Q142 / 155 · WBJEE 2026 · Physics
A ray of light travelling in air is incident on one face of a parallel glass slab of thickness $t$ a refractive index $\mu$ at an angle of incidence $i$. Total time spent by the ray inside the slab is
- A$\dfrac{\mu^2 t}{c\sqrt{1 - \mu^2\sin i}}$
- B$\dfrac{\mu t}{c\sqrt{\mu^2 - \sin^2 i}}$
- Correct answerC$\dfrac{\mu^2 t}{c\sqrt{\mu^2 - \sin^2 i}}$
- D$\dfrac{t}{c\sqrt{\mu^2 - \sin^2 i}}$
Solution
By Snell's law, $\sin i = \mu \sin r$, which gives $\sin r = \dfrac{\sin i}{\mu}$.
The distance travelled by the ray inside the glass slab is $d = \dfrac{t}{\cos r}$.
The speed of light inside the glass slab is $v = \dfrac{c}{\mu}$.
The total time spent by the ray inside the slab is $T = \dfrac{d}{v} = \dfrac{t}{\cos r} \times \dfrac{\mu}{c} = \dfrac{\mu t}{c \cos r}$.
Using $\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - \dfrac{\sin^2 i}{\mu^2}} = \dfrac{\sqrt{\mu^2 - \sin^2 i}}{\mu}$.
Substituting $\cos r$ in the expression for $T$, we get $T = \dfrac{\mu t}{c \left( \dfrac{\sqrt{\mu^2 - \sin^2 i}}{\mu} \right)} = \dfrac{\mu^2 t}{c \sqrt{\mu^2 - \sin^2 i}}$.
Answer: $\dfrac{\mu^2 t}{c\sqrt{\mu^2 - \sin^2 i}}$
The distance travelled by the ray inside the glass slab is $d = \dfrac{t}{\cos r}$.
The speed of light inside the glass slab is $v = \dfrac{c}{\mu}$.
The total time spent by the ray inside the slab is $T = \dfrac{d}{v} = \dfrac{t}{\cos r} \times \dfrac{\mu}{c} = \dfrac{\mu t}{c \cos r}$.
Using $\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - \dfrac{\sin^2 i}{\mu^2}} = \dfrac{\sqrt{\mu^2 - \sin^2 i}}{\mu}$.
Substituting $\cos r$ in the expression for $T$, we get $T = \dfrac{\mu t}{c \left( \dfrac{\sqrt{\mu^2 - \sin^2 i}}{\mu} \right)} = \dfrac{\mu^2 t}{c \sqrt{\mu^2 - \sin^2 i}}$.
Answer: $\dfrac{\mu^2 t}{c\sqrt{\mu^2 - \sin^2 i}}$
Hence, the correct answer is option: 'C'.