A light ray is incident on a glass slab of thickness 4√3 cm and refractive index √2. The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of ray after passing through glass slab is ____cm. (Given sin 15° = 0.25)

To find the lateral displacement of the light ray after passing through the glass slab, we can follow these steps:

Step 1: Understand the given variables
Thickness of the glass slab, $t=4\sqrt{3}\text{cm}$
Refractive index of the glass slab, $\mu =\sqrt{2}$
Let the refractive index of air be ${\mu }_{\text{air}}=1$.
Given: $\sin \left({15}^{\circ }\right)=0.25$

Step 2: Find the angle of incidence
The problem states that the angle of incidence ($i$) is equal to the critical angle (${\theta }_c$) for the glass-air interface.
The relation for the critical angle is:
$\sin \left({\theta }_c\right)=\frac{1}{\mu }$
Substituting $\mu =\sqrt{2}$:
$\sin \left(i\right)=\sin \left({\theta }_c\right)=\frac{1}{\sqrt{2}}$
Therefore, the angle of incidence is:
$i={45}^{\circ }$

Step 3: Find the angle of refraction
Using Snell's law at the first interface (air to glass):
$1\cdot \sin \left(i\right)=\mu \cdot \sin \left(r\right)$
$\sin \left({45}^{\circ }\right)=\sqrt{2}\cdot \sin \left(r\right)$
$\frac{1}{\sqrt{2}}=\sqrt{2}\cdot \sin \left(r\right)$
$\sin \left(r\right)=\frac{1}{2}$
Therefore, the angle of refraction is:
$r={30}^{\circ }$

Step 4: Calculate the lateral displacement
The formula for lateral displacement ($d$) of a ray passing through a glass slab is:
$d=\frac{t\cdot \sin (i-r)}{\cos \left(r\right)}$
Substitute the values:
$i-r={45}^{\circ }-{30}^{\circ }={15}^{\circ }$
$d=\frac{4\sqrt{3}\cdot \sin \left({15}^{\circ }\right)}{\cos \left({30}^{\circ }\right)}$
Since $\cos \left({30}^{\circ }\right)=\frac{\sqrt{3}}{2}$ and $\sin \left({15}^{\circ }\right)=0.25=\frac{1}{4}$, we have:
$d=\frac{4\sqrt{3}\cdot \frac{1}{4}}{\frac{\sqrt{3}}{2}}$
Simplifying the expression:
$d=\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}=2\text{cm}$

Thus, the lateral displacement of the ray after passing through the glass slab is 2 cm.