The angle of a prism is 60. If the refractive index of the material of the prism is square root 2, the angle of minimum deviation is?

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Jasmine Grover

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The angle of minimum deviation in a prism is the angle at which a ray of light passing through the prism emerges parallel to the base of the prism. It is denoted by the symbol "δm".

The angle of a prism is given as 60 degrees. The refractive index of the material of the prism is given as √2.

Using the formula for the angle of minimum deviation, which is given by:

sin[(A+δm)/2]/sin(A/2) = n

where A is the angle of the prism and n is the refractive index of the prism material.

Substituting the given values, we get:

sin[(60+δm)/2]/sin(60/2) = √2

Simplifying this equation, we get:

sin[(30+δm)/2] = 1/√2

Taking the inverse sine of both sides, we get:

(30+δm)/2 = 45

Solving for δm, we get:

δm = 30 degrees

Therefore, the angle of minimum deviation for the prism is 30 degrees.

The angle of a prism is 60∘ and its refractive index is √2. The angle of minimum deviation suffered by a ray of light in passing through it is 30 degrees.

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