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Angle of minimum deviation for a prism o...

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)`

A

`21^(@)`

B

`42^(@)`

C

`60^(@)`

D

`82^(@)`

Text Solution

Verified by Experts

The correct Answer is:
D
(d) Refractive index of prism, µ`=(sin""((A+delta_(m))/2))/(sin(A/2))`
substituting `A=delta_(m)and µ=1.5,`
`1.5=(sinA)/(sin((A/2)))`
`(1.5)/2=cos""A/2`
`7.5=cos""A/2`
We get, A=`82^(@)`.
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