Radius of curvature of an equiconvex lens 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 ?

Radius of curvature of an equiconvex lens is `0.2 m`. Its refractive i

1.	Radius of curvature of an equiconvex lens 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 ?
Answer» Here, `R_(1) = 0.2 m = 20 cm`, `R_(2) = -20 cm, mu = 1.5` , `f = ? D = 0.2 m = 20cm, 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) = (10)/(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`

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Radius of curvature of an equiconvex lens 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 ?

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