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Q.

A vessel is filled with water $(μ = \frac{4}{3})$ up to a height 20 cm. A concave mirror of radius of curvature 40 cm is placed 15 cm above the water level. The position of final image of an object placed 20 cm below the water surface will be

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a

45 cm below the water surface

b

60 cm above the water surface

c

45 cm above the water surface

d

60 cm below the water surface

answer is A.

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Detailed Solution

The object distance for the mirror $= - (15 + \frac{20}{4 / 3}) = - 30 cm$
For mirror, $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\Rightarrow \frac{1}{v} + \frac{1}{- 30} = \frac{1}{- 20} \Rightarrow v = - 60 cm$
The distance from water surface equals 60 – 15 = 45 cm

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