A plane wavefront is incident on a water surface at an angle of incidence $60^{\circ}$ then it gets…
A plane wavefront is incident on a water surface at an angle of incidence $60^{\circ}$ then it gets refracted at $45^{\circ} .$ The ratio of width of incident wavefront to that of refracted
wavefront will be $\left[\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin 60^{\circ}=\frac{\sqrt{3}}{2}, \cos 60^{\circ}=\frac{1}{2}\right]$
$\frac{\sqrt{3}}{2}$
$2 \sqrt{3}$
$\frac{1}{\sqrt{2}}$
$\sqrt{2}$
Solution
In the case of a plane wavefront incident on a water surface at an angle of incidence, the refracted wavefront will experience a change in width. This change is due to the change in the speed of light as it moves from air (or another medium) into water.
When the wavefront refracts, the width of the wavefront is altered by the refractive index of the medium. The relationship between the incident and refracted wavefront widths is given by the ratio of the speed of light in the two media (related to their refractive indices). The ratio is the inverse of the refractive index of the second medium (water, in this case).
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