Two light beams fall on a transparent material block at point 1 and 2 with angle θ1 and θ2, respectively, as shown in figure.

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, \(\mathrm{~d}=4 \sqrt{3} \mathrm{~cm}\) 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 \(n_{1},\) then the thickness of the block is _______cm.

Answer 1

Answer is: 6

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

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

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

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

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

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