A ray of light enters a glass slab of thickness $t=10\text{ cm}$ and refractive index $\mu =1.5$ at an angle of incidence $i=60^{\circ }$. Find the lateral shift $s$.

Solution

1. Given the thickness $t=10\text{ cm}$, refractive index $\mu =1.5$, and angle of incidence $i=60^{\circ }$, use Snell's law $\sin i=\mu \sin r$ to find the angle of refraction $r$: $$\begin{array}{rl}\sin r& =\frac{\sin 60^{\circ }}{1.5}\\ & =\frac{\sqrt{3}/2}{3/2}\\ & =\frac{1}{\sqrt{3}}\end{array}$$.
2. Calculate $\cos r$ using the identity $$\begin{array}{rl}\cos r& =\sqrt{1-{\sin }^{2}r}\\ & =\sqrt{1-1/3}\\ & =\sqrt{2/3}\end{array}$$.
3. Apply the lateral shift formula $$\begin{array}{rl}s& =\frac{t\sin (i-r)}{\cos r}\\ & =t(\sin i-\cos i\tan r)\end{array}$$.
4. Substitute the values: $$\begin{array}{rl}s& =10\times (\sin 60^{\circ }-\cos 60^{\circ }\times \frac{\sin r}{\cos r})\\ & =10\times (\frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{1}{\sqrt{2}})\\ & =10\times (0.866-0.353)\approx 6.4\text{ cm}\end{array}$$.
5. Hence the answer is (B).