• # question_answer A student sees the top edge and the bottom centre $C$ of a pool simultaneously from an angle $\theta$ above the horizontal as shown in the figure. The refractive index of water which fills up to the top edge of the pool is $\frac{4}{3}.\operatorname{if}\frac{h}{x}=\frac{7}{4},$ then $\theta$is A) $\frac{2}{7}$ B) $\frac{8}{3\sqrt{45}}$ C) $\frac{8}{3\sqrt{53}}$ D) $\frac{8}{21}$

 Ray diagram for pool is as shown below. Using ${{n}_{1}}.\sin i={{n}_{2}}.\sin \,r,$we have $1\times \sin (90{}^\circ -\theta )=\frac{4}{3}\sin \,r$ ? (i) Also,     $\tan r=\frac{x}{2h}=\frac{4}{7\times 2}=\frac{2}{7}$$\Rightarrow$     $\sin r=\frac{2}{\sqrt{53}}$ Substituting sin $r$in Eq. (i), we have $\cos \,\theta =\frac{4}{3}\times \frac{2}{\sqrt{53}}=\frac{8}{3\sqrt{53}}$