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Asked in GUJCET 2021 · Deviation and minimum deviation
Given: a prism of refractive index μ in air with δₘ=A.
Idea: at minimum deviation the ray is symmetric and μ=(sin ((A+δₘ)/2))/(sin (A/2)).
Put δₘ=A: μ=(sin A)/(sin (A/2)).
Now sin A=2 sin (A/2)cos (A/2), so the sines cancel and μ=2 cos (A/2).
Hence cos (A/2)=μ/2.
Taking the inverse cosine and doubling, A=2 cos⁻¹(μ/2).
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