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NEET 1994 (AIPMT) · Total Internal Reflection

Total Internal Reflection question from NEET 1994 (AIPMT)

Q64NEET 1994 · AIPMTMedium

Total Internal Reflection

A small source of light is 4 m below the surface of water of refractive index 5/3. In order to cut off all the light, coming out of water surface, minimum diameter of the disc placed on the surface of water is

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Answer: (a)

Radius of the circle of light: $R = \frac{h}{\sqrt{n^2-1}}$ .
$\sqrt{\frac{25}{9}-1} = \frac{4}{3}$ , so $R = \frac{4}{4/3} = 3$ m.
Minimum diameter $= 2R = 6$ m.
So the answer is (a) 6 m.

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