A ray of light incident normally on one of the faces of a right-angled isosceles prism is found to be totally reflected as shown.
(a)    What is the minimum value of the refractive index of the material of the prism?
(b)    When the prism is immersed in water trace the path of the emergent ray for the same incident ray indicating the values of all the angles. (μ of water = 4/3).

(a)
ABC is the section of the prism right angled at B.
A and C are equal angles i.e., A = C = 45°. 



The ray PQ is normally incident on the face AB. Hence it is normally refracted and the ray QR strikes the face AC at an angle of incidence 45°.
It is given that the ray does not undergo refraction but is totally reflected at the face AC. This gives a maximum value for the critical angle as 45°. 
Therefore,
                       $\mathrm{Sin} \mathrm{C} = \sin 45° = \frac{1}{\sqrt{2}}$
Since,                   $\mathrm{\mu } = \frac{1}{\sin \mathrm{C}} = \sqrt{2}$  

$\Rightarrow$                        $\mathrm{\mu } = \frac{1}{\sin 45°}$ 

i.e.,                  ${\mathrm{\mu }}_{\min } = \sqrt{2}$
Thus, the minimum value of refractive index $= \sqrt{2}.$ 

(b) When the prism is immersed in water the critical angle for the glass-water interface is given by
$\sin {\mathrm{C}}_1 = \frac{4/3}{\sqrt{2}} = \frac{4}{3\sqrt{2}}$
$\therefore$ ${\mathrm{C}}_1 =70.53°$

The angle of incidence at R continues to be 45° and obviously, 45° < 70.53°.
Hence, there is refraction taking place now and the refracted ray is RS.

The angle of refraction r is given by,
            μ_g sin i = μ_w sin r 

 
Refractive index of water w.r.t glass is, 

                $\boxed{\mathrm{g} {\mathrm{\mu }}_{\mathrm{\omega }} = \frac{\sin \mathrm{i}}{\sin \mathrm{r}}}$

                 $\frac{{\mathrm{\mu }}_{\mathrm{\omega }}}{{\mathrm{\mu }}_{\mathrm{g}}} = \frac{\sin \mathrm{i}}{\sin \mathrm{r}}$ 

       $\sqrt{2} \sin 45° = \frac{4}{3} \sin \mathrm{r}$ 

                 $\sin \mathrm{r} = \frac{3\sqrt{2}}{4}\sin 45°$

                        $= \frac{3\sqrt{2}}{4}\times \frac{1}{\sqrt{2}} = \frac{3}{4}$
i.e.,                $\mathrm{r} ={\sin }^{-1}\frac{3}{4} = 48°36\text{'}$ 
The angle of refraction in water = 48°36'.