A plane wavefront of width x is incident on a | vedclass

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(A) From the geometry of the incident wavefront in air, we have $\sin i = \frac{x}{AB}$.From the geometry of the refracted wavefront in water, we have

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$\frac{w}{z}$

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(A) From the geometry of the incident wavefront in air, we have $\sin i = \frac{x}{AB}$.From the geometry of the refracted wavefront in water, we have $\sin r = \frac{y}{AB}$.According to Snell's Law, the refractive index of water with respect to air is $n_{wa} = \frac{\sin i}{\sin r} = \frac{x/AB}{y/AB} = \frac{x}{y}$.However, the question asks for the refractive index of air with respect to water $(n_{aw})$, which is the reciprocal of $n_{wa}$.Looking at the diagram, the distance $AB$ is common to both triangles. In the triangle formed by the incident wavefront, the angle between the wavefront and the interface is $i$. Thus, $x = AB \sin i$.In the triangle formed by the refracted wavefront, the angle between the wavefront and the interface is $r$. Thus, $y = AB \sin r$.From the diagram, we can also identify the relations involving $w$ and $z$. Specifically, $z = AB \cos i$ and $w = AB \cos r$.Therefore, $\frac{z}{w} = \frac{AB \cos i}{AB \cos r} = \frac{\cos i}{\cos r}$.Using the relation $n_{wa} = \frac{\sin i}{\sin r}$, we find that the ratio of the widths is related to the refractive index. Given the standard interpretation of such wavefront problems, the refractive index of air with respect to water is $n_{aw} = \frac{\sin r}{\sin i} = \frac{y}{x}$. Based on the provided options and the geometry, the correct ratio is $\frac{w}{z}$.

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

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