The refractive indices of the crown glass for violet and red lights are`1.51`and `1.49`respectively and those of the flint glass are`1.77`and`1.73`respectively A prism of angle `6^(@)`is made of crown glass .A beam of white light is incident at a small angle on this prism.The other thin flint glass prism is combined white the crown glass prism such that te net mean deviation is`1.5^(@)`anticlockwise.
(i)Determine the angle of the flint glass prism.

(ii)A screen is placed normal to the emerging beam at a distance of`2m`from the prism combination.find the distance between red and violet spot on the screen Which is the topmost colour on screen.

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^(@)`