Consider a hemisphere of radius R with center of curvature at origin O, as shown in figure. Refractive index of material of the hemisphere varies as $\mu =\frac{2R}{2R-x}$, where x is x-coordinate of material point. A ray travelling in air in XY-plane is grazingly incident at O, as shown. Choose the correct option(s).

The figure shows a strip at a distance x of thickness dx, 

As $\mu$ of material increases with x, ray will deviate continuously as shown. By Snell’s law, between O and A,
 

 $\begin{array}{l}1\times \sin {90}^0=\frac{2R}{2R-X}\times \sin \theta \Rightarrow \tan \theta =\frac{2R-X}{\sqrt{{\left(2R\right)}^2-{(2R-X)}^2}}\\ \Rightarrow \frac{dy}{dx}=\frac{2R-X}{\sqrt{{\left(2R\right)}^2-{(2R-X)}^2}}\Rightarrow {\int }_0^{y}dy={\int }_0^x\frac{2R-X}{\sqrt{{\left(2R\right)}^2-{(2R-X)}^2}}dx\\ \Rightarrow Y=\sqrt{{\left(2R\right)}^2-{(2R-X)}^2}\Rightarrow Y^2+{(X-2R)}^2={\left(2R\right)}^2\end{array}$  
Which is equation of circle of radius 2R centered at (2R,0). 

Also, equation of hemispherical surface is,  $X^2+Y^2=R^2\Rightarrow Y^2=R^2-X^2$ 

Putting this value of Y in equation of trajectory, we can find coordinates of point where the ray comes out of hemisphere, as      $R^2-X^2+{(X-2R)}^2=4R^2$

$\Rightarrow \mathrm{X}=\frac{\mathrm{R}}{4}$

$\Rightarrow \mathrm{Y}=\sqrt{{\mathrm{R}}^2-{\mathrm{X}}^2}=\sqrt{{\mathrm{R}}^2-{\left(\frac{\mathrm{R}}{4}\right)}^2}$$=\frac{\sqrt{15}\mathrm{R}}{4}=0.97\mathrm{R}$         
From the diagram, $\sin \delta =\frac{Y}{2R}=\frac{\sqrt{15}}{8}<\frac{1}{2}\Rightarrow 0<\delta <{30}^0$