A ray of light enters a glass slab of thickness $t=10\text{ cm}$ and refractive index $\mu =1.5$ at an angle of incidence $i=45^{\circ }$. Calculate the lateral shift $s$.

Solution

1. Given: thickness $t=10\text{ cm}$, refractive index $\mu =1.5$, and angle of incidence $i=45^{\circ }$.
2. Apply Snell's Law, $\sin (i)=\mu \sin (r)$, to find the angle of refraction: $\sin (r)=\frac{\sin (45^{\circ })}{1.5}\approx 0.471$, which gives $r\approx 28.1^{\circ }$.
3. Use the lateral shift formula $s=\frac{t\sin (i-r)}{\cos (r)}$ to calculate the displacement.
4. Substitute the values: $s=\frac{10\sin (45^{\circ }-28.1^{\circ })}{\cos (28.1^{\circ })}\approx \frac{10\times 0.2907}{0.8821}\approx 3.2\text{ cm}$.
5. Hence the answer is (B).