An equiconvex lens of refractive index 1.5 has a radius of curvature of 30 cm for both of its surfaces. If one of the curved surfaces is silvered, what is the effective focal length of this arrangement?

Step-by-step solution

The silvered lens acts as an equivalent mirror. Light passes through the lens, reflects off the silvered back surface, and passes through the lens again. The equivalent power is given by P_ eq = 2P_l + P_m . First, calculate the focal length of the unsilvered lens ( f_l ): 1 f_l = ( - 1) ( 1 R₁ - 1 R₂ ) = (1.5 - 1) ( 1 30 - 1 -30 ) = 0.5 2 30 = 1 30 cm ⁻¹ The power of the lens is P_l = 1 f_l = 1 30 cm ⁻¹ . The silvered back surface acts as a concave mirror to the light inside the lens. Its radius of curvature is R