An equiconvex lens made of glass of refractive index 1.5 produces a 2 times magnified virtual image of an object. If the distance between the object and the image is 10 cm , the radius of curvature of each surface of the lens is ________ cm .

Step-by-step solution

For a virtual image, the magnification is positive, so m = +2 . Since m = v u , we have v = 2u . Both the object and the virtual image lie on the same side of the lens. The distance between them is given by: |v| - |u| = 10 cm |2u| - |u| = 10 |u| = 10 cm Applying the sign convention for a real object, u = -10 cm and v = -20 cm . Using the lens formula to find the focal length f : 1 f = 1 v - 1 u 1 f = 1 -20 - 1 -10 1 f = -1 + 2 20 = 1 20 f = 20 cm Now, applying the Lens Maker's formula for an equiconvex lens ( R₁ =