Practice question

**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. Shift = t ( 1 - (1/n) ) . t = 8 cm , n = 1.6 . Shift = 8 ( 1 - (1/1.6) ) = 8 ( 1 - 0.625 ) = 8 × 0.375 = 3 cm . Substituting values gives 3 cm, 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.