91Ó°ÊÓ

1. Angle of incidence is,${\mathrm{θ}}_1={50}^0$
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:

$n_1\sin {\mathrm{θ}}_1=n_2\sin {\mathrm{θ}}_2$

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

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

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

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

Therefore, the shift in the angle is

$\mathrm{Δ}\mathrm{θ}={30.507}^0−{30.176}^0$

$\mathrm{Δ}\mathrm{θ}={0.331}^0~{0.33}^0$

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,$\mathrm{Δ}\mathrm{θ}=0^0$