An equiconvex lens made of glass of refractive index 3 2 has focal length f in air. It is completely immersed in water of refractive index 4 3 . The percentage change in the focal length is

Step-by-step solution

Given, ^a _g= 3 2 , f_ air =f ^a _w= 4 3 Using lens Maker's formula, when lens is in air, aligned & 1 f_ air = ( ^a _g-1 ) ( 1 R₁ - 1 R₂ ) 1 f & = ( 3 2 -1 ) ( 1 R₁ - 1 R₂ ) 1 f & = 1 2 ( 1 R₁ - 1 R₂ ) aligned When lens is immersed in water, then 1 f_w = ( ^w _g-1 ) ( 1 R₁ - 1 R₂ ) Dividing Eq. (i) by Eq. (ii), we get aligned f_w f & = 1 2 ( _g-1 ) f_w & = f 2 ( _g _w -1 ) = f 2 ( 3 2 4 3 -1 ) = f 1 4 =4 f f_w & =4 f aligned Percentage change in focal length aligned & = f_w-f f 100 & = 4 f-f f 100=300 % aligned