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) 600 nm (b) `sin^(-1) (3//4)`

Explanation

`n_(1) = 1.20 + (10.8 xx 10^(4))/(lambda^(2))` and
`n_(2) = 1.45 + (1.80 xx 10^(4))/(lambda^(2))` Here , `lambda` is in nm.
(a) The incident ray will not deviate at BC if `n_(1) = n_(2)`
`implies 1.20 + (10.8 xx 10^(4))/(lambda_(0)^(2))` = 1.45 + (1.80 xx 10^(4))/(lambda_(0)^(2)) (lambda = lambda_(0))`
`implies (9 xx 10^(4))/(lambda_(0)^(2)) = 0.25 implies lambda_(0) = (3 xx 10^(2))/(0.5) "or" lambda_(0) = 600nm`
(b) The given system is a part of an equiolateral prism of prism angle `60^(@)` as shown in the figure .
At minimum deviation
`r_(1) = r_(2) = (60^(@))/(2) = 30^(@)` = r (say)
`therefore n_(1) = ("sini")/("sinr")`
`therefore "sini" = n_(1)"sin"30^(@)`
`"sin i" = {1.20 + (10.8 xx 10^(4))/((600)^(2))}((1)/(2)) = (1.5)/(2) = (3)/(4)`
`(lambda = lambda_(0) = 600 "nm")`
`implies i = sin^(-1) (3//4)`