A diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is 4/3 and the diver's eyes are 15 cm below the surface of water. Then the radius of the circle is: (a) 15 × 3 × √(5) cm (b) 15 × 3 √(7) cm (c) (15×√(7))/(3) cm (d) (15× 3)/(√(7)) cm

Answer: (d). The diver sees the outside world through a circle whose edge corresponds to the critical angle: R = htan C = (h)/(√(n²-1)). √(16/9-1) = √(7)/3. R = (15×3)/(√(7)) cm. So the answer is (d).

A diver looking up through the water sees the outside world contained… | JEE Main 2014 (9 Apr, Shift 1)

Q49JEE Main 2014 · 9 APR · SHIFT 1Medium

Total Internal Reflection

A diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is $\frac{4}{3}$ and the diver's eyes are 15 cm below the surface of water. Then the radius of the circle is:

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Answer: (d)

The diver sees the outside world through a circle whose edge corresponds to the critical angle: $R = h\tan C = \frac{h}{\sqrt{n^2-1}}$ .
$\sqrt{\frac{16}{9}-1} = \frac{\sqrt{7}}{3}$ .
$R = \frac{15\times3}{\sqrt{7}}$ cm.
So the answer is (d).

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