A long cylindrical glass rod of refractive index 1.5 has a convex hemispherical end of radius of curvature 20 cm . A small transverse object is placed in air (refractive index 1 ) on the principal axis at a distance of 20 cm from the pole of the hemispherical end. The transverse magnification of the image formed by refraction at the spherical surface is:

Step-by-step solution

Given: Refractive index of air, n₁ = 1 Refractive index of glass, n₂ = 1.5 Radius of curvature, R = +20 cm (convex towards air) Object distance, u = -20 cm Using the formula for refraction at a single spherical surface: n₂ v - n₁ u = n₂ - n₁ R Substituting the values: 1.5 v - 1 -20 = 1.5 - 1 20 1.5 v + 1 20 = 0.5 20 1.5 v = 0.5 20 - 1 20 = - 0.5 20 = - 1 40 v = -60 cm The transverse magnification m for a single spherical surface is given by: m = n₁ v n₂ u m = 1 (-60) 1.5 (-20) = -60 -30 = +2 Answer: +2