JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Geometrical Optics

A ray of ligh is incident at an angle of incidence

6 0^{\circ}

on the glass slab of refractive index

3

. After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is

43 cm

. The thickness of the glass slab is

____

cm.

Answer: 12

Solution

The lateral shift for a parallel-sided glass slab is given by

ℓ = t sin i [1 - \frac{c o s i}{μ ^{2} - s i n ^{2} i}]

. Given:

i = 6 0^{\circ}

μ = 3

ℓ = 43

cm. Substituting:

sin 6 0^{\circ} = \frac{3}{2}

cos 6 0^{\circ} = \frac{1}{2}

, and

μ^{2} - sin^{2} i = 3 - 3/4 = \frac{3}{2}

. Therefore,

43 = t \cdot \frac{3}{2} [1 - \frac{1/2}{3/2}] = t \cdot \frac{3}{2} \cdot \frac{2}{3} = \frac{t 3}{3}

. Solving:

t = 12

cm.

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Integer answer

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