Physics · Geometrical Optics

JEE Main 2025 — 29 January, Morning Shift — Question 42

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 $2, d=4 \sqrt{3} \mathrm{~cm}$ and $\theta_{1}=\theta_{2}=\cos ^{-1}\left(\frac{n_{2}}{2 n_{1}}\right)$ ,

where refractive index of the block $n_{2}>$

refractive index of the outside medium $n_{1}$ , then the thickness of the block is $\qquad$ cm .

Answer: 6

Numerical answer — enter this value.

$n_{1} \sin \left(90-\theta_{1}\right)=n_{2} \sin \theta_{3}$

$\mathrm{n}_{1} \cos \theta_{1}=\mathrm{n}_{2} \sin \theta_{3}$

$\mathrm{n}_{1} \frac{\mathrm{n}_{2}}{2 \mathrm{n}_{1}}=\mathrm{n}_{2} \sin \theta_{3}$

$\frac{1}{2}=\sin \theta_{3}, \theta_{3}=30$

$\tan 30=\frac{\mathrm{d}}{2(\mathrm{t})}$

$\mathrm{t}=\frac{\mathrm{d} \sqrt{3}}{2}=\frac{4 \sqrt{3} \times \sqrt{3}}{2} \mathrm{~cm}=6 \mathrm{~cm}$

Answer key and solution verified before publishing.

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