QJEEAdv23Ph0ee4: JEE Advanced 2023 Physics - Class 12 JEEAdv23Ph0ee4 Solution: A monochromatic light wave is incident normally on a glass slab …

At point A on one end of the slab, the wavefront takes a time $t$ to reach the other end. The time $t$ can be expressed as the ratio of the distance traveled, $d$ , to the speed of the wave in the medium, which is $\frac{c}{n_1}$ (where $c$ is the speed of light in vacuum and $n_1$ is the refractive index of the medium at A). Therefore, the time taken $t$ is given by $t = \frac{d}{\frac{c}{n_1}} = \frac{n_1 d}{c}$ .

Simultaneously, consider the wavefront at point B, located in a different medium with a refractive index $n_2$ . In the same time $t$ , the wavefront at B travels a distance $s_1$ . The distance $s_1$ can be calculated as the product of the speed of light in the medium at B, $\frac{c}{n_2}$ , and the time $t$ , yielding $s_1 = \frac{c}{n_2} \times \frac{n_1 d}{c} = \frac{d n_1}{n_2}$ .

Once the wavefront at B reaches the end of the slab, an additional time $t_1$ elapses. The time $t_1$ can be determined by considering the remaining distance the wavefront must travel in the slab, which is $d - s_1$ , and dividing it by the speed of light in the medium at B. This gives us $t_1 = \left(\frac{d - s_1}{c}\right) n_2 = \left(d - \frac{d n_1}{n_2}\right) \frac{n_2}{c} = \frac{d(n_2 - n_1)}{c}$ .

During this time $t_1$ , the wavefront at A will travel a distance $s_2$ . The distance $s_2$ is simply the speed of light in vacuum multiplied by the time $t_1$ , leading to $s_2 = c t_1 = d(n_2 - n_1)$ .

Finally, the angle $\theta$ can be determined using the tangent function, where $\tan \theta$ is the ratio of the difference in distances traveled by the wavefronts in the two media to the height $h$ of the slab.

Therefore, $\tan \theta = \frac{(n_2 - n_1) d}{h}$ .