# a prism of refractive index n and angle a is placed in minimum deviation position. if the angle of minimum deviation is equal to the angle a, then the value of a is

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## A prism of refractive index mu and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of mu is:

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## A prism of refractive index μ and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of μ is:

**A**sin

−1 ( 2 μ )

**B**sin

−1 2 μ−1

**C**2cos

−1 ( 2 μ )

**D**2cos

−1 ( 8 μ ) Medium Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is C)

We can use the formula: μ=sin( 2 A ) sin( 2 A+D )

where D is angle of minimum deviation and A is angle of prism.

According to given condition A=D

∴ μ= sin( 2 A ) sin( 2 A+A ) μ= sin( 2 A ) sin(A) μ= sin( 2 A ) 2sin( 2 A )cos( 2 A ) μ=2cos( 2 A ) ∴ A=2cos −1 ( 2 μ

) which is the required solution.

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## A prism of refractive index n and length A is placed in minimum deviation position.

A prism of refractive index n and length A is placed in minimum deviation position. If the angle of minimum deviation is ... (d) 2sin^-1[(4-n^2/4)^½]

## A prism of refractive index n and length A is placed in minimum deviation position.

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## Related questions

## A prism of refractive index m and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of m is

mu=sin((2A)/(2))/(sin((A)/(2)))A prism of refractive index m and angle A is placed in the minimum deviation position. If the angle of minimum deviation is A, then the value of A in terms of m is

Home > English > Class 12 > Physics > Chapter >

Experimental Physics

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A prism of refractive index m ...

A prism of refractive index

m m and angle A A

is placed in the minimum deviation position. If the angle of minimum deviation is

A A , then the value of A A in terms of m m is

Updated On: 27-06-2022

( 00 : 21 ) ADVERTISEMENT Text Solution Open Answer in App A sin −1 [ μ 2 ] sin-1[μ2] B sin −1 [ μ 2 −1 − − − − − √ 2 ] sin-1[μ2-12] C 2 cos −1 [ μ 2 ] 2cos-1[μ2] D cos −1 [ μ 2 ] cos-1[μ2] Answer

The correct Answer is C

Solution μ= sin( 2A 2 ) sin( A 2 ) μ=sin(2A2)sin(A2) Answer

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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अपवर्तनांक तथा A प्रिज्म कोण वाले प्रिज्म को न्यूनतम विचलन की स्थिति में रखा गया है। यदि न्यूनतम विचलन कोण A है तो A का मान

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Guys, does anyone know the answer?