In the previous question, the angle made by light ray with +ve x-axis at the upper surface of a slab-air boundary, inside the medium is

The correct option is D $ta{n}^{-1}\left(\frac{1}{2}\right)$

given,

$\mu =\sqrt{1+e^{\frac{2x}{a}}}$

the angle of incidence is $\angle i\simeq {90}^0$

then at any point inside the medium angle of refraction , $\angle r=\angle \theta$

$\mu =\frac{sini}{sinr}=\frac{sin{90}^0}{sin\theta }$

$\Rightarrow \mu sin\theta =1$

$\Rightarrow sin\theta =\frac{1}{\mu }$

then, $tan\theta =\frac{1}{\sqrt{{\mu }^2-1}}$ (1)

here, from the figure , $tan\theta =\frac{dy}{dx}$ (2)

By equating equation 1 and 2

$\frac{dy}{dx}=\frac{1}{\sqrt{{\mu }^2-1}}=e^{-\frac{x}{a}}$

$\int dy=\int e^{-\frac{x}{a}}dx$

by integrating on both sides we get

$y=a{e}^{-\frac{x}{a}}$

when $y=\frac{a}{2}thenx=a\ln 2$

thus coordinates are

$[a\ln 2,\frac{a}{2}]$

at coordinate $[a\ln 2,\frac{a}{2}]$

$tan\theta =\frac{dy}{dx}=\frac{1}{\sqrt{{\mu }^2-1}}=e^{-\frac{x}{a}}$

put value of $x=a\ln 2$

we get $tan\theta =\frac{1}{2}$

$\Rightarrow \theta ={tan}^{-1}\left(\frac{1}{2}\right)$

Option D is correct.