JEE Advanced 2022 · Paper 2 · Physics Q5

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PhysicsInteger+3 / -1Question 23 of 54

Consider a configuration of $n$ identical units, each consisting of three layers. The first layer is a column of air of height $h = \dfrac13$ cm, and the second and third layers are of equal thickness $d = \dfrac{\sqrt3 - 1}{2}$ cm, and refractive indices $\mu_1 = \sqrt{\dfrac32}$ and $\mu_2 = \sqrt3$ , 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 $l = \dfrac{8}{\sqrt3}$ cm, as shown in the figure. The value of $n$ is _____.

A stack of n identical units. Each unit: an air gap of height h, then a layer of thickness d with index μ1, then a layer of thickness d with index μ2. A source O at the top sends a ray at angle θ to the normal; the ray emerges from the bottom of the stack displaced horizontally by l from the vertical through O.

Your answer

Answer and worked solution

Official answer: 4

Angles in each layer. $\mu\sin\theta$ is the same in every layer, equal to $\sin60^\circ=\frac{\sqrt3}2$ :

In $\mu_1=\sqrt{3/2}$ : $\sin\theta_1=\frac1{\sqrt2}$ , so $\theta_1=45^\circ$ .
In $\mu_2=\sqrt3$ : $\sin\theta_2=\frac12$ , so $\theta_2=30^\circ$ .

Sideways shift per unit.
$h\tan60^\circ+d(\tan45^\circ+\tan30^\circ)=\frac{\sqrt3}3+\frac{\sqrt3-1}2\cdot\frac{\sqrt3+1}{\sqrt3}=\frac1{\sqrt3}+\frac1{\sqrt3}=\frac2{\sqrt3}\ \text{cm}.$

Number of units. $n\cdot\dfrac2{\sqrt3}=\dfrac8{\sqrt3}$ , so $n=4$ .

Topics:Refraction at plane surfaces

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JEE Advanced 2022 · Paper 2, Physics question 5