JEE Advanced 2022 — Paper 2 — PHY-I solution

JEE Advanced2022Paper 2PHY-I

Consider a configuration of $n$ identical units, each consisting of three layers. The first layer is a column of air of height $h = \frac{1}{3}$ cm, and the second and third layers are of equal thickness $d = \frac{3 - 1}{2}$ cm, and refractive indices $μ_{1} = \frac{3}{2}$ and $μ_{2} = 3$ , 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 $θ = 60°$ to the normal. For a specific value of $n$ , the ray of light emerges from the bottom of the configuration at a distance $l = \frac{8}{3}$ cm, as shown in the figure. The value of $n$ is _______.

Your answer

Solution

Within one unit, compute the horizontal displacement of the ray.

In air, with incidence at $60°$ over height $h = \frac{1}{3}$ cm: lateral shift $x_{1} = h tan 60° = \frac{1}{3}$ cm.

Refraction into the $μ_{1} = 3/2$ layer using Snell's law: $sin 60° = μ_{1} sin θ_{2} \Rightarrow sin θ_{2} = \frac{3 /2}{3/2} = \frac{1}{2}$ , so $θ_{2} = 45°$ . Lateral shift $x_{2} = d tan 45° = d$ .

Refraction into the $μ_{2} = 3$ layer: $μ_{1} sin θ_{2} = μ_{2} sin θ_{3} \Rightarrow sin θ_{3} = \frac{1}{2}$ , so $θ_{3} = 30°$ . Lateral shift $x_{3} = d tan 30° = \frac{d}{3}$ .

Total per unit: $x_{1} + x_{2} + x_{3} = \frac{1}{3} + d (1 + \frac{1}{3}) = \frac{1}{3} + \frac{3 - 1}{2} \cdot \frac{3 + 1}{3} = \frac{1}{3} + \frac{2}{2 3} = \frac{2}{3}$ cm.

Therefore $n = \frac{l}{x _{1} + x _{2} + x _{3}} = \frac{8/ 3}{2/ 3} = 4$ .

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