Question

In the above situation if $y_0=a/2$, the coordinates of the point where the ray intersects the upper surface of the slabair boundary are

• A. $[a{e}^2,\frac{a}{2}]$
• B. $[aln2,\frac{a}{2}]$
• C. $\left[\frac{a}{2}ln2,\frac{a}{2}\right]$
• D. $[\sqrt{2}a,\frac{a}{2}]$

Answer 1

The correct option is B $[aln2,\frac{a}{2}]$

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

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

Option B is correct.