Question
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 $\frac{4}{3} .$ If $\frac{h}{x}=\frac{7}{4}$, then $\cos \theta$ is

Answer

(c)

Ray diagram for pool is as shown below.

Using $n_1 \cdot \sin i=n_2 \cdot \sin r$, we have

$1 \times \sin \left(90^{\circ}-\theta\right)=\frac{4}{3} \sin r \quad \dots(i)$

$\text { Also, } \tan r=\frac{x}{2 h}=\frac{4}{7 \times 2}=\frac{2}{7}$

$\Rightarrow \sin r=\frac{2}{\sqrt{53}} \overline{ }$

Substituting $\sin r$ in Eq. $(i)$, we have

$\cos \theta=\frac{4}{3} \times \frac{2}{\sqrt{5} \overline{3}}=\frac{8}{3 \sqrt{53}}$

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