Question

(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`).

Answer 1

i) Power `=+2` dioptre.
ii) Here, we have f=`+12cm`, `R_(1) = +10cm, R_(2)=-15cm`.
Refractive index of air is taken as unity.
We use the lens formula of Eq.(9.22). The sign convention hasto be applield for f, `R_(1)` and `R_(2)`.
Substituting the values, we have
`1/12=(n-1)(1/10-1/(-15))`
This gives n=1.5
iii) For a glass lens in air, `n_(2) = 1.5`, `n_(1)=1`, `f=+20cm`. Hence, the lens formula gives
`1/20 = 0.5[1/(R_(1))-1/(R_(2))]`
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_(2)]`
Combining these two equations, we find f=`+78.2cm`.