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Geometrical Optics

Section: Physics

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Question : 52 of 56

Marks: +1, -0

A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of ball as (Take g = 10 m/

s^2

[JEE Adv 2009 P1]

9 m/s
12 m/s
16 m/s
21.33 m/s

Solution:

The velocity of the ball when it has fallen through 7.2 m is

V =

\sqrt{2g(7.2)}

= 12 m/s

For a general height h above the water surface, we have

\frac{h'}{h}

=

\frac{4}{3}

⇒ h' =

\frac{4}{3} h

Therefore, V; =

\frac{4}{3} V

=

\frac{4}{3} \times 12

= 16 m/s

where V ′ is the velocity of ball with respect to the fish.

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