An equiconvex glass lens of focal length 20 cm in air and refractive index 1.5 is cut symmetrically into two identical plano-convex lenses. One of these plano-convex halves is completely immersed in a transparent liquid of refractive index 1.6 . The focal length of this plano-convex lens in the liquid is:

Step-by-step solution

For the original equiconvex lens in air, the lens maker's formula is: 1 f = ( _g - 1) ( 1 R - 1 -R ) Given f = 20 cm and _g = 1.5 : 1 20 = (1.5 - 1) ( 2 R ) = 0.5 2 R = 1 R Thus, the radius of curvature R = 20 cm . When the lens is cut into two identical plano-convex lenses, for each half, R₁ = 20 cm and R₂ = . Now, one plano-convex lens is immersed in a liquid of refractive index _l = 1.6 . The new focal length f_ new is given by: 1 f_ new = ( _g _l - 1 ) ( 1 R₁ - 1 R₂ ) 1 f_ new = ( 1.5 1.6 - 1 ) ( 1 20 - 0 ) 1 f