Radius of curvature of an equiconvex lans is `0.2 m`. Its refractive index is `1.5` . Calculate its focal length. If two such lenses are kept separated with common principal axis by a distance of `0.2 m`, what will be the effective focal length of the combination ?

Explanation

Here, `R_1 = 0.2 m = 20 cm`
`R_2 = -20 cm, mu = 1.5 , f = ? D = 0.2 m = 20 cm`,
`F = ?`
As, `(1)/(f) =(mu - 1) ((1)/(R_1) - (1)/(R_2))`
`:. (1)/(f) = (1.5 - 1) ((1)/(20) + (1)/(20)) = 0.5 xx (1)/(10) = (1)/(20)`
`:. f = 20 cm`
Now, `f_1 = f_2 = 20 cm`
As, `(1)/(F) = (1)/(f_1)+(1)/(f_2) - (d)/(f_1 f_2)`
`:. (1)/(F)=(1)/(20)+(1)/(20) - (20)/(20 xx 20) = (1)/(20), F = 20 cm`.