Question

A light ray enters into a medium whose refractive index varies along the x-axis as $n\left(x\right)=n_0\sqrt{1+\frac{x}{4}}$ where $n_0=1$. The medium is bounded by the planes $x=0,x=1$ & $y=0$. If the ray enters at the origin at an angle ${30}^{\circ }$ with x-axis. Then

• A. the equation of trajectory of the light ray is $y=[\sqrt{3+x}-\sqrt{3}]$
• B. the equation of trajectory of the light ray is $y=2[\sqrt{3+x}-\sqrt{3}]$
• C. the coordinate the point at which light ray comes out from the medium is $[1,2(2-\sqrt{3}\left)\right]$
  
• D. the coordinate the point at which light ray comes out from the medium is $[0,2(2-\sqrt{3}\left)\right]$

Answer 1

Answer: (B), (C)

the coordinate the point at which light ray comes out from the medium is $[1,2(2-\sqrt{3}\left)\right]$


Given :-
$n\left(x\right)=n_0\sqrt{1+\frac{x}{4}}$
$n\left(x\right)=\sqrt{\frac{4+x}{4}}$
Now, applying snell`s law at first surface
${\mu }_1\sin i={\mu }_2\sin r$
$\Rightarrow 1\sin i={\mu }_2\sin r$
$\Rightarrow \sin {30}^{\circ }={\mu }_2\sin r\Rightarrow \sin r=\frac{1}{2{\mu }_2}$
or $\tan r=\frac{1}{\sqrt{4{\mu }_2^2-1}}$
$\because {\mu }_2=n\left(x\right)$
$\Rightarrow \tan r=\frac{dy}{dx}=\frac{1}{\sqrt{4{\left(\sqrt{\frac{4+x}{4}}\right)}^2-1}}$
or $\frac{dy}{dx}=\frac{1}{\sqrt{x+3}}$
or ${\int }_0^{y}dy={\int }_0^x\frac{dx}{(x+3{)}^{1/2}}$
$y=2\left[\right(x+3{)}^{1/2}{]}_0^x$
or $y=2(\sqrt{x+3}-\sqrt{3})$
Hence, it is the trajectory of light ray

Now, at $x=1,y=2(2-\sqrt{3})$
at $x=0,y=0$
Co-ordainate of point where light ray come out is
$[1,2(2-\sqrt{3}\left)\right]$