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A ray of light undergoes deviation of 30∘ when incident on an equilateral prism of refractive index √2 . The angle made by the ray inside the prism with the base of the prism is |
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Answer» δ = 30°, A = 60°, μ = √2 Let us find the minimum value of angle of deviation δ that is δm \(\boxed {µ = \frac {sin [\frac {A + δm}2]} {sin (\frac {A}{2})}} \) This is the formula for minimum deviation of prism. Using this, we will find δm √2 = [sin(60° + δm) / 2] / sin(60°/2) √2 = [sin(60° + δm) / 2] / 1/2 [sin(60° + δm) / 2] = √2 × 1/2 [sin(60° + δm) / 2] = 1/√2 = sin45° (60° + δm) / 2 = 45° δm = 90 - 60 = 30° As δm = δ, Hence we conclude that the prism is in minimum deviation position. So the ray inside the prism will be parallel to the base of the prism. Therefore, the angle made by the ray inside the prism with the base of prism is 0°. |
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