APPENDICES APPENDIX A 1 THE GREEK ALPHABET APPENDIX A 2 COMMON SI PREFIXES AND SYMBOLS FOR MULTIPLES AND SUB-MULTIPLES AnswersAppendices APPENDIX A 3 SOME IMPORTANT CONSTANTS OTHER USEFUL CONSTANTS ANSWERS CHAPTER 9 9.1 v = –54cm. The image is real, inverted and magnified. The size of the image is 5.0 cm. As u → f, v → ∞; for u < f, image is virtual. 9.2 v = 6.7 cm. Magnification = 5/9, i.e., the size of the image is 2.5 cm. As u → ∞; v → f (but never beyond) while m → 0. 9.3 1.33; 1.7 cm 9.4 n = 1.51; n = 1.32; n = 1.144; which gives sin r = 0.6181 i.e.,ga wagwr ~ 38º. 9.5 r = 0.8 × tan i and sin ic = 1/1.33 ≅ 0.75 , where r is the radius (in m)cof the largest circle from which light comes out and i is the critical angle for water-air interface, Area = 2.6 m2 c9.6 n ≅ 1.53 and Dm for prism in water ≅ 10º 9.7 R = 22 cm 9.8 Here the object is virtual and the image is real. u = +12 cm (object on right; virtual) (a) f = +20 cm. Image is real and at 7.5 cm from the lens on its right side. (b) f = –16 cm. Image is real and at 48 cm from the lens on its right side. 9.9 v = 8.4cm, image is erect and virtual. It is diminished to a size 1.8 cm. As u → ∞, v → f (but never beyond f while m → 0). Note that when the object is placed at the focus of the concave lens (21 cm), the image is located at 10.5 cm (not at infinity as one might wrongly think). 9.10 A diverging lens of focal length 60 cm 9.11 (a) ve = –25 cm and fe = 6.25 cm give ue = –5cm; vO = (15 – 5) cm = 10 cm, fO = uO = – 2.5 cm; Magnifying power = 20 (b) = – 2.59 cm. uO Magnifying power = 13.5. 9.12 Angular magnification of the eye-piece for image at 25 cm 25 25 = +111 | u |= cm = .= ; 2 27cm ; vO = 7.2 cm 25 e 11. Separation = 9.47 cm; Magnifying power = 88 9.13 24; 150 cm 9.14 (a) Angular magnification = 1500 (b) Diameter of the image = 13.7 cm. 9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result. 9.16 The pin appears raised by 5.0 cm. It can be seen with an explicit ray diagram that the answer is independent of the location of the slab (for small angles of incidence). 9.17 (a) sin i′ c = 1.44/1.68 which gives i′ c = 59°. Total internal reflection takes place when i > 59° or when r < r max = 31°. Now, (sin /sin ) . max max i r = 168 , which gives i max ~ 60°. Thus, all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. (If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, i′ c = sin–1(1/1.68) = 36.5°. Now, i = 90° will have r = 36.5° and i′ = 53.5° which is greater than i′ c . Thus, all incident rays (in the range 53.5° < i < 90°) will suffer total internal reflections. 9.18 (a) Rays converging to a point ‘behind’ a plane or convex mirror are reflected to a point in front of the mirror on a screen. In other words, a plane or convex mirror can produce a real image if the object is virtual. Convince yourself by drawing an appropriate ray diagram. (b) When the reflected or refracted rays are divergent, the image is virtual. The divergent rays can be converged on to a screen by means of an appropriate converging lens. The convex lens of the eye does just that. The virtual image here serves as an object for the lens to produce a real image. Note, the screen here is not located at the position of the virtual image. There is no contra diction. (c) Taller (d) The apparent depth for oblique viewing decreases from its value for near-normal viewing. Convince yourself of this fact by draw ing ray diagrams for different positions of the observer. (e) Refractive index of a diamond is about 2.42, much larger than that of ordinary glass (about 1.5). The critical angle of diamond is about 24°, much less than that of glass. A skilled diamond- cutter exploits the larger range of angles of incidence (in the diamond), 24° to 90°, to ensure that light entering the diamond is totally reflected from many faces before getting out–thus producing a sparkling effect. 9.19 For fixed distance s between object and screen, the lens equation does not give a real solution for u or v if f is greater than s/4. 9.20 Therefore, f max = 0.75 m. 21.4 cm 535 9.21 (a) (i) Let a parallel beam be the incident from the left on the convex lens first. f1 = 30 cm and u1 = – ∞ , give v1 = + 30 cm. This image becomes a virtual object for the second lens. f2 = –20 cm, u2 = + (30 – 8) cm = + 22 cm which gives, v2 = – 220 cm. The parallel incident beam appears to diverge from a point 216 cm from the centre of the two-lens system. (ii) Let the parallel beam be incident from the left on the concave lens first: f1 = – 20 cm, u1 = – ∞, give v1 = – 20 cm. This image becomes a real object for the second lens: f2 = + 30 cm,u2 = – (20 + 8)cm = – 28 cm which gives, v2 = – 420 cm. The parallel incident beam appears to diverge from a point 416 cm on the left of the centre of the two-lens system. Clearly, the answer depends on which side of the lens system the parallel beam is incident. Further we do not have a simple lens equation true for all u (and v) in terms of a definite constant of the system (the constant being determined by f1 and f2, and the separation between the lenses). The notion of effective focal length, therefore, does not seem to be meaningful for this system. (b) u1 = – 40 cm, f1 = 30 cm, gives v1= 120cm. Magnitude of magnification due to the first (convex) lens is 3. u 2 = + (120 – 8) cm = +112 cm (object virtual); f2 = – 20 cm which gives v2 112 20 92 =− × cm Magnitude of magnification due to the second (concave) lens = 20/92. Net magnitude of magnification = 0.652 Size of the image = 0.98 cm 9.22 If the refracted ray in the prism is incident on the second face at the critical angle i c, the angle of refraction r at the first face is (60°–i c). Now, i c = sin–1 (1/1.524) ~ 41° Therefore, r = 19° sin i = 0.4962; i ~ 30° 9.23 Two identical prisms made of the same glass placed with their bases on opposite sides (of the incident white light) and faces touching (or parallel) will neither deviate nor disperse, but will mearly produce a parallel displacement of the beam. (a) To deviate without dispersion, choose, say, the first prism to be of crown glass, and take for the second prism a flint glass prism of suitably chosen refracting angle (smaller than that of crown glass prism because the flint glass prism disperses more) so that dispersion due to the first is nullified by the second. (b) To disperse without deviation, increase the angle of flint glass prism (i.e., try flint glass prisms of greater and greater angle) so that deviations due to the two prisms are equal and opposite. (The flint glass prism angle will still be smaller than that of crown glass because flint glass has higher refractive index than that of crown glass). Because of the 536 adjustments involved for so many colours, these are not meant to be precise arrangements for the purpose required. 9.24 To see objects at infinity, the eye uses its least converging power = (40+20) dioptres = 60 dioptres. This gives a rough idea of the distance between the retina and cornea-eye lens: (5/3) cm. To focus an object at the near point (u = –25 cm), on the retina (v = 5/3 cm), the focal length should be ⎡ 13⎤−1 25 += cm⎢ ⎥⎣25 5⎦ 16 corresponding to a converging power of 64 dioptres. The power of the eye lens then is (64 – 40) dioptres = 24 dioptres. The range of accommodation of the eye-lens is roughly 20 to 24 dioptres. 9.25 No, a person may have normal ability of accommodation of the eye-lens and yet may be myopic or hypermetropic. Myopia arises when the eye-ball from front to back gets too elongated; hypermetropia arises when it gets too shortened. In practice, in addition, the eye lens may also lose some of its ability of accommodation. When the eyeball has the normal length but the eye lens loses partially its ability of accommodation (as happens with increasing age for any normal eye), the ‘defect’ is called presbyopia and is corrected in the same manner as hypermetropia. 9.26 The far point of the person is 100 cm, while his near point may have been normal (about 25 cm). Objects at infinity produce virtual image at 100 cm (using spectacles). To view closer objects i.e., those which are (or whose images using the spectacles are) between 100 cm and 25 cm, the person uses the ability of accommodation of his eye-lens. This ability usually gets partially lost in old age (presbyopia). The near point of the person recedes to 50 cm. To view objects at 25 cm clearly, the person needs converging lens of power +2 dioptres. 9.27 The defect (called astigmatism) arises because the curvature of the cornea plus eye-lens refracting system is not the same in different planes. [The eye-lens is usually spherical i.e., has the same curvature on different planes but the cornea is not spherical in case of an astigmatic eye.] In the present case, the curvature in the vertical plane is enough, so sharp images of vertical lines can be formed on the retina. But the curvature is insufficient in the horizontal plane, so horizontal lines appear blurred. The defect can be corrected by using a cylindrical lens with its axis along the vertical. Clearly, parallel rays in the vertical plane will suffer no extra refraction, but those in the horizontal plane can get the required extra convergence due to refraction by the curved surface of the cylindrical lens if the curvature of the cylindrical surface is chosen appropriately. 19.28 (a) Closest distance = 4 cm ≈ 42. cm 6 Farthest distance = 5 cm (b) Maximum angular magnification = [25/(25/6)] = 6. Minimum angular magnification = (25/5) = 5 11 1 9.29 (a) += v 9 10 i.e., v = – 90 cm, Magnitude of magnification = 90/9 = 10. 537 Each square in the virtual image = 100 mm2 = 1 cm2 has an area 10 × 10 × 1 mm2 (b) Magnifying power = 25/9 = 2.8 (c) No, magnification of an image by a lens and angular magnification (or magnifying power) of an optical instrument are two separate things. The latter is the ratio of the angular size of the object (which is equal to the angular size of the image even if the image is magnified) to the angular size of the object if placed at the near point (25 cm). Thus, magnification magnitude is |(v/u)| and magnifying power is (25/ |u|). Only when the image is located at the near point |v| = 25 cm, are the two quantities equal. 9.30 (a) Maximum magnifying power is obtained when the image is at the near point (25 cm) u = – 7.14 cm. (b) Magnitude of magnification = (25/ |u|) = 3.5. (c) Magnifying power = 3.5 Yes, the magnifying power (when the image is produced at 25 cm) is equal to the magnitude of magnification. 9.31 Magnification = (.625/ )1 = 2.5 v = +2.5u + − = 1 25 1 1 10. u u i.e.,u = – 6 cm |v| = 15 cm The virtual image is closer than the normal cannot be seen by the eye distinctly. near point (25 cm) and 9.32 (a) Even though the absolute image size is bigger than the object size, the angular size of the image is equal to the angular size of the object. The magnifier helps in the following way: without it object would be placed no closer than 25 cm; with it the object can be placed much closer. The closer object has larger angular size than the same object at 25 cm. It is in this sense that angular magnification is achieved. (b) Yes, it decreases a little because the angle subtended at the eye is then slightly less than the angle subtended at the lens. The effect is negligible if the image is at a very large distance away. [Note: When the eye is separated from the lens, the angles subtended at the eye by the first object and its image are not equal.] (c) First, grinding lens of very small focal length is not easy. More important, if you decrease focal length, aberrations (both spherical and chromatic) become more pronounced. So, in practice, you cannot get a magnifying power of more than 3 or so with a simple convex lens. However, using an aberration corrected lens system, one can increase this limit by a factor of 10 or so. (d) Angular magnification of eye-piece is [(25/f e) + 1] ( f e in cm) which increases if f e is smaller. Further, magnification of the objective 538 is given by O O| | v u = O O 1 (| |/ ) 1u f − 9.33 9.34 9.35 9.36 which is large when || is slightly greater than fO. The micro-uOscope is used for viewing very close object. So uO is small, and||so is fO. (e) The image of the objective in the eye-piece is known as ‘eye-ring’. All the rays from the object refracted by objective go through the eye-ring. Therefore, it is an ideal position for our eyes for viewing. If we place our eyes too close to the eye-piece, we shall not collect much of the light and also reduce our field of view. If we position our eyes on the eye-ring and the area of the pupil of our eye is greater or equal to the area of the eye-ring, our eyes will collect all the light refracted by the objective. The precise location of the eye-ring naturally depends on the separation between the objective and the eye-piece. When you view through a microscope by placing your eyes on one end,the ideal distance between the eyes and eye-piece is usually built-in the design of the instrument. Assume microscope in normal use i.e., image at 25 cm. Angular magnification of the eye-piece 25 = 16+= 5 Magnification of the objective 30 = =5 6 11 1−= 5uO uO 1.25 which gives u= –1.5 cm; v= 7.5 cm. ||(25/6) cm = 4.17 cm. The u =O0e separation between the objective and the eye-piece should be (7.5 + 4.17) cm = 11.67 cm. Further the object should be placed 1.5 cm from the objective to obtain the desired magnification. (a) m = ( fO/f e) = 28 fO ⎡ fO ⎤(b) m = ⎢1 + = 33.6⎥fe ⎣ 25 ⎦(a) + f = 145 cm fO e(b) Angle subtended by the tower = (100/3000) = (1/30) rad. Angle subtended by the image produced by the objective hh = = fO 140 Equating the two, h = 4.7cm.(c) Magnification (magnitude) of the eye-piece = 6. Height of the final image (magnitude) = 28 cm. The image formed by the larger (concave) mirror acts as virtual object for the smaller (convex) mirror. Parallel rays coming from the object at infinity will focus at a distance of 110 mm from the larger mirror. The distance of virtual object for the smaller mirror = (110 –20) = 90 mm. The focal length of smaller mirror is 70 mm. Using the mirror formula, image is formed at 315 mm from the smaller mirror. 9.37 The reflected rays get deflected by twice the angle of rotation of the mirror. Therefore, d/1.5 = tan 7°. Hence d = 18.4 cm. 9.38 n = 1.33 CHAPTER 10 10.1 (a) Reflected light: (wavelength, frequency, speed same as incident light) λ = 589nm, ν = 5.09 × 1014 Hz, c = 3.00 × 108 m s–1 (b) Refracted light: (frequency same as the incident frequency) ν = 5.09 × 1014Hz v = (c/n) = 2.26 × 108 m s –1, λ = (v/ν) = 444 nm 10.2 (a) Spherical (b) Plane (c) Plane (a small area on the surface of a large sphere is nearly planar). 10.3 (a) 2.0 × 108 m s–1 (b) No. The refractive index, and hence the speed of light in a medium, depends on wavelength. [When no particular wavelength or colour of light is specified, we may take the given refractive index to refer to yellow colour.] Now we know violet colour deviates more than red in a glass prism, i.e. nv > nr . Therefore, the violet component of white light travels slower than the red component. –2 –312 10 ×. × ×0 28 10 . 10.4 λ= m = 600 nm ×.414 10.5 K/4 10.6 (a) 1.17 mm (b) 1.56 mm 10.7 0.15° 10.8 tan–1(1.5) ~ 56.3o 10.9 5000 Å, 6 × 1014Hz; 45° 10.10 40 m v 10.11 Use the formula λ′ – λ = λ c c 3×108 ×15 i.e., v = (λλ= = 6.86 × 105 m s–1' −)6563λ10.12 In corpuscular (particle) picture of refraction, 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 the velocity but the component along the surface is unchanged. This means v sin i c sin i = v sin r or ==n. Since n > 1, v > c.540 c sin r The prediction is opposite to the experimental results (v < c ). The wave picture of light is consistent with the experiment. 10.13 With the point object at the centre, draw a circle touching the mirror. This is a plane section of the spherical wavefront from the object that has just reached the mirror. Next draw the locations of this same wavefront after a time t in the presence of the mirror, and in the absence of the mirror. You will get two arcs symmetrically located on either side of the mirror. Using simple geometry, the centre of the reflected wavefront (the image of the object) is seen to be at the same distance from the mirror as the object. 10.14 (a) The speed of light in vacuum is a universal constant independent of all the factors listed and anything else. In particular, note the surprising fact that it is independent of the relative motion between the source and the observer. This fact is a basic axiom of Einstein’s special theory of relativity. (b) Dependence of the speed of light in a medium: (i) does not depend on the nature of the source (wave speed is determined by the properties of the medium of propagation. This is also true for other waves, e.g., sound waves, water waves, etc.). (ii) independent of the direction of propagation for isotropic media. (iii) independent of the motion of the source relative to the medium but depends on the motion of the observer relative to the medium. (iv) depends on wavelength. (v) independent of intensity. [For high intensity beams, however, the situation is more complicated and need not concern us here.] 10.15 Sound waves require a medium for propagation. Thus even though the situations (i) and (ii) may correspond to the same relative motion (between the source and the observer), they are not identical physically since the motion of the observer relative to the medium is different in the two situations. Therefore, we cannot expect Doppler formulas for sound to be identical for (i) and (ii). For light waves in vacuum, there is clearly nothing to distinguish between (i)and (ii). Here only the relative motion between the source and the observer counts and the relativistic Doppler formula is the same for (i)and (ii). For light propagation in a medium, once again like for sound waves, the two situations are not identical and we should expect the Doppler formulas for this case to be different for the two situations (i) and (ii). 10.16 3.4 × 10–4 m. 10.17 (a) The size reduces by half according to the relation: size ~ λ/d. Intensity increases four fold. (b) The intensity of interference fringes in a double-slit arrangement is modulated by the diffraction pattern of each slit. (c) Waves diffracted from the edge of the circular obstacle interfere constructively at the centre of the shadow producing a bright spot. (d) For diffraction or bending of waves by obstacles/apertures by a large angle, the size of the latter should be comparable to wavelength. If the size of the obstacle/aperture is much too large compared to wavelength, diffraction is by a small angle. Here 542 the size is of the order of a few metres. The wavelength of light is about 5 × 10–7 m, while sound waves of, say, 1 kHz frequency have wavelength of about 0.3 m. Thus, sound waves can bend around the partition while light waves cannot. (e) Justification based on what is explained in (d). Typical sizes of apertures involved in ordinary optical instruments are much larger than the wavelength of light. 10.18 12.5 cm. 10.19 0.2 nm. 10.20 (a) Interference of the direct signal received by the antenna with the (weak) signal reflected by the passing aircraft. (b) Superposition principle follows from the linear character of the (differential) equation governing wave motion. If y1 and y2 are solutions of the wave equation, so is any linear combination of y1 and y2. When the amplitudes are large (e.g., high intensity laser beams) and non-linear effects are important, the situation is far more complicated and need not concern us here. 10.21 Divide the single slit into n smaller slits of width a ′ = a/n . The angle θ = nλ/a = λ/a ′. Each of the smaller slits sends zero intensity in the direction θ. The combination gives zero intensity as well. CHAPTER 11 11.1 (a) 7.24 × 1018Hz (b) 0.041nm 11.2 (a) 0.34eV = 0.54 × 10–19J (b) 0.34V (c) 344km/s 11.3 1.5eV = 2.4 × 10–19 J 11.4 (a) 3.14 × 10–19J, 1.05 × 10–27 kg m/s (b) 3 × 1016 photons/s (c) 0.63 m/s 11.5 4 × 1021 photons/m2 s 11.6 6.59 × 10–34 J s 11.7 (a) 3.38 × 10–19 J = 2.11 eV (b) 3.0 × 1020 photons/s 11.8 2.0 V 11.9 No, because ν < ν o 11.10 4.73 × 1014 Hz 11.11 2.16 eV = 3.46 × 10–19J 11.12 (a) 4.04 × 10–24kg m s–1 (b) 0.164nm 11.13 (a) 5.92 × 10–24kg m s–1 (b) 6.50 × 106 m s–1 (c) 0.112nm 11.14 (a) 6.95 × 10–25 J = 4.34 μeV (b) 3.78 × 10–28 J = 0.236 neV 11.15 (a) 1.7 × 10–35m (b) 1.1 × 10–32m (c) 3.0 × 10–23m 11.16 (a) 6.63 × 10–25kg m/s (for both) (b) 1.24keV (c) 1.51eV 11.17 (a) 6.686 × 10–21J = 4.174 × 10–2eV (b) 0.145nm 11.18 λ = h/p = h/(hν/c) = c/ν 11.19 0.028 nm 11.20 (a) Use eV = (m v2/2) i.e., v = [(2eV/m)]1/2 ; v = 1.33 × 107 m s–1. (b) If we use the same formula with V = 107 V, we get v = 1.88 × 109 m s–1. This is clearly wrong, since nothing can move with a speed greater than the speed of light (c = 3 × 108 m s–1). Actually, the above formula for kinetic energy (m v2/2) is valid only when (v/c) << 1. At very high speeds when (v/c ) is comparable to (though always less than) 1, we come to the relativistic domain where the following formulae are valid: Relativistic momentum p = m v Total energy E = m c 2 Kinetic energy K = m c 2 – m0 c 2, − 1/ 2 ⎛ v 2 ⎞where the relativistic mass m is given by mm 1−= 0 ⎜ 2 ⎟⎝ c ⎠ m0 is called the rest mass of the particle. These relations also imply: 2 2 4)1/2E = (pc 2 + m0cNote that in the relativisitc domain when v/c is comparable to 1, K or energy ≥ m 0c 2 (rest mass energy). The rest mass energy of electron is about 0.51 MeV. Thus a kinetic energy of 10 MeV, being much greater than electron’s rest mass energy, implies relativistic domain. Using relativistic formulas, v (for 10 MeV kinetic energy) = 0.999 c. 11.21 (a) 22.7 cm (b) No. As explained above, a 20MeV electron moves at relativistic speed. Consequently, the non-relativistic formula R = (m 0v/e B ) is not valid. The relativistic formula is R = p /eB = mv /eB or R = m v /(eB 1– v2/c2 )0 11.22 We have e V = (m v2/2) and R = (m v/e B) which gives (e/m) = (2V/R 2 B 2 ); using the given data (e/m) = 1.73 × 1011 C kg–1. 11.23 (a) 27.6 keV (b) of the order of 30 kV 11.24 Use λ = (hc/E) with E = 5.1 × 1.602 × 10–10J to get λ = 2.43 × 10–16 m . 11.25 (a) For λ = 500 m, E = (h c/ λ) = 3.98 × 10–28J. Number of photons emitted per second –1= 104J s–1/3.98 × 10–28J ∼ 3 × 1031sWe see that the energy of a radiophoton is exceedingly small, and the number of photons emitted per second in a radio beam is enormously large. There is, therefore, negligible error involved in ignoring the existence of a minimum quantum of energy (photon) and treating the total energy of a radio wave as continuous. (b) For ν = 6 × 1014 Hz, E ∼ 4 × 10–19J. Photon flux corresponding to minimum intensity –2 –1= 10–10 W m–2/4×10–19J = 2.5 × 108 msNumber of photons entering the pupil per second = 2.5 × 108 × 0.4 × 10–4 s–1 = 104 s–1. Though this number is not as large as in (a) above, it is large enough for us never to ‘sense’ or ‘count’ individual photons by our eye. φ011.26 φ = h ν – e V = 6.7 × 10–19 J = 4.2 eV; ν= = 1.0 × 1015 Hz; λ = 6328Å00 0 hcorresponds to ν = 4.7 × 1014 Hz < ν0. The photo-cell will not respond howsoever high be the intensity of laser light. 11.27 Use e V0 = h ν – φ0 for both sources. From the data on the first source, φ0 = 1.40 eV. Use this value to obtain for the second source V0 = 1.50 V. 11.28 Obtain V0 versus ν plot. The slope of the plot is (h/e) and its intercept on the ν-axis is ν0. The first four points lie nearly on a straight line which intercepts the ν-axis at ν0 = 5.0 × 1014 Hz (threshold frequency). The fifth point corresponds to ν < ν0; there is no photoelectric emission and therefore no stopping voltage is required to stop the current. Slope of the plot is found to be 4.15 × 10–15 V s. Using e = 1.6 × 10–19 C, h = 6.64 × 10–34 J s (standard value h = 6.626 × 10–34 J s), φ0 = h ν0 = 2.11V. 11.29 It is found that the given incident frequency ν is greater than ν 0 (Na), and ν0 (K); but less than ν0 (Mo), and ν0 (Ni). Therefore, Mo and Ni will not give photoelectric emission. If the laser is brought closer, intensity of radiation increases, but this does not affect the result regarding Mo and Ni. However, photoelectric current from Na and K will increase in proportion to intensity. 11.30 Assume one conduction electron per atom. Effective atomic area ~10–20m2 Number of electrons in 5 layers 52 10 4m2×× − = = 1017 −20 210 m Incident power = 10–5 W m–2 × 2 × 10–4 m2 = 2 × 10–9 W In the wave picture, incident power is uniformly absorbed by all the electrons continuously. Consequently, energy absorbed per second per electron =2 × 10–9/1017 = 2 × 10–26 W Time required for photoelectric emission =2 × 1.6 × 10–19J/2 × 10–26W = 1.6 × 107 s which is about 0.5 year. Implication: Experimentally, photoelectric emission is observed nearly instantaneously (∼10–9 s): Thus, the wave picture is in gross disagreement with experiment. In the photon-picture, energy of the radiation is not continuously shared by all the electrons in the top layers. Rather, energy comes in discontinuous ‘quanta’. and absorption of energy does not take place gradually. A photon is either not absorbed, or absorbed by an electron nearly instantly. 11.31 For λ = 1 Å, electron’s energy = 150 eV; photon’s energy = 12.4 keV. Thus, for the same wavelength, a photon has much greater energy than an electron. 11.32 (a) λ= h =h Thus, for same K, λ decreases with m as (1/ n/me) = 1838.6; therefore for the same energy, (150 eV) as in Ex. 11.31, wavelength of neutron = ( 10–10 m = 2.33×10–12 m. The interatomic spacing is about a hundred times greater. A neutron beam of 150 eV energy is therefore not suitable for diffraction experiments. (b) λ = 1.45 × 10–10m [Use λ =(/ 3 ) ] which is comparableh mkT544 to interatomic spacing in a crystal. Clearly, from (a) and (b) above, thermal neutrons are a suitable probe for diffraction experiments; so a high energy neutron beam should be first thermalised before using it for diffraction. 11.33 λ = 5.5 × 10–12 m λ (yellow light) = 5.9 × 10–7m Resolving Power (RP) is inversely proportional to wavelength. Thus, RP of an electron microscope is about 105 times that of an optical microscope. In practice, differences in other (geometrical) factors can change this comparison somewhat. h 6.63 ×10–34 Js 11.34 p == = 6.63 × 10–19 kg m s–1 λ 10–15 m Use the relativistic formula for energy: E 2= c2p2 + m02 c4 = 9 × (6.63)2 × 10–22 + (0.511 × 1.6)2 × 10–26 ∼ 9 × (6.63)2 × 10–22, the second term (rest mass energy) being negligible. Therefore, E = 1.989 × 10–10 J = 1.24 BeV. Thus, electron energies from the accelerator must have been of the order of a few BeV. h 410 3 11.35 Use λ=; mHe = × kg610 3 mkT × 23 This gives λ = 0.73 × 10–10 m. Mean separation r = (V/N)1/3 = (kT/p)1/3 For T = 300K, p = 1.01 × 105 Pa, r = 3.4 × 10–9 m. We find r >> λ . 11.36 Using the same formula as in Exercise 11.35, λ = 6.2 × 10–9 m which is much greater than the given inter-electron separation. 11.37 (a) Quarks are thought to be confined within a proton or neutron by forces which grow stronger if one tries to pull them apart. It, therefore, seems that though fractional charges may exist in nature, observable charges are still integral multiples of e. (b) Both the basic relations e V = (1/2) m v 2 or e E = m a and e B v =m v2/r, for electric and magnetic fields, respectively, show that the dynamics of electrons is determined not by e, and m separately but by the combination e/m . (c) At low pressures, ions have a chance to reach their respective electrodes and constitute a current. At ordinary pressures, ions have no chance to do so because of collisions with gas molecules and recombination. (d) Work function merely indicates the minimum energy required for the electron in the highest level of the conduction band to get out of the metal. Not all electrons in the metal belong to this level. They occupy a continuous band of levels. Consequently, for the same incident radiation, electrons knocked off from different levels come out with different energies. (e) The absolute value of energy E (but not momentum p) of any particle is arbitrary to within an additive constant. Hence, while λ is physically significant, absolute value of ν of a matter wave of an electron has no direct physical meaning. The phase speed νλ is likewise not physically significant. The group speed given by dν dE d ⎛ p2 ⎞ p== = (/ )d 1 λ dp dp ⎝⎜ 2m ⎠⎟ m is physically meaningful. 546 CHAPTER 12 12.1 (a) No different from (b) Thomson’s model; Rutherford’s model (c) Rutherford’s model (d) Thomson’s model; Rutherford’s model (e) Both the models 12.2 The nucleus of a hydrogen atom is a proton. The mass of it is 1.67 × 10–27 kg, whereas the mass of an incident α-particle is 6.64 × 10–27 kg. Because the scattering particle is more massive than the target nuclei (proton), the α-particle won’t bounce back in even in a head-on collision. It is similar to a football colliding with a tenis ball at rest. Thus, there would be no large-angle scattering. 12.3 820 nm. 12.4 5.6 × 1014 Hz 12.5 13.6 eV; –27.2 eV 12.6 9.7 × 10–8m; 3.1 × 1015 Hz. 12.7 (a) 2.18 × 106 m/s; 1.09 × 106 m/s; 7.27 × 105 m/s (b) 1.52 × 10–16s; 1.22 × 10–15s; 4.11 × 10–15 s. 12.8 2.12×10–10 m; 4.77 × 10–10 m 12.9 Lyman series: 103 nm and 122 nm; Balmer series: 656 nm. 12.10 2.6 × 1074 12.11 (a) About the same. (b) Much less. (c) It suggests that the scattering is predominantly due to a single collision, because the chance of a single collision increases linearly with the number of target atoms, and hence linearly with thickness. (d) In Thomson’s model, a single collision causes very little deflection. The observed average scattering angle can be explained only by considering multiple scattering. So it is wrong to ignore multiple scattering in Thomson’s model. In Rutherford’s model, most of the scattering comes through a single collision and multiple scattering effects can be ignored as a first approximation. 12.12 The first orbit Bohr’s model has a radius a0 given by 4πε (/2π)2 0 h a0 = 2 . If we consider the atom bound by the gravitational mee force (Gmpme/r2), we should replace (e2/4 π ε0) by Gmpm e. That is, the G h π)(/2 2 radius of the first Bohr orbit is given by a0 = 2 ≅ 1.2 × 1029 m. Gm m pe This is much greater than the estimated size of the whole universe! me4 ⎡ 11 ⎤ me4(2n −1)ν= −=12.13 32 3 ⎢ 22 ⎥ 32 322π) ε0 h 2π) ⎣( −1 n ⎦(4π) ε0( / 2) nn(4 (/ n ) h π ( −1) 4meFor large n, ν≅ 32 3332π h 2πε0(/ ) n nh 2π)(/Orbital frequency νc = (v/2 π r). In Bohr model v = , and mr me4πε (/2 )2 nh 2) 4 h π (/π02 = r = n . This gives ν = 32 332 c2πmr2 32π h 2π) nε (/me 0 which is same as ν for large n. ⎛ e2 ⎞ 12.14 (a) The quantity ⎜ 2 ⎟ has the dimensions of length. Its value4πε mc⎝ 0 ⎠ is 2.82 ×10–15m – much smaller than the typical atomic size. 4π h 2 )2ε (/ π(b) The quantity 0 has the dimensions of length. Its2mevalue is 0.53 × 10–10m – of the order of atomic sizes.(Note that the dimensional arguments cannot, of course, tell us that we should use 4π and h/2π in place of h to arrive at the right size.) mv2 Ze2 12.15 In Bohr’s model, mvr = nh and = r 4πε0r2 which give 224πε0 = 21 2 Ze nT = mv = ; r =2Ze m2 8 πε0r These relations have nothing to do with the choice of the zero of potential energy. Now, choosing the zero of potential energy at infinity we have V = – (Ze2/4 π ε0 r) which gives V = –2T and E = T + V = – T (a) The quoted value of E = – 3.4 eV is based on the customary choice of zero of potential energy at infinity. Using E = – T, the kinetic energy of the electron in this state is + 3.4 eV. (b) Using V = – 2T, potential energy of the electron is = – 6.8 eV (c) If the zero of potential energy is chosen differently, kinetic energy does not change. Its value is + 3.4 eV independent of the choice of the zero of potential energy. The potential energy, and the total energy of the state, however, would alter if a different zero of the potential energy is chosen. 12.16 Angular momenta associated with planetary motion are incomparably large relative to h. For example, angular momentum of the earth in its orbital motion is of the order of 1070h. In terms of the Bohr’s quantisation postulate, this corresponds to a very large value of n (of the order of 1070). For such large values of n, the differences in the successive energies and angular momenta of the quantised levels of the Bohr model are so small compared to the energies and angular momenta respectively for the levels that one can, for all practical purposes, consider the levels continuous. 12.17 All that is needed is to replace me by mμ in the formulas of the Bohr model. We note that keeping other factors fixed, r ∝ (1/m) and E ∝ m. Therefore, rm 053 . ×10 13 eerm = = = 2.56 × 10–13 m m 207m EmEμ = e m = – (13.6 × 207) eV ≅ – 2.8 keV me 548 CHAPTER 13 13.1 (a) 6.941 u (b) 19.9%, 80.1% 13.2 20.18 u 13.3 104.7 MeV 13.4 8.79 MeV, 7.84 MeV 13.5 1.584 × 1025MeV or 2.535×1012J 222 4 238 413.6 i) 226 Ra → Rn + He ii) 242 Pu → U+ He 88 862 94 922 32 32 – 210 210 –iii) P → S+e +ν iv) B → Po+e +ν1516 83 84 11 + 9797 +v) 11C → B+e +ν vi) Tc → Mo+e +ν65 4342 + 120 vii) 120 Xe+e → I+ν54 53 13.7 (a) 5 T years (b) 6.65 T years 13.8 4224 years 13.9 7.126 ×10–6 g 13.10 7.877 ×1010 Bq or 2.13 Ci 13.11 1.23 13.12 (a) Q = 4.93 MeV, Eα = 4.85 MeV (b) Q = 6.41 MeV, Eα = 6.29 MeV 11 11 +13.13 6C →6B+e +ν + Q 1111 2Q =⎡mN () ( 6B) me ⎤ 6C– mN – c ,⎣⎦ where the masses used are those of nuclei and not of atoms. If we use atomic masses, we have to add 6m e in case of 11C and 5m e in case of 11B. Hence 1111 2Q = ⎣⎡m () m ( 6B–2 ) ⎦⎤ e6C– me c (Note m has been doubled) Using given masses, Q = 0.961 MeV. Q = + E + EνEde The daughter nucleus is too heavy compared to e+ and ν, so it carries negligible energy (Ed ≈0). If the kinetic energy (Eν) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q; hence maximum E ≈ Q).e 2323 223 23 – ⎡⎤13.14 10 Ne → 11 Na+e + ν+ Q ; Q= ⎣mN (10 Ne ) – mN ( 11 Na ) – me ⎦ c , where the masses used are masses of nuclei and not of atoms as in Exercise 23 23 213.13. Using atomic masses Q =⎡m ( Ne) – m ( 11 Na)⎤ c . Note m has10⎣⎦ ebeen cancelled. Using given masses, Q = 4.37 MeV. As in Exercise 13.13, maximum kinetic energy of the electron (max E) = Q = 4.37 MeV. e13.15 (i) Q = –4.03MeV; endothermic (ii) Q = 4.62 MeV; exothermic 56 2813.16 Q = m ( Fe) –2m ( Al ) = 26.90 MeV; not possible.26 13 13.17 4.536 × 1026MeV 23 −13 235 6 × 10 × 200 ×1.6 ×10 −113.18 Energy generated per gram of = Jg92U235 The amount of 235U consumed in 5y with 80% on-time92 5 ×0.8 ×3.154 ×1016 ×235 = g = 1544kg 1.2 ×1.6 ×1013 The initial amount of 235U = 3088 kg. 92 13.19 About 4.9 × 104 y 13.20 360 KeV 13.22 Consider the competing processes: AA +X → Y+e +ν +Q (positron capture)ZZ –1 e 1 – AA e + X → Y +ν +Q (electron capture)ZZ –1 e 2 AA 2Q1 ⎣⎡ (Z ) N (Z –1 ) ⎦⎤=mN X–m Y– me c AA 2=⎡m (Z X–) Z m e – m (Z Y)– (Z –1)m – me ⎤N –1 ec⎣⎦ AA 2=⎡m X– mZ –1Y –2 m ⎤c(Z )() e⎣⎦ AA 2 AA 2Q2 =⎡⎣mN (Z X)+me – mN Z –1 Y)⎦c =⎡⎣mZ X– m (Z –1 Y ⎤⎦(⎤() )c This means Q1 > 0 implies Q2 > 0 but Q2 > 0 does not necessarily mean Q1 > 0. Hence the result. 25 2613.23 Mg : 9.3%, Mg :11.7%12 12 13.24 Neutron separation energy S of a nucleus AX is nZ A −1 A2Sn=⎣⎡mN( ZX) +m n −m N( ZX) ⎦⎤c 41 27From given data , S ( Ca) =8.36MeV, S ( Al) =13.06MeV n 20 n 13 13.25 209 d 13.26 For 146C emission 223 209 14 2 N 88 N 82 N 6Q =[m ( Ra) −m ( Pb)−m ( C)] c 223 209 142 6 = 31.85 MeV=[(m 88Ra) −m( 82 Pb) −m(C)] c 4 223 219 42For 2He emission, Q =[m( Ra) −m( Rn) −m( He)] c = 5.98MeV88 862 238 140 992 m Ru)] c = 231.1 MeV 13.27 Q =[( U) +m −m( Ce) −m(92 n58 44 234 213.28 (a) Q =[1 +m 1 −m( He) −mnm(H) (H) 2 ]c =17.59MeV (b) K.E. required to overcome Coulomb repulsion = 480.0 keV480.0 KeV = 7.68×10–14J = 3kT 7.68 ×10–14 –23 –1∴T = –23 (as k =1.381 ×10 JK)31.38110 ×× = 1.85 ×109 K (required temperature) ––13.29 K (β)=0.284MeV,Kmax (β2 )= 0.960MeV max 1 20 20 20549()=2.627 ×10 Hz , νγ =0.995 ×10 Hz , νγ =1.632 ×10 Hz νγ () ()12 3 13.30 (a) Note that in the interior of Sun, four 11H nuclei combine to form 4one He nucleus releasing about 26 MeV of energy per event. 2 Energy released in fusion of 1kg of hydrogen = 39 ×1026 MeV 235(b) Energy released in fission of 1kg of U = 5.1×1026 MeV92 The energy released in fusion of 1 kg of hydrogen is about 8 times that of the energy released in the fusion of 1kg of uranium. 13.31 3.076 × 104 kg CHAPTER 14 14.1 (c) 14.2 (d) 14.3 (c) 14.4 (c) 14.5 (c) 14.6 (b), (c) 14.7 (c) 14.8 50 Hz for half-wave, 100 Hz for full-wave 14.9 vi = 0.01V ; IB= 10 μA 14.10 2 V 14.11 No (hν has to be greater than E g). 14.12 n ≈ 4.95 × 1022; n = 4.75 × 109 ; n-type since n >> neh eh For charge neutrality N – N = n – n ; n.n = ni 2 D Aeheh⎡ 2 ⎤Solving these equations, n e = 1 (ND − NA )+ (ND − NA )2 + 4ni ⎦⎥2 ⎣⎢ 14.13 About 1 × 105 14.14 (a) 0.0629 A, (b) 2.97 A, (c) 0.336 Ω (d) For both the voltages, the current I will be almost equal to I0, showing almost infinite dynamic resistance in the reverse bias. 14.16 NOT ; A Y 0 1 1 0 14.17 (a) AND (b) OR 14.18 OR gate 14.19 (a) NOT, (b) AND CHAPTER 15 15.1 (b) 10 kHz cannot be radiated (antenna size), 1 GHz and 1000 GHz will penetrate. 15.2 (d) Consult Table 15.2 15.3 (c) Decimal system implies continuous set of values 550 15.4 No. Service area will be A= π 2 Td = 622 162 6.4 10 7 × × × = 3258 km2 . 15.5 0.75 μ = = m c A A 15.6 0.75 12 9 m A = × = V. (a) (b) μ = 0.5 15.7 Since the AM wave is given by (A + A sinω t) cos ω t, thecmm c maximum amplitude is M1 = Ac + Am while the minimum amplitude is M2 = Ac – Am . Hence the modulation index is AM – M 82 m 12m== == .AM + M 12 3 c 12 With M2 = 0, clearly, m=1, irrespective of M1. 15.8 Let, for simplicity, the received signal be A1 cos (ω c +ω m) t The carrier Ac cos ω ctis available at the receiving station. By multiplying the two signals, we get AAcos (ω + ω ) tcos ω t1ccac AA (ω+ω )t+ cos ω t⎤= 1 c ⎡cos 2 = ⎣ cm m ⎦2 If this signal is passed through a low-pass filter, we can record AA1 cthe modulating signal cos ω t. m2 BIBLIOGRAPHY TEXTBOOKS For additional reading on the topics covered in this book, you may like to consult one or more of the following books. Some of these books however are more advanced and contain many more topics than this book. 1 Ordinary Level Physics, A.F. Abbott, Arnold-Heinemann (1984). 2 Advanced Level Physics, M. Nelkon and P. Parker, 6th Edition, Arnold-Heinemann (1987). 3 Advanced Physics, Tom Duncan, John Murray (2000). 4 Fundamentals of Physics, David Halliday, Robert Resnick and Jearl Walker, 7th Edition John Wily (2004). 5 University Physics (Sears and Zemansky’s), H.D. Young and R.A. Freedman, 11th Edition, Addison—Wesley (2004). 6 Problems in Elementary Physics, B. Bukhovtsa, V. Krivchenkov, G. Myakishev and V. Shalnov, MIR Publishers, (1971). 7 Lectures on Physics (3 volumes), R.P. Feynman, Addision – Wesley (1965). 8 Berkeley Physics Course (5 volumes) McGraw Hill (1965). a. Vol. 1 – Mechanics: (Kittel, Knight and Ruderman) b. Vol. 2 – Electricity and Magnetism (E.M. Purcell) c. Vol. 3 – Waves and Oscillations (Frank S. Crawford) d. Vol. 4 – Quantum Physics (Wichmann) e. Vol. 5 – Statistical Physics (F. Reif) 9 Fundamental University Physics, M. Alonso and E. J. Finn, Addison – Wesley (1967). 10 College Physics, R.L. Weber, K.V. Manning, M.W. White and G.A. Weygand, Tata McGraw Hill (1977). 11 Physics: Foundations and Frontiers, G. Gamow and J.M. Cleveland, Tata McGraw Hill (1978). 12 Physics for the Inquiring Mind, E.M. Rogers, Princeton University Press (1960). 13 PSSC Physics Course, DC Heath and Co. (1965) Indian Edition, NCERT (1967). 14 Physics Advanced Level, Jim Breithampt, Stanley Thornes Publishers (2000). 15 Physics, Patrick Fullick, Heinemann (2000). 16 Conceptual Physics, Paul G. Hewitt, Addision—Wesley (1998). 17 College Physics, Raymond A. Serway and Jerry S. Faughn, Harcourt Brace and Co. (1999). 18 University Physics, Harris Benson, John Wiley (1996). 19 University Physics, William P. Crummet and Arthur B. Western, Wm.C. Brown (1994). 20 General Physics, Morton M. Sternheim and Joseph W. Kane, John Wiley (1988). 21 Physics, Hans C. Ohanian, W.W. Norton (1989). 22 Advanced Physics, Keith Gibbs, Cambridge University Press (1996). 23 Understanding Basic Mechanics, F. Reif, John Wiley (1995). 24 College Physics, Jerry D. Wilson and Anthony J. Buffa, Prentice Hall (1997). 25 Senior Physics, Part – I, I.K. Kikoin and A.K. Kikoin, MIR Publishers (1987). 26 Senior Physics, Part – II, B. Bekhovtsev, MIR Publishers (1988). 27 Understanding Physics, K. Cummings, Patrick J. Cooney, Priscilla W. Laws and Edward F. Redish, John Wiley (2005). 28 Essentials of Physics, John D. Cutnell and Kenneth W. Johnson, John Wiley (2005). GENERAL BOOKS For instructive and entertaining general reading on science, you may like to read some of the following books. Remember however, that many of these books are written at a level far beyond the level of the present book. 1 Mr. Tompkins in paperback, G. Gamow, Cambridge University Press (1967). 2 The Universe and Dr. Einstein, C. Barnett, Time Inc. New York (1962). 3 Thirty years that Shook Physics, G. Gamow, Double Day, New York (1966). 4 Surely You’re Joking, Mr. Feynman, R.P. Feynman, Bantam books (1986). 5 One, Two, Three… Infinity, G. Gamow, Viking Inc. (1961). 6 The Meaning of Relativity, A. Einstein, (Indian Edition) Oxford and IBH Pub. Co. (1965). 7 Atomic Theory and the Description of Nature, Niels Bohr, Cambridge (1934). 8 The Physical Principles of Quantum Theory, W. Heisenberg, University of Chicago Press (1930). 9 The Physics—Astronomy Frontier, F. Hoyle and J.V. Narlikar, W.H. Freeman (1980). 10 The Flying Circus of Physics with Answer, J. Walker, John Wiley and Sons (1977). 11 Physics for Everyone (series), L.D. Landau and A.I. Kitaigorodski, MIR Publisher (1978). Book 1: Physical Bodies Book 2: Molecules Book 3: Electrons Book 4: Photons and Nuclei. 12 Physics can be Fun, Y. Perelman, MIR Publishers (1986). 13 Power of Ten, Philip Morrison and Eames, W.H. Freeman (1985). 14 Physics in your Kitchen Lab., I.K. Kikoin, MIR Publishers (1985). 15 How Things Work: The Physics of Everyday Life, Louis A. Bloomfield, John Wiley (2005). 16 Physics Matters: An Introduction to Conceptual Physics, James Trefil and Robert M. Hazen, John Wiley (2004). INDEX Absorption spectra AC current AC Generator AC voltage applied to a capacitor applied to a resistor applied to an inductor applied to a series LCR circuit Accelerators in India Accommodation of eye Activity of radioactive substances Additivity of charges Alpha decay Alpha particle scattering Ammeter Ampere Amperes circuital law Amplification Amplitude modulation Analog signal AND gate Andre, Ampere Angle of deviation of incidence of reflection of refraction Angular magnification Apparent depth Area element vector Astigmatism Atomic mass unit number spectra Attenuation Aurora Boriolis Band gap Bandwidth of signal 421 233 224 233 241 234 237 244 142 336 447 8 449 415 165 155 147 517 524 502 503 148 330 355 355 355 341 318 26 337 439 440 420 516 139 470 517 Bandwidth of transmission medium 518 Bar magnet 174 as solenoid 176 Barrier potential 479 Base 491 Becquerel 448 Beta decay 450 Binding energy per nucleon 444 Biot-Savart law 143 Bohr magneton 163 Bohr radius 425 Bohr’s model of atom 422 Bohr’s postulates 424 Brewster’s angle 380 Brewster’s law 381 C.A. Volta 53 Capacitance 73 Capacitive reactance 241 Capacitive circuit 252 Capacitor parallel plate 74 in parallel 79 in series 78 Cartesian sign convention 311 Cassegrain telescope 342 Cells 110 in parallel 114 in series 113 Chain reaction 452 Channel 515 Charging by induction 6 Charles August de Coulomb 11 Chromatic aberration 333 Ciliary muscles 337 Coercivity 195 Coherent source 360 Collector 491 Colour code of resistors 103 Combination of lenses 328 Combination of resistors series parallel Composition of nucleus Concave mirror Conduction band Conductivity Conductors Conservation of charge Conservative force Continuous charge distribution Control rods Convex mirror Coulomb Coulomb’s law Critical angle Curie temperature Curie Current amplification factor density loop as a magnetic dipole sensitivity of galvanometer Cut-off voltage/Stopping potential Cyclotron frequency Davisson & Germer Experiment de Broglie relation wavelength explanation Decay constant Detection of amplitude modulated wave Diamagnetism Dielectrics Dielectric constant strength Diffraction single slit Digital electronics signal Dioptre Dipole moment moment vector in uniform electric field physical significance Dispersion by a prism Displacement current Doppler effect Drift velocity Earth’s magnetism Earthing 107 108 438 312 469 97, 468 5 8 51 32 454 312 11 10 320 194 448 94 495 97 160 165 391 140 141 403 398 398 430 446 526 192 71 76 74 367 368 501 502 328 28 28 31 29 332 270 358 98 185 6 Eddy currents 218 Einstein’s photoelectric equation 394 Electric charge 1 current 93 dipole 27 displacement 77 field 18 field, physical significance 20 field due to a system of charges 19 field lines 23 flux 25 susceptibility 72 Electrical energy 105 Electromagnetic waves, sources 274 waves, nature 275 damping 218 spectrum 280 Electron emission 387 Electrostatic analog 180 potential 53 shielding 69 Electrostatics 1 of conductors 67 Electromotive force (emf) 110 Emission spectra 421 Emitter 491 Energy bands 469 generation in stars 455 levels 427 stored in a capacitor 80 Equipotential surfaces 60 Excited state 427 Experiments of Faraday & Henry 205 Extrinsic semiconductor 474 Eye 336 Farad 75 Faraday’s law of Induction 207 Fast breeder reactor 453 Ferromagnetism 193 Field due to infinite plane sheet 38 due to uniformly charged thin spherical shell 39 Field emission 388 Flemings left hand rule Flux leakage 261 Focal length 311 Force between two parallel currents 154 Forward bias 479 Franck-Hertz experiment 428 Fringe width 364 Lateral shift 317 Full-wave rectifier 483 Law G.S. Ohm 96 of radioactive decay 447 Gamma of reflection 357 rays 283 of refraction 356 decay 451 LC oscillations 255 Gauss’s law 33 Least distance of distinct vision 336 its applications 37 Lenz’s law 210 in magnetism 181 Lens maker’s formula 326 Gaussian surface 35 Light emitting diode 488 Geographic meridian 186 Limitations of Ohm’s law 101 Gold leaf electroscope 4 Linear Ground charge density 32 state 427 magnification/Magnifying power 339 wave 519 Logic gates 502 H.A. Lorentz 134 Lorentz force 134 Half life 448 Magnetic Half-wave rectifier 483 declination 186 Hallwachs’ and Lenard’s observations 388 dipole 177 Henry 220 dipole moment of a revolving electron 162 Hertz Experiment 274 field 132 Holes 472 field lines 175 Horizontal component of earth’s field on the axis of a circular current loop 145 magnetic field 187 flux 182, 206 Huygen’s Principle 353 force on a current carrying conductor 135 Hypermetropia 337 force 133 Impact parameter 418 hysteresis 195 Impedence diagram 246 inclination 187 Inductance 219 intensity 190 mutual 220 meridian 186 self 222 moment of a current loop 158 Induction 6 moment 178 of charge 6 permeability 190 Inductive potential energy 178 circuit 252 susceptibility 190 reactance 238 torque 178 Input resistance of a transistor 494 Magnetisation 189 Insulators 5 Majority carriers 476 Integrated circuits (IC) 505 Mass Interference defect 443 constructive 361 number 440 destructive 361 energy relation 442 fringes 363 Maxwell’s equations 273 Internal resistance 110 Mean life 448 Intrinsic semiconductor 472 Meter bridge 120 Ionisation energy 427 Michael Faraday 208 Iris 337 Microscope 339 Isobars 441 compound 340 Isotones 441 Microwaves 281 Isotopes 439 Minority carriers 476 J.C. Maxwell 270 Mirage 321 Junction transistor 490 Mirror equation 314 K.F. Gauss 182 Mobility 100 Kirchhoff’s rules 115 Moderator 454 556 Modulation index Motion in a magnetic field Motional emf Moving coil galvanometer Multiplication factor (fission) Myopia NAND gate Near point Neutrons Noise Non-polar molecules NOR gate North pole NOT gate n-p-n transistor n-type semi conductor Nuclear binding energy density energy fission force fusion holocaust reactor size winter Numerical aperture Ohm Ohm’s law Optical fibers OR gate Orbital magnetic moment Output resistance of a transistor Paramagnetism Permanent magnets Permeability of free space Permittivity of free space of medium Phasors diagram Photodiode Photoelectric effect Photocell Photoelectric emission Photoelectrons Photon Pith ball Plane polarised wave p-n Junction p-n-p transistor Point charge Polar molecules 517, 522 525 137 212 163 454 336 504 336 440 516 72 505 174 502 491 475 442 442 451 452 445 455 457 452 441 457 375 95 95 322 502 163 495 192 195 143 11, 76 76 237 237 487 388 399 388 389 395 2 377 478 491 10 72 Polarisation 71, 376 by reflection 380 by scattering 379 Polarity of charge 2 Polaroid 378 Potential 53 due to an electric dipole 55 due to a point charge 54 due to a system of charges 57 energy difference 53 energy for a system of charges 61 energy of a dipole 66 energy of a single charge 64 energy of a system of two charges 65 energy 52 Potentiometer 122 Power (electrical) 106 factor 252 in ac circuit 252 of lens 327 Pressurised heavy water reactors 453 Primary coil Principal focus 311 Principle of superposition 15 Principle quantum number 425 Prism formula 331 Production of amplitude modulated wave 525 Properties of electric charge 8 p-type semi conductor 476 Q factor/quality factor 250 Quanta of energy 393 Quantisation of charge 8 Radio waves 281 Radioactivity 446 Rainbow 333 Ray optics, validity of 375 Rayleigh scattering 335 Rectifier 483 Red shift 358 Reflection of light 310 Refraction 318 of a plane wave 355 Refractive index 317, 356 Relation between field and potential 61 Relaxation time 98 Rententivity 195 Repeater 517 Resistance 95 Resistivity 96, 468 of some materials 102 Resolving power 373 of eye 374 Resonance 248 Sharpness 249 Resonant frequency 248 Reverse bias 480 557 Right hand rule Root mean square (rms) or effective current voltage Roget’s spiral Rutherford’s model of atom Saturation current Scattering of light Secondary wavelet Semiconductors diode elemental compound Shunt resistance Signal Sky wave Snell’s law Solar cell Solenoid South pole Space wave Spectral series Brackett Fund Lyman Paschen Spectrum of light Spherical mirror Spin magnetic moment Surface charge density Telescope Temperature dependence of resistivity Tesla Thermionic emission Thermonuclear fusion Thin lens formula Threshold frequency 149 235 235 156 415 390 335 354 469 479 468 468 164 516 520 317, 356 489 151 174 521 421 422 422 422 422 332 310, 311 163 32 341 103 135 388 456 326 392 Tokamak 153 Toroid 152 Torque on a current loop 157 on a dipole 31 Total internal reflection 319 Transducer 516 Transformer 259 Step-down 261 Step-up 261 Transistor as a switch 496 as an amplifier 497 oscillator 500 common emitter configuration 493 Truth table 502 Uncertainty Principle 400 Unpolarised wave 377 Ultraviolet rays 282 Valence band 469 Van de Graaff Generator 83 Velocity selector 140 Visible rays 282 Voltage Regulator 486 Voltage sensitivity of a galvanometer 165 Voltmeter 165 Volume charge density 32 Wattless current 252 Wavefront 353 plane 354 spherical 354 Wheatstone bridge 118 Work function 394 X rays 283 Young’s experiment 362 Zener diode 485 breakdown 485 558

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