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 refractive index of air with respect to water in terms of distances ' \( w \) ' and ' \( z \) ' is (AD = w, CB = z)
Show Hint
- Refractive index = ratio of velocities
- Use geometry of wavefront (triangle relations)
Concept:
From wavefront construction and Snell’s law:
\[
\frac{v_{\text{air}}}{v_{\text{water}}} = \frac{\sin i}{\sin r}
\]
Also, refractive index of air with respect to water:
\[
n_{\text{air/water}} = \frac{v_{\text{air}}}{v_{\text{water}}}
\]
Step 1: Relation from geometry
From the diagram:
\[
\sin i = \frac{x}{z}, \quad \sin r = \frac{y}{w}
\]
Step 2: Apply Snell’s law
\[
\frac{\sin i}{\sin r} = \frac{x/z}{y/w}
= \frac{xw}{yz}
\]
But from wavefront construction:
\[
\frac{x}{y} = 1
\]
Step 3: Simplify
\[
n_{\text{air/water}} = \frac{w}{z}
\]
Since refractive index asked is air with respect to water:
\[
\Rightarrow \frac{z}{w}
\]
