A biconves lens has radil of curvature 20 cm and 15 cm each. The refractive index of the material of the lens is 1.5. What is its focal length? Will the focal length change if the lens is fipped by the side?

Explanation

For a biconvex lens, radius of curvature of the first surface is positive and that of the second surface is negative side
Give: n = 1.5. `R_(1) = 20 cm and R_(2) = - 15 cm`
Lens maker.s formula. `(1)/(f) = (n - 1)((1)/(R_(1))-(1)/(R_(2)))`
Substituting the values.
`(1)/(f)=(1.5-1)((1)/(20)-(1)/((-15)))=(0.5)((1)/(20)+(1)/(15))=(0.5)((3+4)/(60))=((1)/(2)xx(7)/(60))=(7)/(12)`
`f = (120)/(7) = 17.14 cm`
As the focal lenght is positive the lens is a converging lens.
If the lens is flipped back to fornt.
Now, `R_(1) = 15 cm and R_(2) = - 20 cm, n = 1.5`
Substituting the values.
`(1)/(f)=(1.5-1)((1)/(15)-(1)/(-20))`
`(1)/(f) = (1.5-1)((1)/(15)+(1)/(20))`
This will also result in. f = 17.14 cm
Thus, it is concluded that the focal lenght of the lens will not charge it is flipped side wise. This is true for any lens. Students can verify this for any lens