A short linear pin of length 2.0 mm lies along the principal axis of a convex spherical boundary separating air (refractive index 1 ) and glass (refractive index 1.5 ). The radius of curvature of the spherical boundary is 10 cm . If the center of the pin is at a distance of 30 cm from the pole in air, the length of the image formed inside the glass is:

Step-by-step solution

Given: Refractive index of air, n₁ = 1 Refractive index of glass, n₂ = 1.5 Radius of curvature, R = +10 cm Object distance, u = -30 cm Length of the object, du = 2.0 mm Using the formula for refraction at a single spherical surface: n₂ v - n₁ u = n₂ - n₁ R Substituting the values: 1.5 v - 1 -30 = 1.5 - 1 10 1.5 v + 1 30 = 0.5 10 = 1.5 30 1.5 v = 1.5 30 - 1 30 = 0.5 30 v = 90 cm For a short linear object along the principal axis, the longitudinal magnification m_L is obtained by differentiating the refraction formul