Light Numerical

Q1. An object is placed 20 cm in front of a plane mirror. What is the distance between the object and its image?
Answer: 40 cm
In a plane mirror, image distance equals object distance.
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Q2. A concave mirror has a focal length of 15 cm. Where should an object be placed to get an image at 30 cm in front of the mirror?
Answer: 30 cm in front of the mirror
Use the mirror formula: 1/f = 1/v + 1/u. Here, f = -15 cm, v = -30 cm.
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Q3. The magnification produced by a spherical mirror is -0.5. What is the nature of the image?
Answer: Real, inverted, and diminished
Negative magnification indicates a real and inverted image. A magnitude less than 1 means diminished.
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Q4. A convex mirror forms an image that is one-third the size of the object. If the object is placed 9 cm in front of it, what is the focal length?
Answer: -4.5 cm
Magnification (m) = +1/3 = -v/u. Find v, then use the mirror formula. The focal length of a convex mirror is negative.
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Q5. Light travels from air (n=1) into glass with a refractive index of 1.5. If the angle of incidence is 30ยฐ, what is the angle of refraction? (Use sin 30ยฐ = 0.5)
Answer: Approximately 19.5ยฐ
Use Snell’s Law: nโ‚ sin i = nโ‚‚ sin r. (1)sin30 = (1.5)sin r.
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Q6. The speed of light in water is 2.25 x 10โธ m/s. What is the refractive index of water?
Answer: 1.33
Refractive index n = c/v. c = 3 x 10โธ m/s.
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Q7. A ray of light enters from air to glass. If the angles of incidence and refraction are 45ยฐ and 28ยฐ respectively, find the refractive index of the glass.
Answer: Approximately 1.5
n = sin i / sin r. Use sin 45ยฐ โ‰ˆ 0.707 and sin 28ยฐ โ‰ˆ 0.469.
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Q8. The refractive index of diamond is 2.42. What is the speed of light in diamond?
Answer: 1.24 x 10โธ m/s
v = c / n = (3 x 10โธ) / 2.42.
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Q9. A convex lens has a focal length of 10 cm. An object is placed 15 cm from it. Where is the image formed?
Answer: 30 cm on the other side of the lens
Use the lens formula: 1/f = 1/v – 1/u. For a convex lens, f = +10 cm, u = -15 cm.
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Q10. A concave lens has a focal length of 20 cm. An object is placed 60 cm from it. What is the image distance?
Answer: -15 cm (on the same side as the object)
Use the lens formula. For a concave lens, f is negative (f = -20 cm).
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Q11. The power of a lens is +2.0 D. What is its focal length?
Answer: +0.5 m or +50 cm
Power (P) = 1/f (in meters). So, f = 1/P.
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Q12. A lens has a focal length of -25 cm. What is its power?
Answer: -4.0 D
First convert f to meters: -0.25 m. Then P = 1/(-0.25).
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Q13. An object 2 cm high is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. If the object distance is 15 cm, what is the height of the image?
Answer: -4 cm (4 cm, inverted)
First find v using the lens formula (Answer to Q9). Then, magnification m = v/u = hแตข/hโ‚’.
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Q14. A convex lens forms a real image twice the size of the object. If the object distance is 12 cm, find the focal length.
Answer: 8 cm
For a real image, m = -2 = v/u. So v = -2 * (-12) = +24 cm. Now use the lens formula.
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Q15. Two thin lenses of power +5 D and -2 D are placed in contact. What is the combined power?
Answer: +3 D
Powers of lenses in contact add up: P = Pโ‚ + Pโ‚‚.
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Q16. Light travels from glass (n=1.5) to air (n=1). What is the critical angle for glass?
Answer: Approximately 41.8ยฐ
sin c = nโ‚‚/nโ‚ = 1/1.5.
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Q17. A person uses a lens of power -1.5 D. What is the defect of vision, and what is the far point of the eye?
Answer: Myopia (Nearsightedness), Far point = 66.7 cm
For a corrective concave lens, f = 1/P gives f = -0.667 m. The far point is at this distance.
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Q18. An object is placed at a distance of 30 cm from a concave mirror of focal length 20 cm. At what distance will the image be formed?
Answer: -60 cm (in front of the mirror)
Use the mirror formula: 1/f = 1/v + 1/u, with f = -20 cm, u = -30 cm.
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Q19. The magnification of a mirror is +3. What does this indicate about the image?
Answer: Virtual, erect, and magnified (3 times)
Positive magnification indicates a virtual and erect image.
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Q20. If the angle between an incident ray and a plane mirror is 25ยฐ, what is the angle of reflection?
Answer: 65ยฐ
The angle of incidence is measured from the normal. Here, the angle with the mirror is 25ยฐ, so the angle with the normal is 90ยฐ – 25ยฐ = 65ยฐ.
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Q21. A convex mirror used on a truck has a focal length of -3 m. If a car is 6 m behind the truck, where is its image?
Answer: -2 m (behind the mirror)
Use the mirror formula: u = -6 m, f = -3 m. Find v.
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Q22. The refractive index of water is 4/3. What is the speed of light in water?
Answer: 2.25 x 10โธ m/s
v = c / n = (3 x 10โธ) / (4/3).
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Q23. A lens forms a virtual image at 10 cm when an object is placed 5 cm from it. What is its focal length and power?
Answer: f = +10 cm, P = +10 D
For a virtual image by a convex lens, v is negative (v = -10 cm, u = -5 cm). Use the lens formula.
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Q24. In a prism, if the angle of incidence is 30ยฐ and the angle of emergence is 45ยฐ, what is the angle of deviation for a small-angle prism? (Assume base angles are small)
Answer: 15ยฐ
For a thin prism, the angle of deviation ฮด โ‰ˆ i + e – A. A common assumption leads to ฮด โ‰ˆ 15ยฐ.
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Q25. A candle flame 3 cm high is placed 30 cm from a concave mirror of focal length 20 cm. Find the height of the image.
Answer: -6 cm (6 cm high, inverted)
First find v using the mirror formula (like Q18), then use magnification m = -v/u = hแตข/hโ‚’.
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Q26. The power of a spectacle lens is +2.5 D. What eye defect is corrected, and what is the near point of the eye?
Answer: Hypermetropia (Farsightedness), Near point = 40 cm (beyond 25 cm)
Convex lens. f = 1/2.5 = 0.4 m = 40 cm. The lens forms an image of an object at 25 cm at the eye’s near point.
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Q27. A ray of light strikes a glass slab (n=1.5) at 60ยฐ. What is the angle of refraction inside the slab?
Answer: Approximately 35.3ยฐ
Use Snell’s Law: 1 * sin 60ยฐ = 1.5 * sin r. sin 60ยฐ = โˆš3/2 โ‰ˆ 0.866.
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Q28. An object is placed at the centre of curvature of a concave mirror (f = 15 cm). What is the image distance?
Answer: -30 cm (at the centre of curvature)
Centre of curvature, u = -2f = -30 cm. For this position, the image is formed at the same point.
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Q29. A converging lens has a focal length of 12 cm. At what distance should an object be placed to get an image magnified 3 times?
Answer: Two possible answers: u = -16 cm (for a real image) or u = -8 cm (for a virtual image)
For a real, inverted image: m = -3 = v/u => v = -3u. Use the lens formula. For a virtual, erect image, use m = +3.
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Q30. Light takes 8 minutes and 20 seconds to reach Earth from the Sun. What is the distance between the Sun and the Earth?
Answer: 1.5 ร— 10ยนยน m
Time = 500 seconds. Distance = Speed ร— Time = (3 ร— 10โธ m/s) ร— 500 s.
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