A prism of refractive index `n_(1)` & another prism of reactive index `n_(2)` are stuck together without a gap as shown in the figure.The angle of the prisms are as shown. `n_(1)&n_(2)`depend on`lambda`,the wavelength of light according to `n_(1)=1.20+(10.8xx10^(4))/(lambda^(2)) & n_(2)=1.45+(1.80xx10^(4))/(lambda^(2))` where `lambda` is in nm.

(i)Calculate the wavelength `lambda_(0)`for which rays incident at any angle on the interface`BC`pass through without bending at that interface.
(ii) for light of wavelength `lambda_(0)`,find the angle of incidencei on face`AC`such that the deviation produced by the combination of prism is minimum.

Answer

A, B, C, D

Explanation

(a) The rays of wavelength `lamda_(0)` incident at any angle of the interface BC will pass through without bending, provided the refractive indices `n_(1)` and `n_(2)` have the same value for the wavelength `lamda_(0)`. Equating the expressions of `n_(1)` and `n_(2)`, we get
`1.20+(10.8xx10^(-4))/(lamda_0^2)=1.45+(1.80xx10^(-4))/(lamda_0^2)` ..(where `lamda_0` is in nm)
`lamda_(0)=((9.0xx10^(4))/(0.25))^((1)/(2))=600nm`
(b) For the wavelength 600nm, the combination of prism acts as a single prism shaped like an isosceles triangle (ABE). At the minimum deviation, the ray inside the prism will be parallel to the base. Hence, the angle of refraction on the face AC will be `r=30^circ`.
Now `sini=nsinr=nsin30^circ=(n)/(2)`....(1)
The value of n at 600nm is
`n=1.20+(10.8xx10^(4))/((600)^(2))=1.50` ...(2)
From (1) and (2),
the angle of incidence is `i=sin^(-1)((3)/(2))`