Question

Ray incidenting on glass ($\mu=1.5$) at angle $sin^{-1}\frac{3}{4}$ is rotated by $\sqrt{21}$ degrees. The refracted ray rotates by

• A. ($\frac{3}{2})^\circ$
• B. ($\frac{7}{3})^\circ$
• C. ($\frac{\sqrt{7}}{2})^\circ$
• D. $2^\circ$

Answer 1

Answer: (B)

($\frac{7}{3})^\circ$

Answer 2

Snell's Law: $\sin i = \mu \sin r$. Differentiating with respect to $i$: $\cos i \, di = \mu \cos r \, dr$. Thus, $\frac{dr}{di} = \frac{\cos i}{\mu \cos r}$. For small changes, $\Delta r \approx \frac{dr}{di} \Delta i$. Given: $\mu = 1.5 = \frac{3}{2}$, initial $i_1 = \sin^{-1}\frac{3}{4}$. $\sin i_1 = \frac{3}{4}$, $\cos i_1 = \sqrt{1 - (\frac{3}{4})^2} = \frac{\sqrt{7}}{4}$. From Snell's Law: $\frac{3}{4} = \frac{3}{2} \sin r_1 \implies \sin r_1 = \frac{1}{2}$, so $r_1 = 30^\circ$. $\cos r_1 = \frac{\sqrt{3}}{2}$. $\frac{dr}{di}\bigg|_{i_1, r_1} = \frac{\cos i_1}{\mu \cos r_1} = \frac{\sqrt{7}/4}{(3/2) \cdot (\sqrt{3}/2)} = \frac{\sqrt{7}}{3\sqrt{3}}$. Given $\Delta i = \sqrt{21}^\circ$. $\Delta r \approx \frac{\sqrt{7}}{3\sqrt{3}} \cdot \sqrt{21}^\circ = \frac{\sqrt{7}}{3\sqrt{3}} \cdot \sqrt{3}\sqrt{7}^\circ = \frac{7}{3}^\circ$.