Question

A beam of white light enters a transparent material. Wavelengths for which the index of refraction is n are refracted at angle ${\theta }_2$. Wavelengths for which the index of refraction is $n+\delta n$, where $\delta n<<n$ , are refracted at angle ${\theta }_2+\delta \theta$.

a). Show that the angular separation of the two wavelengths, in radians, is $\delta \theta =-\left(\frac{\delta n}{n}\right)\tan {\theta }_2$

b). A beam of white light is incident on a piece of glass at 30.0°. Deep violet light is refracted 0.28° more than deep red light. The index of refraction for deep red light is known to be 1.552. What is the index of refraction for deep violet light?

Short answer

a. The statement is proven below.

b. The index of refraction for deep violet light is $n_{violet}=1.574$.

Step-by-step solution

Part (a) Step 1: Given information

We need to show that the angular separation of the two wavelength in radian is $\delta \theta =-\left(\delta n/n\right)\tan {\theta }_2$.

Part (a) step 2: simplify

From Snell's law ,we have:$$\begin{gathered} \sin {\theta }_1=n\sin {\theta }_{2}and\sin {\theta }_1 \\ =(n+\delta n)\sin ({\theta }_2+\delta \theta ) \\ Now, \\ n\sin {\theta }_2=(n+\delta n)(\sin {\theta }_2\cos \delta \theta +\cos {\theta }_2\sin \delta \theta ) \\ n\sin {\theta }_2=(n+\delta n)(\sin {\theta }_2+\cos {\theta }_2\delta n) \\ n\sin {\theta }_2=n\sin {\theta }_2+n\cos {\theta }_2\delta \theta +\delta n\sin {\theta }_2+\delta n\delta \theta \cos {\theta }_2 \\ n\cos {\theta }_2\delta \theta =-\delta n\sin {\theta }_2 \\ \delta \theta =-\left(\frac{\delta n}{n}\right)\tan {\theta }_2. \end{gathered}$$

Here, $\delta \theta$is the angular separation,${\theta }_2$is refractive angle,$n$is the refractive index.

Part (b) step 1: Given information

We have given that:

Angle of deviation $\delta \theta =-0.{28}^o$,

refractive index of red light $n_{red}=1.552$

and incident angle${\theta }_1={30}^o$.

We need to find the index of refraction for deep violet light.

Part (b) Step 2: simplify

From the formula, for calculating the value for ${\delta }_n$:

$$\begin{gathered} \delta n=-\left(\frac{\delta n}{n}\right)\tan {\theta }_2 \\ \delta n=-\frac{n}{\tan {\theta }_2}\delta \theta \end{gathered}$$

Now, calculating $\tan {\theta }_2$:

$$\begin{gathered} n_{red}\sin {\theta }_2=n\sin 30° \\ {\theta }_2={\sin }^{-1}\left(\frac{\sin 30°}{1.552}\right) \\ {\theta }_2=18.794° \\ \tan {\theta }_2=0.3403 \end{gathered}$$

Calculating${\delta }_n$:

$$\begin{gathered} \delta n=-\left(\frac{\delta n}{n}\right)\tan {\theta }_2 \\ \delta n=-\frac{n}{\tan {\theta }_2}\delta \theta (substitutevaluesinequation.) \\ \delta n=-\frac{1.552}{0.3403}\left(-0.28°\right)\left(\frac{\pi rad}{180°}\right) \end{gathered}$$

Therefore the index of refraction for deep violet light;

$$\begin{gathered} n_{violet}=n_{red}+\delta n(substitutevaluesinequation.) \\ {n}_{violet}=1.552+0.022 \\ {n}_{violet}=1.574. \end{gathered}$$

Here, $n_{violet}$is the refractive index of violet light.