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A ray of light is incident at $$45^\circ$$ on to a transparent slab of thickness 10 cm made of a material of refractive index $$\sqrt{2}$$. Find the lateral displacement (in cm) of the ray as it emerges from the slab.
Correct Answer: 2.99
$$\sin r = \frac{\sin 45^\circ}{\sqrt{2}} = \frac{1}{2}$$, so $$r = 30^\circ$$. The lateral displacement is $$\frac{t \sin(i - r)}{\cos r} = \frac{10 \sin 15^\circ}{\cos 30^\circ} = \frac{10 \times 0.2588}{0.8660} \approx 2.99\text{ cm}$$.
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