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 $4/3$. If $h/x=7/4$ then $cos\theta$ is:

• A. $\frac{2}{7}$
• B. $\frac{8}{\sqrt[3]{345}}$
• C. $\frac{8}{3\sqrt{53}}$
• D. $\frac{8}{21}$

Answer 1

Answer: (C)

$\frac{8}{3\sqrt{53}}$

Answer is C.

According to Snell's law,
$sin(90-\theta )=\mu sinr$.
$cos\theta =\frac{4}{3}\times \frac{x/2}{\sqrt{h^2+\frac{x^2}{4}}}$
$cos\theta =\frac{2}{3}\frac{1}{\sqrt{\frac{h^2}{x^2}+\frac{1}{4}}}=\frac{2}{3}\frac{1}{\sqrt{\frac{49}{16}+\frac{1}{4}}}=\frac{2}{3}\frac{4}{\sqrt{53}}=\frac{8}{3\sqrt{53}}$.
Hence,$cos\theta =\frac{8}{3\sqrt{53}}$.