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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 cm8.8 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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