The focal length of a biconvex lens is 20 cm and its refractive index is 1.5. If the radii of curvatures of two surfaces of lens are in the ratio 1:2, then the larger radius of curvature is (in cm)

Options

A. `0.045` m, `0.09` m
B. `0.09` m, `0.18` m
C. `0.04` m, `0.08` m
D. `0.06` m, `0.12` m

Answer

A

Explanation

According to lens maker.s formula,
`(1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))`
For biconvex lens `R_(1)=+R_(1),R_(2)=-R_(2)`
`(1)/(R_(1))-(1)/(R_(2))=(1)/(0.06xx0.5)=(100)/(3)` ...(i)
Now, `(R_(1))/(R_(2))=(1)/(2)` or `R_(2)=2R_(1)`
On substituting it in eq. (i), we get
`(3)/(2R_(1))=(100)/(3),R_(1)=(9)/(200)=0.045` m
`R_(2)=2R_(1)=2xx0.045` m = `0.09` m