14.3 Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated 341 Reprint 2025-26 Physics by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Which of the following statements is true? (a) (Eg)Si < (Eg)Ge < (Eg)C (b) (Eg)C < (Eg)Ge > (Eg)Si (c) (Eg)C > (Eg)Si > (Eg)Ge (d) (Eg)C = (Eg)Si = (Eg)Ge 14.4 In an unbiased p-n junction, holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them. (b) they move across the junction by the potential difference. (c) hole concentration in p-region is more as compared to n-region. (d) All the above. 14.5 When a forward bias is applied to a p-n junction, it (a) raises the potential barrier. (b) reduces the majority carrier current to zero. (c) lowers the potential barrier. (d) None of the above. 14.6 In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency. Reprint 2025-26 Notes Reprint 2025-26 Physics APPENDICES APPENDIX A 1 THE GREEK ALPHABET APPENDIX A 2 COMMON SI PREFIXES AND SYMBOLS FOR MULTIPLES AND SUB-MULTIPLES Reprint 2025-26 AppendicesAnswers APPENDIX A 3 SOME IMPORTANT CONSTANTS OTHER USEFUL CONSTANTS 345 Reprint 2025-26 Physics ANSWERS CHAPTER 9 9.1 v = –54 cm. 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 nga = 1.51; nwa = 1.32; ngw = 1.144; which gives sin r = 0.6181 i.e., r ~ 38°. 9.5 r = 0.8 × tan ic and sin ci = 1/1.33 ≅ 0.75 , where r is the radius (in m) of the largest circle from which light comes out and ic is the critical angle for water-air interface, Area = 2.6 m2 9.6 n ≅ 1.53 and Dm for prism in water ≅ 10° 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 = 8.4 cm, 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 = –5 cm; vO = (15 – 5) cm = 10 cm, fO = uO = – 2.5 cm; Magnifying power = 20 (b) uO = – 2.59 cm. Magnifying power = 13.5. 9.12 Angular magnification of the eye-piece for image at 25 cm 25 25   1  11; | u e |= cm = 2 .27cm ; vO = 7.2 cm 2.5 11 Separation = 9.47 cm; Magnifying power = 88 9.13 24; 150 cm 9.14 (a) Angular magnification = 1500 346 (b) Diameter of the image = 13.7 cm. Reprint 2025-26 Answers

9.27 (a) m = ( fO/fe) = 28 f O  f O  (b) m = 1 + = 33.6 f e  25  349 Reprint 2025-26 Physics 9.28 (a) fO + fe = 145 cm (b) Angle subtended by the tower = (100/3000) = (1/30) rad. Angle subtended by the image produced by the objective h h = = f O 140 Equating the two, h = 4.7 cm. (c) Magnification (magnitude) of the eye-piece = 6. Height of the final image (magnitude) = 28 cm. 9.29 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. 9.30 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. 9.31 n = 1.33 CHAPTER 10 10.1 (a) Reflected light: (wavelength, frequency, speed same as incident light) l = 589 nm, n = 5.09 ´ 1014 Hz, c = 3.00 ´ 108 m s–1 (b) Refracted light: (frequency same as the incident frequency) n = 5.09 ´ 1014Hz v = (c/n) = 2.26 × 108 m s–1, l = (v/n) = 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. 1.2 10 – 2  0.28 10 – 3 10.4  m = 600 nm 4 14. 10.5 K/4 10.6 (a) 1.17 mm (b) 1.56 mm 10.7 0.15° 350 10.8 tan–1(1.5) ~ 56.3o Reprint 2025-26 Answers

8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to 213 a 230 V ac supply with a (angular) frequency of 300 rad s–1. Reprint 2025-26 Physics (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. FIGURE 8.6 8.3 What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500m? 8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? 8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? 8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? 8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? 8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine, B0,w, k, and l. (b) Find expressions for E and B. 8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hn (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation? 8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s–1.] Reprint 2025-26