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(C) Let the width of the incident wavefront be $w_i$ and the width of the refracted wavefront be $w_r$.From the geometry of wavefronts,the width of a

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$\frac{1}{\sqrt{2}}$

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(C) Let the width of the incident wavefront be $w_i$ and the width of the refracted wavefront be $w_r$.From the geometry of wavefronts,the width of a wavefront is related to the beam width by $w = L \cos 	heta$,where $L$ is the length of the wavefront segment on the interface.For the incident wavefront,$w_i = L \cos i$,where $i = 60^{\circ}$.For the refracted wavefront,$w_r = L \cos r$,where $r = 45^{\circ}$.The ratio of the width of the incident wavefront to the refracted wavefront is $\frac{w_i}{w_r} = \frac{\cos i}{\cos r}$.Substituting the values: $\frac{w_i}{w_r} = \frac{\cos 60^{\circ}}{\cos 45^{\circ}} = \frac{1/2}{1/\sqrt{2}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.

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