(i) If `f = + 0.5 m`, what is the power of the lens ?
(ii) The radii of curvature of the faces of a double convex lens are `9 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 the focal length in water ? (Refractive index of air-water `= 1.33`, refractive index of air-glass `= 1.5`).

Explanation

(i) `P = (1)/(f) = (1)/(0.5) = + 2 D`
(ii) Here, `R_1 = 9 cm, R_2 = -15 cm`,
`f = + 12 cm, mu = ?`
From lens maker's formula,
`(1)/(f)=(mu - 1)((1)/(R_1) - (1)/(R_2))`
`(1)/(12) = (mu - 1)((1)/(9) + (1)/(15)) = (mu - 1) xx (8)/(45)`
`mu - 1 = (45)/(12 xx 8) = (15)/(32)`
`mu = 1 + (15)/(32) = (47)/(32) = 1.5`
(iii) Here, `f_1 = + 20 cm, f_w = ?`
`mu_w = 1.33, mu_g = 1.5`
As `(1)/(f_a) = ((mu_g)/(mu_a) -1)((1)/(R_1) - (1)/(R_2))`
`:. (1)/(20)=((1.5)/(1) -1)((1)/(R_1) - (1)/(R_2))`
`(1)/(R_1)-(1)/(R_2)=(1)/(10)`
Again as `(1)/(f_w) = ((mu_g)/(mu_w) - 1)((1)/(R_1) -(1)/(R_2))`
`:. (1)/(f_w) = ((1.5)/(1.33) - 1) xx (1)/(10) = (0.17)/(1.33) xx (1)/(10)`
`f_w = (1.33 xx 10)/(0.17) = 78.23 cm`.