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Asked in GUJCET 2024 · Deviation and minimum deviation
Given: a thin prism with A = 4° and n = 1.6.
Idea: for a thin prism the general minimum-deviation relation n = (sin ((A+δₘ)/2))/(sin (A/2)) simplifies, because for small angles sin θ ≈ θ, to n = (A+δₘ)/A.
Rearranged, δₘ = A(n-1).
δₘ = 4°×(1.6-1) = 4°× 0.6
δₘ = 2.4°.
For a thin prism this small deviation is the same for every angle of incidence, which is why no angle of incidence is quoted.
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