Two optical media have a plane boundary between them. Suppose ${\theta }_{icr}$   is the critical angle of incidence of a beam and ${\theta }_i$ is the angle of incidence at which the refracted beam is perpendicular to the reflected one (the beam is assumed to come from an optically denser medium) If  $\frac{\sin {\theta }_{icr}}{\sin {\theta }_1}=n$. Find the relative refractive index of these media. If  $\frac{{\mu }_{denser}}{{\mu }_{rarrer}}=\frac{1}{\sqrt{n^2-x}}Findx\mathrm{.........}$

Let two optical medium  of R.I $n_{1}and{n}_2$  respectively be such that  $n_1>n_2$. When angle of incidence is  ${\theta }_{icr}$ ( see figure) from the law of reflection $n_1\sin {\theta }_{icr}=n_2$  …………………. (1)
When the angel of incidence is ${\theta }_i,$ from the law of refraction at the interface of medium 1 and 2 
$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

 
But in accordance with the problem  ${\theta }_2=(\frac{\pi }{2}-{\theta }_1)\mathrm{....................}\left(2\right)$
So $n_1\sin {\theta }_1=n_2\sin {\theta }_2$  
Dividing Equation (1) by Equation (2)
$\frac{\sin {\theta }_{1cr}}{\sin {\theta }_1}=\frac{1}{\cos {\theta }_1}$ 
So  $\cos {\theta }_1=\frac{1}{\eta }and\sin {\theta }_1=\sqrt{\frac{{\eta }^2-1}{\eta }}\mathrm{..................}\left(3\right)$
But  $\frac{n_1}{n_2}=\frac{\cos {\theta }_1}{\cos {\theta }_2}so\frac{n_1}{n_2}=\frac{1}{\eta }\frac{\eta }{\sqrt{{\eta }^2-1}}\left(u\sin gEquation3\right)$
Thus  $\frac{n_1}{n_2}=\frac{1}{\sqrt{{\eta }^2-1}}=1.25$