A convex lens of refractive index 1.5 and focal length f=18 ~cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is f . The value of is _ _ _ _ . (refractive index of water =4 / 3 )

Step-by-step solution

Using the lensmaker's equation, in air: 1 f_ air = (n - 1) ( 1 R₁ - 1 R₂ ) where (n-1) = 0.5 for glass with n = 1.5 . In water: 1 f_ water = ( n n_ water - 1 ) ( 1 R₁ - 1 R₂ ) = ( 1.5 4/3 - 1 ) ( 1 R₁ - 1 R₂ ) = 1 8 ( 1 R₁ - 1 R₂ ) Since ( 1 R₁ - 1 R₂ ) = 2 f_ air = 2 18 , We get 1 f_ water = 1 8 2 18 = 1 72 , so f_ water = 72 cm. The difference is f_ water - f_ air = 72 - 18 = 54 cm. Therefore = 54 18 = 3 .