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Question

# A concave mirror has radius of curvature of 40 cm. It is at the bottom of a glass that has water filled up to 5 cm (see figure) If a small particle is floating on the surface of water, its image as seen, from directly above the glass, is at a distance d in cm from the surface of water. The value of d is close to [Refractive index of water = 4/3]

A
8.8
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B
9.0
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C
9.00
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D
9
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E
8.80
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F
8.7
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G
8.70
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H
8.75
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Solution

## In the given case, u=−5 cm Focal length, f = −R2=−402=−20 cm Now, using mirror formula, ⇒v=+203cm For the light getting refracted at water surface, this image will acts as an object. So, distance of object, d′=5 cm+203cm=353 cm (below the surface). Let's assume final image at distance d after refraction. dd′=μ2μ1 ⇒d=d′(μ2μ1)=(353 cm)⎛⎜ ⎜ ⎜⎝143⎞⎟ ⎟ ⎟⎠ =353×34 cm=354 cm =8.75 cm≈8.8 cm Ans. 8.75 cm Why this question? Note: For refraction of rays at the water surface, the image formed by the mirror will act as an object.

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