For a thin equiconvex lens, a graph of linear magnification ( m ) versus image distance ( v ) is plotted. The graph is found to be a straight line which passes through the coordinates v = 30 cm and m = -0.5 . If the radius of curvature of both the surfaces of the lens is 20 cm , the value of 10 is _____ , where is the refractive index of the material of the lens.

Step-by-step solution

Using the lens formula: 1 v - 1 u = 1 f Multiplying both sides by v : 1 - v u = v f Since linear magnification m = v u , we get: m = 1 - v f Given that the graph passes through v = 30 cm and m = -0.5 : -0.5 = 1 - 30 f 30 f = 1.5 f = 20 cm Now, using the Lens Maker's formula for an equiconvex lens ( R₁ = R , R₂ = -R ): 1 f = ( - 1) ( 1 R₁ - 1 R₂ ) = ( - 1) ( 2 R ) Substituting the known values ( f = 20 cm , R = 20 cm ): 1 20 = ( - 1) ( 2 20 ) - 1 = 0.5 = 1.5 Therefore, 10 = 10 1.5 = 15 . Answer: 15