Question
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 $\mathrm{n}_{2}>$ refractive index of the outside medium $n_{1}$, then the thickness of the block is
____________ cm .
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Answer

(6)
Sol.
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$\mathrm{n}_{1} \sin \left(90-\theta_{1}\right)=\mathrm{n}_{2} \sin \theta_{3}$
$\mathrm{n}_{1} \cos \theta_{1}=\mathrm{n}_{2} \sin \theta_{3}$
$\mathrm{n}_{1} \frac{\mathrm{n}_{2}}{2 \mathrm{n}_{1}}=\mathrm{n}_{2} \sin \theta_{3}$
$\frac{1}{2}=\sin \theta_{3}, \theta_{3}=30$
$\tan 30=\frac{\mathrm{d}}{2(\mathrm{t})}$
$\mathrm{t}=\frac{\mathrm{d} \sqrt{3}}{2}=\frac{4 \sqrt{3} \times \sqrt{3}}{2} \mathrm{~cm}=6 \mathrm{~cm}$

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