Question

Two parallel beams of light $P$ and $Q$ (separated by $d$) containing radiation of wave length $4000\mathring{A}$ and $5000\mathring{A}$ (Which are mutually conherent in each wave length separately) are incident normally an a prism as shown in the figure. The refractive index of the prism as a function of wave length is given by the relation, $n\left(\lambda \right)=1.2+\left(b/{\lambda }^2\right)$, where $\lambda$ is in $\mathring{A}$ and $b$ is a constant. The value of $b$ is such that the condition for the total internal reflection for the face $AC$ is just satisfied for one length and not satisfied for the other.
(a)Find the value of $b$
(b)Find the deviation for the beams transmitted through the face $AC$

Answer 1

(a) The refractive index $n_1$ for $\lambda =4000\mathring{A}$ is equal to
$n_1=1.2+\frac{b}{(4000{)}^2}$
The refractive index $n_2$ for $\lambda =5000\mathring{A}$ is equal to
$n_1=1.2+\frac{b}{(5000{)}^2}$
At the total internal reflection, the critical angle
$\theta ={\theta }_c={\sin }^{-1}\left(0.8\right)$
$\Rightarrow$ Therefore, the refractive index for total internal
reflection, $n=\frac{1}{\sin {\theta }_c}=\frac{10}{8}=\frac{5}{4}$
Since $n_1>n_2({\theta }_c{)}_1<({\theta }_c{)}_2$
For this Problem the wavelength one satisfies the condition of just total internal reflaction, the other wavelength gets transmitted.
$\Rightarrow n_1=\frac{5}{4}1.2+\frac{b}{(4000{)}^2}=\frac{5}{4}$
$\Rightarrow b=0.8\times {10}^6{\mathring{A}}^2$
(b) Similarly ${\Delta }_2=r_2-l_2$ where $i_2=\theta$ & $r_2$ can be given as
$\frac{\sin l_2}{\sin l_2{r}_2}=\frac{1}{n_2}$
$\Rightarrow r_2={\sin }^{-1}\left[n_2\sin \theta \right]$
Putting $n_2=1.2+\frac{b}{(5000{)}^2}=1.2+\frac{0.8\times {10}^6}{(5000{)}^2}$
$=1.232$ and $\sin \theta =0.8$
We obtain $r_2={80}^{o}16$
$\Rightarrow {\delta }_2={80}^{o}16-{\sin }^{-1}0.8\cong {27}^o$