(i) If f = 0.5 m for a glass lens, what is the power of the lens? (ii) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of glass? (iii) A convex lens has 20 cm focal length in air. What is focal length in water? (Refractive index of air-water = 1.33, refractive index for air-glass = 1.5.)

Explanation

`rArr` (i) Focal lenght is positive hence lens is convex. Power of lens,
`P = 1/f = 1/0.5`
`therefore P = +2D`
(ii) Here f = + 12 cm `, R_1 = + 10 cm , R_2 =- 15 cm `
n=1
`therefore` Accroding to Lens maker.s formula,
`therefore 1/f = (n-1)[ 1/R_1 - 1/R_2]`
`therefore 1/12 = (n-1) [1/10 - 1/(-15)]`
`therefore 1/12 = (n-1) [(3+2)/(30)]`
`therefore 1/12 = (n-1) [(3+2)/(30)]`
`therefore 1/12 = (n-1) [ 1/6]`
`therefore 1/12 = n-1` ,
`therefore n=1 + 1/2 = 3/2`
(iii) Here, rafractive index of air `n_a = 1.0`
refreactive index of glass lens `n_g = 1.5`
refractive index of water `n_1 = 1.33`
`rArr` From lens maker.s formula for lens in air ,
`1/f_a = ((n_g - n_a)/(n_a))[1/R_1 - 1/R_2]` ...... (1)
From lens maker.s formula for lens in water,
`1/(f_l) = ((n_g - nl)/(nl))[1/R_1 - 1/R_2]`....... (2)
`rArr` By taking ratio of euqation (1) and (2),
`(f_l)/(f_a) = ((n_g -n_a)/(n_g-n_l) (n_l)/(n_a))`
`therefore (f_l)/(20) = ((1.5-1.0)/(1.5-1.33)) (1.33/1.0)` ,
`therefore f_l = 20 xx (0.5)/(0.17) xx 1.33`
`therefore f_l = 78.235` cm
`therefore f_l ~~ 78.2 cm `
`therefore ` Focal length of lens will be 78.2 cm when it is placed in water.