Refractive index of prism is √2. What should be angle of incidence for a light ray such that the emerging ray grazes out the surface.

The correct option is 90°.

Given Data:
Refractive index of the prism, $\mu =\sqrt{2}$
From the given image, the prism is a right-angled isosceles triangle with base angles of 45° and 45°, which means the angle of the prism is:
$A=\mathrm{180°}-\mathrm{45°}-\mathrm{45°}=\mathrm{90°}$

Step 1: Determine the critical angle ($C_{\text{ic}}$ or ${\theta }_c$) of the prism surface
When a light ray emerges grazing the second surface, the angle of emergence at the second surface is $e=\mathrm{90°}$. Therefore, the angle of incidence at the second surface, $r_2$, must be equal to the critical angle ${\theta }_c$.

Using Snell's Law at the second surface:
$sin\left({\theta }_c\right)=\frac{1}{\mu }=\frac{1}{\sqrt{2}}$

Thus,
$r_2={\theta }_c=\mathrm{45°}$

Step 2: Find the angle of refraction at the first surface ($r_1$)
We know that for a prism, the apex angle $A$ is given by the relation:
$A=r_1+r_2$

Substituting the values of $A=\mathrm{90°}$ and $r_2=\mathrm{45°}$:
$\mathrm{90°}=r_1+\mathrm{45°}$

So, the angle of refraction at the first surface is:
$r_1=\mathrm{45°}$

Step 3: Calculate the angle of incidence ($i$) at the first surface
Using Snell's Law at the first surface:
$1·sin\left(i\right)=\mu ·sin\left(r_1\right)$

Substituting $\mu =\sqrt{2}$ and $r_1=\mathrm{45°}$:
$sin\left(i\right)=\sqrt{2}·sin\left(\mathrm{45°}\right)$
$sin\left(i\right)=\sqrt{2}·\frac{1}{\sqrt{2}}=1$

Therefore, the required angle of incidence is:
$i=\mathrm{90°}$