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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

`41^(@)`24'

B

`80^(@)`

C

`60^(@)`

D

`82^(@)`48'

Text Solution

Verified by Experts

The correct Answer is:
D
`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.`
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