The focal length \( f \) of a lens with two spherical surfaces is determined by the radii of curvature \( R_1 \) and \( R_2 \) using the lens maker’s formula:\[\frac{1}{f} = \left( \mu - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]In this formula, \( \mu \) represents the refractive index of the lens material, \( R_1 \) is the radius of curvature of the first surface, and \( R_2 \) is the radius of curvature of the second surface.For a biconvex lens, where both surfaces share the same radius of curvature (with \( R_1 = +R \) and \( R_2 = -R \)), the formula simplifies upon substitution:\[\frac{1}{f} = \left( \mu - 1 \right) \left( \frac{1}{R} - \frac{1}{-R} \right)\]\[\frac{1}{f} = \left( \mu - 1 \right) \left( \frac{2}{R} \right)\]This leads to the focal length expression:\[f = \frac{R}{2(\mu - 1)}\]To determine the condition where the focal length \( f \) equals the radius of curvature \( R \), we set \( f = R \):\[R = \frac{R}{2(\mu - 1)}\]Solving for \( \mu \) yields:\[2(\mu - 1) = 1\]\[\mu - 1 = \frac{1}{2}\]\[\mu = \frac{3}{2}\]Therefore, a refractive index of \( \mu = \frac{3}{2} \) is required for the focal length to equal the radius of curvature.When one surface of the biconvex lens is made plane, its radius of curvature becomes infinite (\( R_2 = \infty \)). Applying the lens maker’s formula with \( R_1 = +R \) and \( R_2 = \infty \):\[\frac{1}{f} = \left( \mu - 1 \right) \left( \frac{1}{R} - \frac{1}{\infty} \right)\]\[\frac{1}{f} = \left( \mu - 1 \right) \frac{1}{R}\]This results in the focal length:\[f = \frac{R}{\mu - 1}\]Substituting \( \mu = \frac{3}{2} \) into this equation gives:\[f = \frac{R}{\frac{3}{2} - 1}\]\[f = \frac{R}{\frac{1}{2}} = 2R\]Consequently, if one surface is made plane, the new focal length of the lens will be \( 2R \).