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. Focal length: f = 18 cm . Object distance: u = -36 cm . Lens formula: (1/v) - (1/u) = (1/f) . (1/v) - (1/-36) = (1/18) ⇒ (1/v) + (1/36) = (1/18) ⇒ (1/v) = (1/18) - (1/36) = (2 - 1/36) = (1/36) . v = 36 cm (real image). Substituting values gives 36 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f