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.

Answer

(i) `lambda_(0)= 6000 nm, n=1.5`
(ii) `I= sin^(-1)(0.75)=48.59^(@) 32, 1.25 m`

Explanation

`n_(1)=1.20+(1.80xx10^(4))/(lambda_(0)^(2))` and `n_(2)=1.45 + (1.80xx10^(4))/(lambda_(0)^(2))`
Here `lambda` is in `nm`.

(i) The incident ray will not deviate at `BC` if
`n_(1)=n_(2)`
`rArr 1.20 + (10.8xx10^(4))/(lambda_(0)^(2))=1.45+(1.80xx10^(4))/(lambda_(0)^(2)) , (lambda=lambda_(0))`
`rArr (9xx10^(4))/(lambda_(0)^(2))=0.25`
or `lambda_(0) =(3xx10^(2))/(0.5)`
or `lambda_(0)=600 nm`
(ii) The given system is a part of an equilateral prism of prism `60^(@)` as shown in figure At minimum deviation,
`r_(1) = r_(2)=(60^(@))/(2)=30^(@)=r` (say)
`therefore n_(1) =(sin i)/(sin r)`
`therefore sin i = n_(1). sin 30^(@)`
`sin i ={1.20+(10.8xx10^(4))/((600)^(2))}((1)/(2))=(1.5)/(2) =(3)/(4) (lambda=lambda_(0)600 nm)`
or `i =sin^(-1)(3//4)`