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(D) Let the width of the incident beam be $w_i$ and the width of the refracted beam be $w_r$. According to Snell's Law, $n_1 \sin i = n_2 \sin r$. Giv

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$\sqrt{3} : \sqrt{2}$

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(D) Let the width of the incident beam be $w_i$ and the width of the refracted beam be $w_r$. According to Snell's Law, $n_1 \sin i = n_2 \sin r$. Given $n_1 = 1$ (air), $n_2 = 1.414 = \sqrt{2}$, and $i = 45^o$. $1 \cdot \sin 45^o = \sqrt{2} \cdot \sin r$ implies $\frac{1}{\sqrt{2}} = \sqrt{2} \cdot \sin r$ implies $\sin r = \frac{1}{2}$ implies $r = 30^o$. From the geometry of the wavefronts, the width of the beam is related to the distance between rays by $w = d \cos 	heta$, where $d$ is the distance between the points of incidence on the interface. Thus, $w_i = d \cos i$ and $w_r = d \cos r$. The ratio of the width of the refracted beam to the incident beam is $\frac{w_r}{w_i} = \frac{d \cos r}{d \cos i} = \frac{\cos 30^o}{\cos 45^o}$. $\frac{w_r}{w_i} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{2} \cdot \sqrt{2} = \frac{\sqrt{3}}{\sqrt{2}}$. Therefore, the ratio is $\sqrt{3} : \sqrt{2}$.

$A$ light beam is incident on a denser medium whose refractive index is $1.414$ at an angle of incidence $45^o$. Find the ratio of the width of the refracted beam in the medium to the width of the incident beam in air.

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