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 $
BETA
