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Question
A concave mirror has a 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 from the surface of water. The value of d is dose to: (Refractive index of water = 1.33)

Options
11.7 cm
6.7 cm
13.4 cm
8.8 cm
MCQ
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Solution
8.8 cm
Explanation:

There is water between the object and the mirror.
Applying mirror formula,
`1/v + 1/u = 1/f`
Or `1/v + 1/(-5) = 1/(-20)`
Or `1/v = 1/5 - 1/20`
`1/v = (4 - 1)/20`
`1/v = 3/20`
Or v = `20/3` cm
This distance in water is equivalent to distance in air
= `20/3 xx n`
= `20/3 xx 1.33`
= 8.8 cm
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