Class 12 Physics Chapter 9: Ray Optics and Optical Instruments worksheet with answer key
A 20-mark worksheet on Ray Optics and Optical Instruments for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (d) 5 cm, away from the mirror
f = −10 cm. At u = −30 cm, 1/v = −1/10 + 1/30 = −1/15, v = −15 cm. At u = −20 cm, 1/v = −1/10 + 1/20 = −1/20, v = −20 cm. The image moves from 15 cm to 20 cm in front of the mirror: 5 cm away from it, as the object moves 10 cm towards it.
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Answer: (d) 9/8
n(glass, water) = n(glass, air) × n(air, water) = (3/2) × (3/4) = 9/8. 8/9 is the index of water with respect to glass.
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Answer: (a) the wavelength of the light
n21 is a property of the pair of media and depends on the wavelength of light, not on the angle of incidence. Because it differs for different colours, white light splits into its colours (dispersion).
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Answer: (d) retraces its path
A ray directed towards the centre of curvature meets the mirror along the normal (the radius), so the angle of incidence is zero and the ray is reflected back along its own path.
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Answer: (a) 3.0 cm
In normal adjustment the tube length is fo + fe = 93 cm and the magnifying power is fo/fe = 30, so fo = 30 fe and 31 fe = 93 cm, fe = 3.0 cm (fo = 90 cm). 3.1 cm comes from dividing the length by 30, forgetting that the length also contains fe; 90 cm is the objective.
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Light inside diamond meets a face at an angle greater than 24° far more often than inside glass, since the critical angle is so small. So light entering a cut diamond is totally reflected again and again at its faces and emerges only from a few faces, in bright concentrated beams; this multiple total internal reflection makes it sparkle. In glass, with ic ≈ 41°, much of the light escapes at the first faces it meets.
- small critical angle of diamond means most rays inside meet faces beyond ic (1)
- repeated total internal reflection; light leaves through a few faces, bright (1)
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At minimum deviation r = A/2 = 30°. sin i = n sin r = 1.5 × 0.5 = 0.75, so i = 48.6°. Dm = 2i − A = 97.2° − 60° = 37.2°.
- r = A/2 = 30°, sin i = n sin r = 0.75, i = 48.6° (1)
- Dm = 2i − A = 37.2° (1)
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A real image is formed where rays from a point actually meet after reflection or refraction; it can be received on a screen. A virtual image is formed where the rays only appear to diverge from when produced backwards; it cannot be received on a screen. A concave mirror forms a real (inverted) image of an object beyond its focus, and a virtual (erect, magnified) image of an object between its focus and its pole.
- real: rays actually meet; virtual: rays only appear to diverge from the point (1)
- examples: object beyond F gives real image; object between F and P gives virtual image (1)
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(a) For the plane face R = ∞. 1/f = (n − 1)(1/R1 − 1/R2) = 0.5 × (1/15 − 0) = 1/30, f = +30 cm. P = 1/0.30 m = +3.3 D. (b) A real image of the same size has m = v/u = −1, so v = −u. 1/v − 1/u = 1/f gives −2/u = 1/30, u = −60 cm: the object is 60 cm (that is, 2f) in front of the lens, and the image 60 cm behind it.
- f = +30 cm from the lens maker's formula with R2 = ∞ (1)
- P = +3.3 D (1)
- m = −1 gives u = −60 cm (2f), v = +60 cm (1)
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The image is real, so it is inverted: m = −v/u = −3, v = 3u. Both are in front of the mirror, with the image farther, so |v| − |u| = 80 cm: 3|u| − |u| = 80, |u| = 40 cm, |v| = 120 cm. With signs u = −40 cm, v = −120 cm. 1/f = 1/v + 1/u = −1/120 − 1/40 = −4/120, f = −30 cm: a concave mirror of focal length 30 cm.
- m = −3 gives v = 3u (1)
- |v| − |u| = 80 cm: u = −40 cm, v = −120 cm (1)
- 1/f = 1/v + 1/u, f = −30 cm (1)
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Convex lens: u1 = −30 cm, f1 = +20 cm: 1/v1 = 1/20 − 1/30 = 1/60, v1 = +60 cm; m1 = v1/u1 = −2, so this image is 4.0 cm tall, real and inverted. For the concave lens, the rays converging towards this image form a virtual object. Parallel light leaves a lens when its object is at a focal point; for f2 = −10 cm, 1/v2 = 1/f2 + 1/u2 = 0 needs u2 = +10 cm, that is the intermediate image must lie 10 cm beyond the concave lens. So the concave lens is placed 60 − 10 = 50 cm from the convex lens.
- v1 = +60 cm, m1 = −2 (1)
- u2 = +10 cm (virtual object at the focal point) for parallel emergent light (1)
- separation 50 cm; intermediate image 4.0 cm, real, inverted (1)