Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism of given prism. Then, the angle is prism is….
`(sin 48^(@)36'=0.75)`

Options

A. `41^(@)`24'
B. `80^(@)`
C. `60^(@)`
D. `82^(@)`48'

Answer

D

Explanation

`n = (sin ((A + delta_(m))/(2)))/(sin A//2) " and " delta_(m) = A `
`rArr " " n= 1 cos A//2 rArr (A)/(2) = cos^(-1) ((1.5)/(2))`
`rArr " " A = 2 xx 41.4 rArr A = 82^(@)48.`