JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Geometrical Optics

Two light beams fall on a transparent material block at point 1 and 2 with angle

θ_{1}

and

θ_{2}

, respectively, as shown in figure. After refraction, the beams intersect at point 3 which is exactly on the interface at other end of the block. Given : the distance between 1 and

2, d = 43 cm

and

θ_{1} = θ_{2} = cos^{- 1} (\frac{n _{2}}{2 n _{1}})

, where refractive index of the block

n_{2} >

refractive index of the outside medium

n_{1}

, then the thickness of the block is

____

cm .

Answer: 6

Solution

Let the refractive index of the block be

n_{2}

and of the surrounding medium be

n_{1}

, with

n_{2} > n_{1}

.

At point 1, the ray strikes the top surface at angle of incidence

θ_{1}

measured from the surface (not the normal), with

cos θ_{1} = \frac{n _{2}}{2 n _{1}}

as given. Applying Snell's law across the top surface, referencing the surface,

n_{1} cos θ_{1} = n_{2} sin θ_{3}

where

θ_{3}

is the angle the refracted ray makes with the normal inside the block. Substituting

cos θ_{1} = \frac{n _{2}}{2 n _{1}}

n_{1} \cdot \frac{n _{2}}{2 n _{1}} = n_{2} sin θ_{3} ⟹ sin θ_{3} = \frac{1}{2} ⟹ θ_{3} = 30°

By symmetry, the two refracted rays (from points 1 and 2, separated by

d = 43

cm) travel toward each other and meet at point 3 on the bottom face, directly below the midpoint of 1 and 2. Each refracted ray makes an angle

θ_{3} = 30°

with the vertical (the normal to the horizontal interfaces), so, with

t

the thickness of the block,

tan 30° = \frac{d /2}{t} ⟹ t = \frac{d}{2 tan 30°} = \frac{d 3}{2} = \frac{4 3 \times 3}{2} = 6 cm

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Integer answer

Practise more: all JEE Main PYQs on Geometrical Optics →

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Geometrical Optics

Solution

More JEE Main questions on Geometrical Optics