Question 79 of 295

Two light beams fall on a transparent material block at point 1 and 2 with angle \(\theta_1\) and \(\theta_{2^{\prime}}\) 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 \(\theta_1=\theta_2=\cos ^{-1}\left(\frac{\mathrm{n}_2}{2 \mathrm{n}_1}\right)\), where refractive index of the block \(\mathrm{n}_2\gt\) refractive index of the outside medium \(\mathrm{n}_1\), then the thickness of the block is ________ cm.

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Answer: 6

$\begin{aligned} & n_1 \times \frac{n_2}{2 n_1}=n_2 \sin r \\ & \sin r=\frac{1}{2}\end{aligned}$
$\begin{aligned} & r=30^{\circ} \\ & \tan r=\left(\frac{d / 2}{t}\right) \\ & t=\frac{d}{2 \tan r}=\frac{d \sqrt{3}}{2}=\frac{(4 \sqrt{3}) \sqrt{3}}{2} \\ & =6 \mathrm{~cm}\end{aligned}$