Question

The two surfaces of a biconvex lens are of radius of curvature \( R \) each. Obtain the condition under which its focal length \( f \) is equal to \( R \). If one of the two surfaces of this lens is made plane, what will be the new focal length of the lens?

Hint:

For a biconvex lens with equal radii of curvature, the focal length depends on the refractive index of the material. When one surface is made plane, the focal length doubles.

Answer 1

For a lens with two spherical surfaces, the focal length \( f \) is related to the radii of curvature \( R_1 \) and \( R_2 \) of the surfaces by the lens maker’s formula: \[ \frac{1}{f} = \left( \mu - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where: - \( \mu \) is the refractive index of the material of the lens, - \( R_1 \) is the radius of curvature of the first surface, - \( R_2 \) is the radius of curvature of the second surface. For a biconvex lens, both surfaces have the same radius of curvature, \( R_1 = +R \) and \( R_2 = -R \). Substituting these values into the lens maker’s formula: \[ \frac{1}{f} = \left( \mu - 1 \right) \left( \frac{1}{R} - \frac{1}{-R} \right) \] \[ \frac{1}{f} = \left( \mu - 1 \right) \left( \frac{2}{R} \right) \] \[ f = \frac{R}{2(\mu - 1)} \] Now, to find the condition under which the focal length \( f \) is equal to \( R \), set \( f = R \): \[ R = \frac{R}{2(\mu - 1)} \] Solving for \( \mu \): \[ 2(\mu - 1) = 1 \] \[ \mu - 1 = \frac{1}{2} \] \[ \mu = \frac{3}{2} \] Thus, for the focal length to be equal to the radius of curvature, the refractive index of the material must be \( \mu = \frac{3}{2} \). Next, when one of the surfaces of the biconvex lens is made plane, the radius of curvature of that surface becomes infinite, i.e., \( R_2 = \infty \). Using the lens maker’s formula again with \( R_1 = +R \) and \( R_2 = \infty \): \[ \frac{1}{f} = \left( \mu - 1 \right) \left( \frac{1}{R} - \frac{1}{\infty} \right) \] \[ \frac{1}{f} = \left( \mu - 1 \right) \frac{1}{R} \] \[ f = \frac{R}{\mu - 1} \] Substitute \( \mu = \frac{3}{2} \) into this equation: \[ f = \frac{R}{\frac{3}{2} - 1} \] \[ f = \frac{R}{\frac{1}{2}} = 2R \] Thus, when one surface is made plane, the new focal length of the lens will be \( 2R \).