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If minimum deviation is deltam, for pris...

If minimum deviation is `delta_m`, for prism then refractive index of material of prism is ......

A

`mu=(sin(A+deltam))/((2)/(sinA//2))`

B

`mu=(sin((A+deltam)/(2)))/(sinA//2)`

C

`mu=(2sin(A+deltam))/(sinA)`

D

`mu=(sin(A+deltam))/(sinA)`

Text Solution

Verified by Experts

The correct Answer is:
B

`mu=(sin((A+deltam)/(2)))/(sinA//2)`
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Knowledge Check

  • Minimum deviation of prism having refractive index mu and small angle of prism A is shown by .....

    A
    `delta_m=(mu-1)A`
    B
    `delta_m=A(mu+1)`
    C
    `delta=(sin((A+delta_m)/(2)))/(sintheta/2)`
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  • 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)

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