A man of height 2m  is standing on level road where because of temperature variation the refractive index of air is varying as $\mu =\sqrt{1+ay},$  where  y is height from road. If  $a=2\times {10}^{-6}{\text{m}}^{-1}.$ The maximum distance, in km, till which he can see on the road is

Since,  $\theta =90°-i$
 $\Rightarrow \tan \theta =\cot i \Rightarrow \frac{dy}{dx}=\cot i$
According to Snell’s Law at  $O\text{ and }P,$  we have  ${\mu }_0\sin i_0={\mu }_p\sin i_p$
Since  $\mu =\sqrt{1+ay}$
$\Rightarrow \text{ At }y=0,\mu =1\Rightarrow \sin (90°)=\left(\sqrt{1+ay}\right)\sin i$

$\Rightarrow \sin i=\frac{1}{\sqrt{1+ay}} \Rightarrow \cot i=\sqrt{ay}=\frac{dy}{dx} \Rightarrow \underset{0}{\overset{y}{\int }}\frac{dy}{\sqrt{ay}}=\underset{0}{\overset{x}{\int }}dx \Rightarrow x=2\sqrt{\frac{y}{a}}$

Substituting,  $y=2\text{ m and }a=2\times {10}^{-6}{m}^{-1},$ we get  $x_{\max }=2000\text{ m }=2\text{ km}$