A plane wavefront of width x , is incident on an air-water interface and the corresponding refracted wavefront has a width y as shown in figure. The refreactive index of air with respect to water in t

Question

A plane wavefront of width x , is incident on an air-water interface and the corresponding refracted wavefront has a width y as shown in figure. The refreactive index of air with respect to water in terms of distances w and z is

TestHub · Accepted Answer

Let the length of the interface between the incident and refracted wavefronts be $L$ .

From the geometry of the given figure, the angle of incidence $i$ is the angle between the incident wavefront and the interface. Similarly, the angle of refraction $r$ is the angle between the refracted wavefront and the interface.

In the right-angled triangle in air, we have:
$sin i = \frac{z}{L}$

In the right-angled triangle in water, we have:
$sin r = \frac{w}{L}$

According to Snell's law:
$μ_{a} sin i = μ_{w} sin r$

Substituting the values of $sin i$ and $sin r$ :
$μ_{a} (\frac{z}{L}) = μ_{w} (\frac{w}{L})$

$μ_{a} z = μ_{w} w$

The refractive index of air with respect to water is given by the ratio of the refractive index of air to the refractive index of water:
$μ_{a w} = \frac{μ _{a}}{μ _{w}} = \frac{w}{z}$

Answer: $\frac{w}{z}$

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