JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Geometrical Optics

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} c m

, and the second and third layers are of equal thickness

d = \frac{3 - 1}{2} c m

, 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 = 6 0^{\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 = \frac{8}{3} c m

, as shown in the figure. The value of

n

is ____.

Answer: 4

Solution

Apply Snell's law at each interface:

At air-medium 1:

1 \cdot sin 60° = \frac{3}{2} sin θ_{1}

, giving

θ_{1} = 45°

At medium 1-medium 2:

\frac{3}{2} sin 45° = 3 sin θ_{2}

, giving

θ_{2} = 30°

Lateral displacement per unit:

Δ x = h tan 60° + d tan 45° + d tan 30° = \frac{1}{3} 3 + \frac{3 - 1}{2} + \frac{3 - 1}{2 3} = \frac{2 3}{3}

Total displacement:

n \cdot \frac{2 3}{3} = \frac{8}{3}

Solving:

n = 4

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Integer answer

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JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Geometrical Optics

Solution

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