JEE MAIN - Physics (2025 - 29th January Morning Shift - No. 20)
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, $\mathrm{d}=4 \sqrt{3} \mathrm{~cm}$ 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 $\mathrm{n}_1$, then the thickness of the block is ______ cm .
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Explanation
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$$\begin{aligned} & \mathrm{n}_1 \sin \left(90-\theta_1\right)=\mathrm{n}_2 \sin \theta_3 \\ & \mathrm{n}_1 \cos \theta_1=\mathrm{n}_2 \sin \theta_3 \\ & \mathrm{n}_1 \frac{\mathrm{n}_2}{2 \mathrm{n}_1}=\mathrm{n}_2 \sin \theta_3 \\ & \frac{1}{2}=\sin \theta_3, \theta_3=30 \\ & \tan 30=\frac{\mathrm{d}}{2(\mathrm{t})} \\ & \mathrm{t}=\frac{\mathrm{d} \sqrt{3}}{2}=\frac{4 \sqrt{3} \times \sqrt{3}}{2} \mathrm{~cm}=6 \mathrm{~cm} \end{aligned}$$
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