Question

A prism of refractive index $n_1$ and another prism of refractive index $n_2$ are stuck together with a gap as shown in the figure. The angles of the prism are as shown. $n_1$ and $n_2$ depend on $\lambda$ the wavelength of light according to :
$n_1=1.20+\frac{10.8\times 10^4}{{\lambda }^2}and$
$n_2=1.45+\frac{1.80\times 10^4}{{\lambda }^2}$
where $\lambda$ is in nm.
(a) Calculate the wavelength ${\lambda }_0$ for which rays incident at any angle on the interface BC pass through without bending at that interface.
(b) For light of wavelength ${\lambda }_0$, find the angle of incidence i on the face AC such that the deviation produced by the combination of prisms is minimum.

Answer 1

$n_1=1.20+\frac{10.8\times 10^4}{{\lambda }_0^2}and{n}_2=1.45+\frac{1.80\times 10^4}{{\lambda }^2}$ Here, $\lambda$ is in nm (a) The incident ray will not deviate at B C if $n_1=n_2$,br> $\Rightarrow 1.20+\frac{10.8\times 10^4}{{\lambda }_0^2}and{n}_2=1.45+\frac{1.80\times 10^4}{{\lambda }_0^2}$ $\Rightarrow \frac{9\times 10^4}{{\lambda }_0^2}=0.25or{\lambda }_0=\frac{3\times 10^2}{0.5}$ $or{\lambda }_0=600nm$ (b) The given system is a part of an equilateral prism of prism angle $60^{\circ }$ as shown in figure. At minimum deviation. $r_1=r_2=\frac{60^{\circ }}{2}=30^{\circ }=r$(say) $\therefore n_1=\frac{sini}{sinr}$ $\therefore sini=n_1.sin30^{\circ }$ $sini=\{1.20+\frac{10.8\times 10^4}{(600{)}^2}\}\left(\frac{1}{2}\right)=\frac{1.5}{2}=\frac{3}{4}$ $or$$i=si{n}^{-1}$(3/4)