The refractive indices of the crown flass for the blue and red lights are 1.51 and 1.49, repectively, and those of the flint glass are 1.77 and 1.73, respectively. An isosceles prism of angle `6^@` is made of crown glass. A beam of white light is incident at a smalll angle on this prism. The other flint glass isosceles prism is combined with the crown glass prism such that there is no deviation of the incident light. Determing the angle of the flint glass prism. Calculate the net dispersion of the combined system.

Answer

(i) `4^(@) (ii) -0.04^(@)`

Explanation

When angle of prism is small and angle of incidence is also small , the deviation is given by the `delta = (mu- 1) A`

Net deviation by the two prism is zero .
So , `delta_(1) + delta_(2) = 0 "or" (mu_(1) - 1) A_(1) + (mu_(2) - 1)A_(2) = 0 .... (i)`
Here `mu_(1) "and" mu_(2)` are the refractive indices for crown and flint glasses respectively.
Hence , `mu_(1) = (1.51 + 1.49)/(2) = 1.5`
and `mu_(2) = (1.77 + 1.73)/(2) = 1.75`
`A_(1)` = Angle of prism for crown glass = `6^(@)`
Substituting the values in Eq.(i) we get
`(1.5 - 1) (6^(@)) + (1.75 - 1) A_(2) = 0 `
This gives `A_(2) = -4^(@)` . Hence ,angle of flint glass prism is `4^(@)` (Negative sign shows that flint glass prism is inverted with respect to the crown glass prism ).
(ii) Net dispersion due to the two prism is
=`(mu_(b_(1)) - mu_(r_(1)))A_(1) + (mu_(b_(2)) - mu_(r_(2)))A_(2)`
=`(1.51 - 1.49) (6^(@)) + (1.77 - 1.73) (-4^(@)) = -0.04^(@)`