A biconvex lens is cut into two plano-convex lenses by a plane perpendicular to its principal axis. The powers of these two pieces in air are 3 D and 5 D . If the original uncut lens is immersed in a liquid of refractive index 4 3 , what will be its power? (Given the refractive index of the lens material is 1.5 )

Step-by-step solution

Let the radii of curvature of the original biconvex lens be R₁ and R₂ . When cut by a plane perpendicular to the principal axis, the lens is divided into two plano-convex lenses with radii of curvature ( R₁ , ) and ( , R₂ ). The powers of these pieces in air are: P₁ = ( _g - 1) ( 1 R₁ ) = 3 D P₂ = ( _g - 1) ( 1 -R₂ ) = 5 D The power of the original uncut lens in air is the sum of the powers of the two pieces: P_ air = P₁ + P₂ = 3 + 5 = 8 D Using the lens maker's formula for the original lens in air: P_ air = ( _g -