Optics Question 801

Question: 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:

Options:

A) $ 15\times 3\times \sqrt{5}cm $

B) $ 15\times 3\sqrt{7}cm $

C) $ \frac{15\times \sqrt{7}}{3}cm $

D) $ \frac{15\times 3}{\sqrt{7}}cm $

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Given, $ u=\frac{4}{3}h=15cm $

$ \frac{\sin 90{}^\circ }{\sin C}=\mu \Rightarrow \sin C=\frac{1}{\mu }=\frac{R}{\sqrt{R^{2}+h^{2}}}=\frac{3}{4} $

$ \Rightarrow 16R^{2}=9R^{2}+h^{2}\text{ or, }7R^{2}\text{ =9}{{h}^{2}} $

$ \text{or, }R=\frac{3}{\sqrt{7}}h=\frac{3}{\sqrt{7}}\times 15cm $



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