An equiconvex lens made of glass (refractive index 1.5 ) has a power of +5 D in air. If this lens is completely immersed in water (refractive index 4 3 ), what will be its new focal length?

Step-by-step solution

The focal length of the lens in air is given by: f_a = 100 P_a = 100 5 = 20 cm Using the Lens Maker's formula in air: 1 f_a = ( _g - 1) ( 1 R₁ - 1 R₂ ) For an equiconvex lens, R₁ = R and R₂ = -R . 1 20 = (1.5 - 1) ( 2 R ) 1 20 = 0.5 2 R R = 20 cm When the lens is immersed in water, the relative refractive index of glass with respect to water is _ gw = _g _w . The new focal length f_w is: 1 f_w = ( _g _w - 1 ) ( 2 R ) 1 f_w = ( 1.5 4/3 - 1 ) ( 2 20 ) = ( 9 8 - 1 ) ( 1 10 ) 1 f_w = ( 1 8 ) 1 10 = 1 80 f_w = +80 cm An