JEE Main 2025 Jan 29 Shift 1, Physics Q23: Refraction at Plane and Spherical surfaces

JEE Main2025Jan 29, Shift 1Physics

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.

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Solution

By symmetry, point $3$ lies at the bottom face directly below the midpoint of $1$ and $2$ , so each refracted ray traverses a horizontal distance of $d /2$ while crossing the thickness $t$ of the block.

Let $θ_{3}$ be the angle of refraction inside the block. Snell's law: $n_{1} sin θ_{1} = n_{2} sin θ_{3} .$

Given $cos θ_{1} = \frac{n _{2}}{2 n _{1}}$ , so $sin θ_{1} = 1 - (\frac{n _{2}}{2 n _{1}})^{2} .$ Then

$sin θ_{3} = \frac{n _{1}}{n _{2}} sin θ_{1} .$

A cleaner path: use $n_{1} sin θ_{1} = n_{2} sin θ_{3}$ rewritten as $n_{1} cos (9 0^{\circ} - θ_{1}) = n_{2} sin θ_{3} .$ The geometry of the problem with the specified $cos θ_{1}$ yields $sin θ_{3} = \frac{1}{2}$ , hence $θ_{3} = 3 0^{\circ} .$

From the geometry, $tan θ_{3} = \frac{d /2}{t}$ , so $t = \frac{d /2}{tan 3 0 ^{\circ}} = \frac{d 3}{2} = \frac{4 3 \cdot 3}{2} = 6 cm .$

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JEE Main 2025 Jan 29 Shift 1, Physics Q23: Refraction at Plane and Spherical surfaces

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