A tank contains water. A fish is near the bottom as shown at a depth of 20 cm below the free surface of water. Refractive index of water is 4/3. A convex lens of focal length 10 cm is kept horizontally 5 cm above the free surface of water. At what height above the lens a horizontal screen is to be held so as to form a sharp image of the fish ?

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20 cm

15 cm

40 cm

30 cm

answer is B.

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Detailed Solution

Hint: The apparent depth of fish is $20 / (4 / 3) = 15 cm$

U= -15-5 = -20 cm

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\frac{1}{10} = \frac{1}{v} + \frac{1}{20}$

$v = 20$

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