question_answer

A fish in a lake at a depth h sees the outside world through a circular aperture at the surface. The radius of the aperture is

A)  \[\frac{\mu }{\sqrt{{{h}^{2}}-1}}\]          

B)         \[\mu \sqrt{{{h}^{2}}-1}\]

C)  \[h\sqrt{{{\mu }^{2}}-1}\]          

D)         \[\frac{h}{\sqrt{{{\mu }^{2}}-1}}\]  

Correct Answer: D

Solution :

 Referring figure,\[C=\frac{r}{h}\] where, \[C=\] critical angle \[r=h\tan C=h(\sin C/\cos C)=\frac{h\sin C}{\sqrt{1-{{\sin }^{2}}C}}\] We know,\[\sin C=\frac{1}{\mu }\] \[\therefore \]  \[r=\frac{h/\mu }{\sqrt{1-\frac{1}{{{\mu }^{2}}}}}=\frac{h}{\sqrt{{{\mu }^{2}}-1}}\]