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 of prism is ...... `(sin 48^@36. = 0.75)`

Options

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

Answer

B, D

Explanation

`eta=(sin((A+delta_m)/(2)))/(sin""A/2)=(sinA)/(sin""A/2)[because delta_m=A]`
`1.5=(2sin""A/2os""A/2)/(sin""A/2)`
`0.75=cos""A/2`
`0.75=sin(90^@-A/2)`
`therefore 48^@36.=90^@-A/2`
`therefore A/2=90^@-48^@36.`
`therefore A/2=41^@24.`
`therefore A=82^@48.`