Question 42: Ray Optics

Question

A transparent block A having refractive index $\mu = 1.25$ is surrounded by another medium of refractive index $\mu = 1.0$ as shown in the figure. A light ray is incident on the flat face of the block with incident angle $	heta$ as shown in the figure. What is the maximum value of $	heta$ for which light suffers total internal reflection at the top surface of the block?

Accepted Answer

$$\begin{aligned}
& \mathrm{r}+	heta_{\mathrm{C}}=90^{\circ} \
& \mu_1 \sin 	heta=\mu_2 \operatorname{sinr} \
& \sin 	heta=\frac{\mu_2}{\mu_1} \sin \left(90-	heta_{\mathrm{C}}ight) \
& \sin 	heta=\frac{\mu_2}{\mu_1} \cos 	heta_{\mathrm{C}} \
& \sin 	heta_{\mathrm{C}}=\frac{\mu_1}{\mu_2}
\end{aligned}$$
$$\begin{aligned}
& \sin 	heta=\frac{\mu_2}{\mu_1} \sqrt{1-\frac{\mu_1^2}{\mu_2^2}} \
& \sin 	heta=\sqrt{\frac{\mu_2^2-\mu_1^2}{\mu_1^2}}=\sqrt{\frac{\frac{25}{16}-1}{1}} \
& \sin 	heta=\frac{3}{4} \
& 	heta=\sin ^{-1}\left(\frac{3}{4}ight)
\end{aligned}$$

Answer

$	an^{-1}(4/3)$

Answer

$\sin^{-1}(3/4)$

Answer

$	an^{-1}(3/4)$

Answer

$\cos^{-1}(3/4)$

Ray Optics

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