JEE Advanced 2022 Paper 2 · Q05 · Geometrical Optics

Question

Consider a configuration of $n$ identical units, each consisting of three layers. The first layer is a column of air of height $h = \dfrac{1}{3}\,\text{cm}$, and the second and third layers are of equal thickness $d = \dfrac{\sqrt{3}-1}{2}\,\text{cm}$, with refractive indices $\mu_1 = \sqrt{3/2}$ and $\mu_2 = \sqrt{3}$, respectively. A light source $O$ is placed on the top of the first unit, as shown in the figure. A ray of light from $O$ is incident on the second layer of the first unit at an angle of $\theta = 60^\circ$ to the normal. For a specific value of $n$, the ray of light emerges from the bottom of the configuration at a distance $l = \dfrac{8}{\sqrt{3}}\,\text{cm}$ as shown in the figure. The value of $n$ is _____ .

[Figure: n identical stacked units, each with three layers (air of height h, then medium mu1 of thickness d, then medium mu2 of thickness d). A ray from source O on top, hits the first interface at 60 degrees and zig-zags down, finally exiting at horizontal distance l from O.]

SolveKar AI · Accepted Answer

Apply Snell's law at each interface. From air ($\mu=1$) into $\mu_1 = \sqrt{3/2}$: $\sin 60^\circ = \sqrt{3}/2$, so $\sin r_1 = (\sqrt{3}/2)/\sqrt{3/2} = 1/\sqrt{2}$, $r_1 = 45^\circ$. From $\mu_1$ into $\mu_2 = \sqrt{3}$: $\sqrt{3/2}\sin 45^\circ = \sqrt{3}\sin r_2$, so $\sin r_2 = 1/2$, $r_2 = 30^\circ$. Horizontal displacement per unit: $\Delta x = h	an 60^\circ + d	an 45^\circ + d	an 30^\circ$. With $h = 1/3$ and $d = (\sqrt{3}-1)/2$: $\Delta x = (1/3)\sqrt{3} + ((\sqrt{3}-1)/2)(1) + ((\sqrt{3}-1)/2)(1/\sqrt{3}) = \sqrt{3}/3 + (\sqrt{3}-1)/2 + (\sqrt{3}-1)/(2\sqrt{3})$. Simplifying: $\Delta x = \sqrt{3}/3 + (\sqrt{3}-1)/2 + (3-\sqrt{3})/6 = (2\sqrt{3} + 3(\sqrt{3}-1) + (3-\sqrt{3}))/6 = (2\sqrt{3} + 3\sqrt{3} - 3 + 3 - \sqrt{3})/6 = 4\sqrt{3}/6 = 2\sqrt{3}/3 = 2/\sqrt{3}\,	ext{cm}$. Then $n = l/\Delta x = (8/\sqrt{3})/(2/\sqrt{3}) = 4$.

JEE Advanced 2022 Paper 2 · Q05 · Geometrical Optics

Solution