameter, b 1 Ze2 cot . 40 2 1 mv2 2 28. Among alpha, beta and gamma radiations, which get affected by the electric field? Ans. Alpha and beta radiations. 29. Why Thomson’s atomic model failed? Ans. This is because this model could not explain -ray scattering. 30. What do you mean by ground state atom? Ans. An unexcited atom is called ground state atom.
234 Physics—XII 31. What is the ratio of radii of the orbits corresponding to first excited state and ground state in a hydrogen atom? Ans. rn n2; r2 22 =4 r1 12 32. Write the expression for Bohr’s radius in hydrogen atom. Ans. Bohr’s radius = 1 h2 = h20 k 42me2 me2 33. The radius of innermost electron orbit of a hydrogen atom is 5.3 × 10–11 m. What is the radius of orbit in the second excited state? Ans. r = n2 × 5.3 × 10–11 m Radius of orbit in the second excited state (n = 3) is: r = 32 × 5.3 × 10–11 m = 4.77 × 10–10 m SHORT ANSWER TYPE–I QUESTIONS (2 Marks) 1. What is the physical significance of impact parameter? Ans. The impact parameter decides the angle of scattering. When the impact parameter is small, the electrostatic force is large. So, the scattering is also large. When the impact parameter is zero, the electrostatic force is maximum and the scattering angle is also maximum i.e., 180°. When the impact parameter is large, the electrostatic force is weak and the angle of scattering is also small. 2. Most mass of the atom is with positive charge. In the case of hydrogen atom, what fraction of atomic mass is with positive charge? Ans. In a hydrogen atom, an electron (negative charge) revolves around the proton (positive charge). The ratio of mass of proton and the mass of electron is 1.67 1027 i.e., 9.1 1031 1835. So, the fraction of the atomic mass with the positive charge is 1835 . 1836 3. Two energy levels in an atom are separated by 2.3 eV. What is the frequency of radiation emitted when the atom transits from upper to lower level? Ans. E = h = 2.3 eV = 2.3 × 1.6 × 10–19 J or = 2.3 1.6 1019 Hz = 5.6 × 1014 Hz 6.6 1034 4. Which level of the doubly ionised lithium (Li++) has the same energy as the ground state energy of the hydrogen atom? Compare the orbital radii of the two levels. Ans. En Z2/n2. For Li++, Z = 3. Therefore, (n = 3) state of Li++ has the same energy as (n = 1) state of hydrogen. Now rn n2 . Therefore, r3 (Li++) = 3r1 (H). Z 5. Determine the speed of the electron in the n = 3 orbit of He+. Is the non-relativistic approximation valid? Ans. v = n , r= 402 n2 Ze2 mr Ze2m That is v = 40n
Atoms 235 For n = 3, Z = 2, v = 1.46 × 106 m s–1. Thus, v 0.005, non-relativistic approximation c is valid because v < < 1. c 6. What path should an electron follow in Rutherford model? Use classical electromagnetic theory. Ans. According to classical electromagnetic theory, an accelerated charge should emit electromagnetic radiation. Consequently, the energy and radius should also continuously go on decreasing. Thus, the electron shall follow a spiral path and finally fall into the nucleus. 7. How did de Broglie equation lead to the quantisation condition laid down by Bohr? Ans. In the year 1925, E. Schrodinger postulated that the circumference of a stationary Bohr orbit of radius r is equal to integral number of times the wavelength of the matter wave. Figure shows three waves in the circumference of the orbit. 2r = n = nh or mvr = nh mv 2 which is the quantisation condition laid down by Bohr. 8. An electron revolves in a circular orbit around a nucleus of charge Ze. How is the electron velocity related to the radius of its orbit? Ans. mvr = nh , v= nh r 1 n2h2 2 2mr Here, k 42mZe2 9. The wavelengths of some of the spectral lines obtained in hydrogen spectrum are 9546 Å, 6463 Å and 1216 Å. Which one of these wavelengths belongs to Lyman series? Ans. The Lyman series lies in the ultraviolet region. So, the wavelength of any line in the Lyman series has to be less than the wavelength in the visible region i.e., less than 3900 Å. Clearly, 1216 Å belongs to the Lyman series. 10. Define (i) distance of closest approach and (ii) impact parameter. Ans. (i) It is distance of the centre of the nucleus from the point where the a-particle loses the whole of its kinetic energy. (ii) It is the perpendicular distance of the velocity vector of the alpha particle from the centre line of the nucleus when the potential energy of the alpha particle is zero. 11. In Bohr’s atomic model, the radius of the first electron orbit of a hydrogen atom is x metre. What is the radius of second orbit? Ans. rn n2 r2 = 4 or r2 = 4r1 = 4x metre. r1 12. Write two limitations of Bohr’s atomic model. Ans. 1. It is not clear as to why only circular orbits are considered. 2. The fine structure of spectral lines of hydrogen could not be explained.
236 Physics—XII 13. Define ionisation energy. What is its value for a hydrogen atom? Ans. Ionisation energy is the energy required to remove an electron completely from the outermost shell of the atom. The ionisation energy required to remove an electron from ground state of hydrogen atom is E = E – E1 = 0 – (–13.6) eV = 13.6 eV 14. Why is the classical (Rutherford) model for an atom (of electron orbiting around the nucleus) not able to explain the atomic structure? [CBSE 2012 Comptt.] Ans. The electrons revolving around the nucleus are continuously accelerated. Since an accelerated charge emits energy therefore the radius of the circular path of a revolving electron should go on decreasing. The electron should then ultimately fall into the nucleus. Thus, the classical model could not explain the atomic structure. 15. Define the distance of closest approach. An -particle of kinetic energy ‘K’ is bombarded on a thin gold foil. The distance of the closest approach is ‘r’. What will be the distance of closest approach for an -particle of double the kinetic energy? [CBSE 2017] Ans. The distance from the nucleus, where kinetic energy of -particle is completely converted into potential energy is known as the distance of closest approach. r = 1 2 z e2 or r 1 4 0 K K If kinetic energy will be double, then the distance of closest approach will become half. 16. Write two important limitations of Rutherford nuclear model of the atom. [CBSE 2017] Ans. The two important limitations of Rutherford nuclear model of the atom are: (i) This model cannot explain the stability of matter. (ii) It cannot explain the characteristic line spectra of atoms of different elements. 17. An alpha particle is scattered through an angle of 10° on passing through a thin copper foil. If energy of the particle is 5 MeV, what is the impact parameter? Given: Z of copper = 29. Ans. 1 m vi2 = 5 MeV = 5 × 106 × 1.6 × 10–19 J = 8 × 10–13 J, = 10°, Z = 29 2 Impact parameter, b= 1 Z e2 cot 40 2 1 mvi2 2 or b = 9 × 109 29 (1.6 1019 )2 cot 5 m= 9 29 1.6 1.6 11.4286 × 10–16 m 8 1013 8 = 0.955 × 10–13 m 18. The number of particles scattered at 60° is 100 per minute in an -particle scattering experiment, using gold foil. Calculate the number of particles per minute scattered at 90° angle. Ans. We know that the number N of particles scattered at an angle is directly proportional to 1 . sin4 2
Atoms 237 N = k1 sin4 2 In the first case, 100 = k 1 k or k = 100 = 6.25 sin4 60 sin4 30 16 2 The number of particles scattered at 90° is given by N = 6.25 × 1 = 6.25 × 4 = 25 minutes 90 sin4 2 19. In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of an alpha particle before it comes momentarily to rest and reverses its direction? Given: Kinetic energy of a-particle = 7.7 MeV, Z = 79 for gold. Ans. Let r0 be the centre-to-centre distance between the alpha particle and the gold nucleus when the -particle is at its stopping point. Now, r0 = 1 (Ze)(2e) = 9 109 79 2(1.6 1019 )2 m 40 Ek 7.7 106 1.6 1019 = 2.95 × 10–14 m = 29.5 fm This is a small distance on the atomic scale, well under 10–9 m (atomic dimensions). However, this is not a small distance on the nuclear scale. It is considerably larger than the sum of the radii of the gold nucleus and the alpha particle. Thus the alpha particle reverses its motion without ever actually touching the gold nucleus. 20. An alpha particle having kinetic energy of 7.68 MeV is projected towards the nucleus of copper (Z = 29). Calculate its distance of nearest approach. Ans. At the distance of nearest approach, P.E. = K.E. k (Ze)(2e) = 7.68 MeV = 7.68 × 1.6 × 10–13 J r0 or r0 = k (Ze) 2e m= 9 109 29 2 (1.6 1019 )2 m 7.68 1.6 1013 7.68 1.6 1013 = 1.09 × 10–14 m 21. In a head-on collision between an alpha particle and a gold nucleus, the minimum distance of approach is 4 × 10–14 m. Calculate the energy of the alpha particle. Atomic number of gold = 79. Ans. At the minimum distance of approach r0, K.E. = k (Ze) (2e) 9 109 79 2 (1.6 1019 )2 J r0 4 1014 = 9 109 79 2 (1.6 1019 )2 MeV [ 1 MeV = 1.6 × 10–13 J] 4 1014 1.6 1013 = 5.688 MeV
238 Physics—XII 22. The radius of the first electron orbit of a hydrogen atom is 5.3 × 10–11 m. What is the radius of the second orbit? Ans. We know that rn n2 r2 = 2 2 =4 or r2 = 4r1 r1 1 or r2 = 4 × 5.3 × 10–11 m = 2.12 × 10–10 m 23. The total energy of an electron in the first excited state of the hydrogen atom is – 3.4 eV. What is the potential energy of the electron in this state? Ans. Ep = – ke2 and E = – ke2 r 2r Now, Ep = 2 or Ep = 2E E Ep = 2(– 3.4) eV = – 6.8 eV 24. The total energy of an electron in the first excited state of the hydrogen atom is – 3.4 eV. What is the kinetic energy of the electron in this state? Ans. Ek = ke2 , E = – ke2 2r 2r In the first excited state, Ek = – E = – (– 3.4) eV = 3.4 eV 25. Calculate the shortest wavelength of Balmer series. Or Calculate the wavelength of the Balmer series limit. Given: R = 10970000 m–1. Ans. The last spectral line of Balmer series has the shortest wavelength or highest frequency. This line is obtained when the electron jumps from last orbit (n2 = ) to second orbit (n1 = 2). 1 =R 1 1 or 1R or = 4 22 2 4 R or = 4 m = 3.646 × 10–7 m = 3646 × 10–10 m = 3646 Å 10970000 It will lie in the ultraviolet region. 26. A muon is a particle that has the same charge as an electron but is 200 times heavier than it. If we had an atom in which the muon revolves around a proton instead of an electron, what would be the magnetic moment of the muon in the ground state of such an atom? Ans. = e 2m M L or M = e h eh = 1.6 1019 6.63 1034 7 A m2 = 4.6 × 10–26 A m2 2m 2 4m 4 22 200 9.1 1031 27. In a hydrogen atom, electron moves from second excited state to first excited state and then from first excited state to ground state. Find ratio of wavelengths obtained. Ans. 1 = R 1 1 R 5 3 2 22 32 36
Atoms 239 1 = R 1 1 R 3 2 1 12 22 4 3 2 = 36 3 27 2 1 54 5 28. An electron in hydrogen atom makes transition from an excited state of energy – 0.85 eV to its ground state. Find out energy of photon emitted in this transition. Ans. Energy of photon = E2 – E1 = [– 0.85 – (– 13.6)] eV = 12.75 eV. l1 l3 29. Find the relation between the three wavelengths 1, 2 and l2 3 from the energy level diagram shown in the figure. [CBSE 2016] Ans. E3 = E1 + E2 or hc = hc hc or 1 11 3 = 12 . 3 1 2 3 1 2 1 2 30. Which state of the triply ionised beryllium (Be+++) has the same orbital radius as that of the ground state of hydrogen? Compare the energies of the two states. Ans. rn = 402 n2 i.e., rn n2 Ze2m Z For hydrogen Z = 1, for Be+++, Z = 4. The (n = 2) state of Be+++ has the same radius as (n = 1) state of hydrogen. Now En Z2/n2. Therefore, E2 (Be+++ ) = 4. E1(H) 31. Why the mass of the nucleus has no significance in scattering in Rutherford’s experiment? Ans. The cause of scattering in Rutherford’s experiment is the charge on the nucleus. The scattering occurs due to the electric field of the charge on the nucleus. So, scattering is independent of the mass of the nucleus. 32. The Lyman series of hydrogen spectrum lies in the ultraviolet region. Why? Ans. This is because of the large energies involved in the transitions. As an illustration, the energy radiated corresponding to the first line of Lyman series is [– 3.4 – (– 13.6)] eV i.e., 10.2 eV. Such high energy photons are in the UV region of the spectrum. 33. The potential energy of an electron, in Bohr’s theory of hydrogen atom, is negative. Moreover, its numerical value is greater than the kinetic energy. What is the physical significance of these facts? Ans. The negative value of potential energy implies that the electron is bound to the nucleus. The greater numerical value of potential energy implies that the electron cannot leave the atom on its own. It would require a supply of external energy for doing so. 34. Find the series limit of Paschen series of hydrogen spectrum. Ans. 1 = R 1 1 = R 32 9 = 9 9 m = 8204 Å R = 1.097 107
240 Physics—XII 35. What is the shortest wavelength present in the Paschen series of spectral lines? Ans. The shortest wavelength of the spectral line (series limit) of Paschen series is given by 1 R 1 1 R min . 32 2 9 or min. = 9 9 m= 9 107 1010 Å = 8204.2 Å R 1.097 107 1.097 36. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level? Ans. h = E2 – E1 or = E2 E1 = 2.3 eV = 2.3 1.6 1019 J = 5.55 × 1014 Hz h 6.63 1034 J s 6.63 1034 J s 37. The ground state energy of hydrogen atom is – 13.6 eV. What are the kinetic and potential energies of the electron in this state? Ans. Energy, E = – ke2 ; Kinetic energy, Ek = ke2 ; Potential energy, Ep = ke2 2r 2r r Clearly, Ek = – E and Ep = 2 E Now, E = – 13.6 eV Ek = – (– 13.6 eV) = 13.6 eV Ep = 2 (– 13.6 eV) = – 27.2 eV 38. In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104 m s–1. (Mass of earth = 6.0 × 1024 kg) Ans. Using Bohr’s postulate m vnrn = nh 2 or n= 2m vnrn = 2 3.142 6 1024 3 104 1.5 1011 = 2.559 × 1074 h 6.63 10 34 SHORT ANSWER TYPE–II QUESTIONS (3 Marks) 1. A hydrogen atom contains one electron. But the spectrum of hydrogen atom has many lines. Why? Ans. In the ground state of the hydrogen atom, an electron occupies the lowest energy level. When the atom is excited, the electron jumps to some higher energy level. But, within 10–8 second, the electron jumps to the lowest energy level. However, it may so happen that the electron may first jump to some lower energy level and then to the lowest energy level. When an electron jumps from higher energy level to lower energy level, a spectral line is produced. Since a large number of energy levels are available therefore a large number of electron-transitions are possible. This explains as to why the spectrum of hydrogen atom contains a large number of lines.
Atoms 241 2. Formulate a table giving longest and shortest wavelengths of different spectral series. Ans. Name of the spectral Lower Upper Wavelength series State State (in Å) 1 2 1216 Lyman 1 (series limit) 912 2 3 6563 Balmer 2 (series limit) 3646 Paschen 3 4 18751 3 (series limit) 8220 3. In Bohr’s theory of hydrogen atom, the quantum number of orbits, radii of the orbits of electrons, the angular momenta and energies are represented by n, r, L and E respectively. Draw Bar graphs (separately for each) showing the variation of r, L and E with n. Choose convenient scales on y-axis for different quantities. Ans. 16 r μ n2 4 L = nh 2p r 12 3 8 L2 41 1234 n 1234 n (b) (a) n 1 234 –4 Eμ– 1 n2 E (c) –8 –12 –14 –16
242 Physics—XII 4. The energy level diagram of an element is given. A C –0.85 eV Identify, by doing necessary calculations, which B D –1.5 eV transition corresponds to the emission of a spectral line of wavelength 102.7 nm. –3.4 eV Ans. Energy of emitted photon, E = hc = 6.626 1034 3 108 eV = 12.1 eV –13.6 eV 102.7 109 1.6 1019 This energy corresponds to the transition D for which energy change = – 1.5 – (– 13.6) = 12.1 eV 5. Calculate the ratio of energies of photons produced due to transition of electron of hydrogen atom from its, (i) Second permitted energy level to the first level, and (ii) Highest permitted energy level to the second permitted level. Ans. The energy of electron in nth orbit of hydrogen atom is given by En = 13.6 eV n2 (i) E = E2 – E1 = – 13.6 13.6 = 13.6 13.6 = (13.6 – 3.4) eV = 10.2 eV 22 12 14 (ii) E = E – E2 = 13.6 13.6 = 0 + 3.4 = 3.4 eV 2 22 Required ratio is 10.2eV 3 . 3.4 eV 1 6. (i) In hydrogen atom, an electron undergoes transition from 3rd excited state to the first excited state and then to the ground state. Identify the spectral series to which these transitions belong. (ii) Find out the ratio of the wavelengths of the emitted radiations in the two cases. Ans. (i) The given transitions belong to : 1. Balmer series, 2. Lyman series (ii) 1 = R 1 1 = 3 R, 1 = R 1 1 3 R B 4 16 16 L 1 4 4 B = 16 3 4 L 34 7. In the ground state of hydrogen atom, its Bohr radius is given as 5.3 × 10–11 m. The atom is excited such that the radius becomes 21.2 × 10–11 m. Find (i) the value of the principal quantum number and (ii) the total energy of the atom in this excited state. Ans. (i) r2 = 21.2 1011 4; n2 2 =4 or n2 = 2n1 r1 5.3 1011 n1 So, the value of the principal quantum number in the excited state is 2. (ii) E = 13.6 = – 3.4 eV 22 So, the total energy of the atom in the excited state is – 3.4 eV.
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