Question 225: Ray Optics

Question

Two light beams fall on a transparent material block at point 1 and 2 with angle $	heta_1$ and $	heta_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 $	heta_1=	heta_2=\cos ^{-1}\left(\frac{n_2}{2 n_1}ight)$. 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 .

Accepted Answer

$$\begin{aligned}
& \mathrm{n}_1 \sin \left(90-	heta_1ight)=\mathrm{n}_2 \sin 	heta_3 \
& \mathrm{n}_1 \cos 	heta_1=\mathrm{n}_2 \sin 	heta_3 \
& \mathrm{n}_1 \frac{\mathrm{n}_2}{2 \mathrm{n}_1}=\mathrm{n}_2 \sin 	heta_3 \
& \frac{1}{2}=\sin 	heta_3, 	heta_3=30 \
& 	an 30=\frac{\mathrm{d}}{2(\mathrm{t})} \
& \mathrm{t}=\frac{\mathrm{d} \sqrt{3}}{2}=\frac{4 \sqrt{3} 	imes \sqrt{3}}{2} \mathrm{~cm}=6 \mathrm{~cm}
\end{aligned}$$

Ray Optics

Integer / Numeric Answer Type