Practice question
Question
In a prism, what condition results in the minimum deviation of light?
Explanation
**Prism minimum deviation** δ_m satisfies n = sin[(A+δ_m)/2]/sin(A/2), A prism angle (degrees), n refractive index, δ_m minimum deviation. For A=60°, n=1.45, sin[(60+δ_m)/2]=1.45×sin30°=0.725, (60+δ_m)/2=46.5°, 60+δ_m=93°, δ_m=33°, illustrating n increase raises δ_m. In a prism, minimum deviation occurs when the angle of incidence equals the angle of emergence. This symmetry ensures the refracted ray inside the prism is parallel to the base, minimizing the deviation angle. Substituting values gives Angle of incidence equals angle of emergence, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.
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