Dispersion in a window pane. In the Figure (a), a beam of white light is incident at angle$\mathit{\theta }\mathbf{=}\mathbf{50}\mathbf{°}$on a common windowpane (shown in cross section). For the pane’s type of glass, the index of refraction for visible light ranges from$\mathbf{1}\mathbf{.524}$at the blue end of the spectrum to $\mathbf{1}\mathbf{.509}$at the red end. The two sides of the pane are parallel. (a) What is the angular spread of the colors in the beam when the light enters the pane and(b) What is the angular spread of the colors in the beam when it emerges from the opposite side? (Hint: When you look at an object through a window pane, are the colors in the light from the object dispersed as shown in Figure (b)?)

a.

b.

1. Angle of incidence is,${\theta }_1={50}^0$
2. Index of refraction for blue is$n_{2b}=1.524$
3. Index of refraction for red is $n_{2r}=1.509$

We can use the Snell’s law for blue end of the spectrum and redend of the spectrumto find the angular spread of the colors in the beam when the light enters the pane and when it emerges from the opposite side.

Formula:

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

According to Snell’s law, for the blue end of the spectrum.

$\begin{array}{c}{n}_{air}\sin {\theta }_1=n_{2b}\sin {\theta }_{2b}\\ {\theta }_{2b}={\sin }^{-1}\left(\frac{n_{air}\sin {\theta }_1}{n_{2b}}\right)\\ {\theta }_{2b}={\sin }^{-1}\left(\frac{\left(1.0\right)\sin {50}^0}{1.524}\right)\\ {\theta }_{2b}={30.176}^0\end{array}$

Now, according to Snell’s law forth red end of the spectrum,

$\begin{array}{c}{n}_{air}\sin {\theta }_1=n_{2r}\sin {\theta }_{2r}\\ {\theta }_{2r}={\sin }^{-1}\left(\frac{n_{air}\sin {\theta }_1}{n_{2r}}\right)\\ {\theta }_{2r}={\sin }^{-1}\left(\frac{\left(1.0\right)\sin {50}^0}{1.509}\right)\\ {\theta }_{2r}={30.507}^0\end{array}$

Therefore, the shift in the angle is

$\Delta \theta ={30.507}^0-{30.176}^0$

$\Delta \theta ={0.331}^0~{0.33}^0$

Therefore, the angular spread of the colors in the beam when the light enters the pane is ${0.33}^0$

The angle of incidence and angle of emergence for both rays are same.Hence,$\Delta \theta =0^0$