A ray of light suffers minimum deviation while passing through a prism of refracting angle A . If the angle of incidence `i` = Refractive angle and n =` sqrt3` , find the angle of the prism .

Options

A. `45^(@)`
B. `50^(@)`
C. `55^(@)`
D. `60^(@)`

Answer

D

Explanation

Given : `i` = A and when the ray suffers minimum deviation, i=e `therefore i= e = A `
`therefore` Using i+ e = `A + delta_(m)`
`therefore i+ e - A = delta_(m) therefore delta_(m) = A + A - A = A `
When the ray suffers minimum deviation , `r = (A)/(2)`
Thus , `n = (sin ((A + delta_(m))/(2)))/(sin ((A)/(2)))`
`therefore sqrt3 = (sin ((A + A)/(2)))/(sin ((A)/(2)))= (sin [2((A)/(2))])/(sin ((A)/(2)))`
`therefore sqrt3 = ( 2 sin ((A)/(2)) cos ((A)/(2)))/( sin ((A)/(2)))`
`therefore cos ((A)/(2)) = (sqrt3)/(2) = cos 30^(@)`
`therefore (A)/(2) = 30^(@) therefore A = 60^(@)`