Saved Bookmarks
| 1. |
A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when angle of incidence is 3/4thth of the angle of prism. Calculate the speed of light in the prism. |
|
Answer» Solution :For equilateral triangular prism A = 60° and the angle of incidence for minimum deviation STATE `i=3/4 A = 45^(@)`. As in minimum deviation position i = e `therefore D_(m) = i+e-A=2i -A = 2 xx 45^(@) - 60^(@) = 30^(@)` `therefore n=(sin(A+D_(m))/2)/(sin(A/2)) =(sin 45^(@))/(sin 30^(@)) =(1//sqrt(2))/(1//2) = sqrt(2)` But, `n=c/v`, hence, SPEED of light in the PRIM `v=c/n = (3 xx 10^(8))/sqrt(2) =2.12 xx 10^(8) ms^(-1)` |
|