A prism of angle 45° produces a miniμm deviation of 25°. Which value is closest to the refractive index of the prism?
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A prism of angle 45° produces a miniμm deviation of 25°. Which value is closest to the refractive index of the prism?

G1 Physics · Accepted Answer

At miniμm deviation, μ=sin[(A+δₘ)/2]/sin(A/2). Substituting A=45° and δₘ=25° gives μ≈1.499.