A convex lens of refractive index 1.55, with both the surfaces of the same radius of curvature has a focal length of 20 cm. The radius of curvature of the surface will be:

20 cm 22 cm 24 cm 26 cm

22 cm

$ \frac{1}{F} = (\mu -1)(\frac{1}{R} - \frac{1}{-R}) = \frac{2(\mu -1)}{R}$ $R = 2F (\mu -1) = 40\times 0.55 cm = 22cm$