A biconvex lens of refractive index 1.5 and a biconcave lens of refractive index 1.2 are placed in contact. All the curved surfaces of both lenses have the same radius of curvature of 30 cm . If this entire combination is completely immersed in a transparent liquid of refractive index 1.8 , the equivalent focal length of the combination will be:

Step-by-step solution

For the biconvex lens, the radii of curvature are R₁ = +30 cm and R₂ = -30 cm . The focal length f₁ of the biconvex lens in the liquid of refractive index 1.8 is given by: 1 f₁ = ( ₁ _m - 1 ) ( 1 R₁ - 1 R₂ ) 1 f₁ = ( 1.5 1.8 - 1 ) ( 1 30 - 1 -30 ) = ( 5 6 - 1 ) ( 2 30 ) = (- 1 6 ) ( 1 15 ) = - 1 90 cm ⁻¹ For the biconcave lens, the radii of curvature are R₁ = -30 cm and R₂ = +30 cm . The focal length f₂ of the biconcave lens in the liquid is: 1 f₂ = ( ₂ _m - 1 ) ( 1 R₁ - 1 R₂ ) 1 f₂ = ( 1.2 1.8 - 1 ) ( 1 -30 - 1 30