question_answer

A point source of light is kept at a depth of h in water of refractive index 4/3. The radius of the circle at the surface of water through which light emits is

A)  \[\frac{3}{\sqrt{7}}h\]                  

B)  \[\frac{\sqrt{7}}{3}h\]

C)  \[\frac{\sqrt{3}}{7}h\]                  

D)  \[\frac{7}{\sqrt{3}}h\]

Correct Answer: A

Solution :

                 \[\frac{\sin {{90}^{o}}}{\sin C}=\mu \] \[\sin \,C=\frac{1}{\mu }\Rightarrow \frac{R}{\sqrt{{{R}^{2}}+{{h}^{2}}}}=\frac{3}{4}\] Squaring, \[16\,{{R}^{2}}=9{{R}^{2}}+9{{h}^{2}}\] \[7{{R}^{2}}=9{{h}^{2}}\Rightarrow R=\frac{3}{\sqrt{7}}h\]