Class 10 Science Chapter 9: Light: Reflection and Refraction worksheet with answer key
A 20-mark worksheet on Light: Reflection and Refraction 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: (a) P only
The pole is the centre of the reflecting surface and lies on the mirror. F and C lie on the principal axis in front of a concave mirror, not on its surface.
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Answer: (d) appear to diverge from a point on the principal axis behind the mirror
A convex mirror reflects parallel rays so that they appear to come from a point on the principal axis behind the mirror; that point is its principal focus.
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Answer: (d) 28 cm
The bright spot is the real image of the distant Sun, formed at the principal focus, so f ≈ 14 cm. For a mirror of small aperture R = 2f = 28 cm.
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Answer: (b) 55°
Angles of incidence and reflection are measured from the normal. The ray makes 90° − 35° = 55° with the normal, so i = 55° and, since the angle of incidence equals the angle of reflection, r = 55°.
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Answer: (d) virtual, erect and enlarged
With the object between the pole and the focus, a concave mirror forms an enlarged, virtual and erect image behind the mirror; that is why dentists use it to see large images of teeth.
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(i) The angle of incidence equals the angle of reflection. (ii) The incident ray, the normal to the mirror at the point of incidence and the reflected ray all lie in the same plane. Yes: the laws apply to all reflecting surfaces, including spherical mirrors.
- first law: angle of incidence equals angle of reflection (0.5)
- second law: incident ray, normal and reflected ray in one plane (1)
- states that the laws hold for spherical mirrors too (0.5)
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P = P₁ + P₂ = +5 + (−3) = +2 D. f = 1/P = 1/2 m = 0.5 m = 50 cm. The power is positive, so the combination converges light.
- net power +2 D by adding powers (1)
- f = 50 cm, converging (1)
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It is a concave mirror. Held close, his face is between the pole and the focus, so the image is virtual, erect and enlarged. At arm's length his face is beyond the focus, where a concave mirror forms a real, inverted image.
- names concave mirror (0.5)
- close: object between P and F gives virtual, erect, enlarged image (1)
- far: object beyond F gives real, inverted image (0.5)
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By Snell's law n = sin i / sin r = 0.6/0.375 = 1.6. v = c/n = (3 × 10⁸)/1.6 = 1.875 × 10⁸ m/s. Since 1.6 > 1.52, the block is optically denser than crown glass.
- n = sin i / sin r = 1.6 (1)
- v = c/n = 1.875 × 10⁸ m/s (1)
- denser, as 1.6 > 1.52 (1)
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The image is virtual and erect, so m = h′/h = +30/180 = +1/6. Also m = −v/u, so v = −u/6. Then 1/v + 1/u = −6/u + 1/u = −5/u = 1/f = 1/30, so u = −150 cm. v = −u/6 = +25 cm. Tenzin is 1.5 m in front of the mirror; his image is 25 cm behind it.
- m = +1/6 and v = −u/6 (1)
- mirror formula gives u = −150 cm (1)
- v = +25 cm, behind the mirror (1)
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Beyond 2F₁: image between F₂ and 2F₂, diminished. At 2F₁: image at 2F₂, the same size. Between F₁ and 2F₁: image beyond 2F₂, enlarged. In all three the image is real and inverted.
- beyond 2F₁: between F₂ and 2F₂, diminished (1)
- at 2F₁: at 2F₂, same size; between F₁ and 2F₁: beyond 2F₂, enlarged (1)
- real and inverted throughout (1)
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