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

9.5 A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)

9.20 (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, u 2 = + (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= 120 cm. Magnitude of magnification due to the first (convex) lens is 3. u 2 = + (120 – 8) cm = +112 cm (object virtual); 112 × 20 f2 = – 20 cm which gives v2 = − cm 92 Magnitude of magnification due to the second (concave) 347 Reprint 2025-26 Physics lens = 20/92. Net magnitude of magnification = 0.652 Size of the image = 0.98 cm 9.21 If the refracted ray in the prism is incident on the second face at the critical angle ic, the angle of refraction r at the first face is (60°–ic). Now, ic = sin–1 (1/1.524) ~ 41° Therefore, r = 19° sin i = 0.4962; i ~ 30° 1 1 1 9.22 (a) + = v 9 10 i.e., v = – 90 cm, Magnitude of magnification = 90/9 = 10. Each square in the virtual image has an area 10 × 10 × 1 mm2 = 100 mm2 = 1 cm2 (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. 9.23 (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. 9.24 Magnification = ( 6.25 / 1) = 2.5 v = +2.5u 1 1 1    2.5u u 10 i.e.,u = – 6 cm |v| = 15 cm The virtual image is closer than the normal near point (25 cm) and cannot be seen by the eye distinctly. 9.25 (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 Reprint 2025-26 Answers 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/fe) + 1] ( fe in cm) which increases if fe is smaller. Further, magnification of the objective v O 1 = is given by | u O | (| u O |/ f O ) − 1 which is large when |u O | is slightly greater than fO. The micro- scope is used for viewing very close object. So |u O | 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.