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 10cm and 15cm . Its focal length is 12cm. What is the refractive index of glass ?
iii. A convec 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

i. P = ` (1)/(f) = (1)/(0.5) = 2 (D) `
Here, we have f = + 12 cm , `R_(1) = + 10 cm R_(2) = - 15 cm `
We use the lens formula `(1)/(f) = (n - 1) [ (1)/(R_(1)) - (1)/(R_(2)) ] `
The sign convention has to be applied for f, `R_(1) and R_(2)`.
Substituting the values, we have `(1)/(12) =(n - 1) ((1)/(10) - (1)/(-15)) = (n - 1) ((1)/(10)+ (1)/(15)) = (n + 1) ((1)/(10) + (1)/(15)) therefore n = 1.5 `
iii. for a glass lens in air , `n_(2) = 1.5, n_(1), f = + 20 `cm.
The lens formula gives, `(1)/(20) = 0.5 [(1)/(R_(1)) - (1)/(R_(2)) ] " " `(1)
For the same glass lens in water, `n_(2) = 1.5 , n_(1) = 1.33.`
Therefore, `(1.33)/(f) = (1.5 - 1.33) [ (1)/(R_(1)) - (1)/(R_(1)) ]` Substituting in (1), `(133)/(f ) = (1.5 - 1.33) xx (1)/(10)`
Hence , f = +78.2 cm