Two light beams fall on a transparent material block at point 1 and 2… | JEE Main 2025 (29 Jan, Shift 1)

Q17JEE Main 2025 · 29 JAN · SHIFT 1Difficult

Refraction, Glass Slab & Apparent Depth

Two light beams fall on a transparent material block at point 1 and 2 with angle $\theta_1$ and $\theta_2$ , 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, $d = 4\sqrt{3}\text{ cm}$ and $\theta_1 = \theta_2 = \cos^{-1}\left(\frac{n_2}{2n_1}\right)$ . where refractive index of the block $n_2 >$ refractive index of the outside medium $n_1$ , then the thickness of the block is ______ cm.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 6

Angle with surface is $\theta$ , so angle of incidence $i = 90^\circ - \theta$ and $\sin i = \cos\theta = \frac{n_2}{2n_1}$ .
Snell: $n_1\sin i = n_2\sin r \Rightarrow n_1 \cdot \frac{n_2}{2n_1} = n_2\sin r \Rightarrow \sin r = \frac{1}{2},\ r = 30^\circ$ .
By symmetry point 3 lies below the midpoint of 1 and 2: $\tan r = \frac{d/2}{t}$ .
$t = \frac{2\sqrt{3}}{\tan 30^\circ} = 2\sqrt{3} \times \sqrt{3} = 6$
So the thickness is $6\text{ cm}$ .

Report a mistake

Refraction, Glass Slab & Apparent Depth question from JEE Main 2025 (29 Jan, Shift 1)

More questions on Refraction, Glass Slab & Apparent Depth