An object is placed at `A(OA gt f)`. Here, f is the focal length of the lens. The image is formed at B. A perpendicular is erected at O and C is chosen such that `/_BCA = 90^(@)`. Let OA = a, OB = b and OC = c. Then the value of f is

Explanation

From `1.v -1/u =1/f` we have `1/b + 1/a=1/f`
or `f=(ab)/(a+b)to (1)`
further `AC^(2) + BC^(2) =AB^(2)`
or `(a^(2)+c^(2)) + (b^(2) + c^(2)) = (a+b)^(2)`
`ab=c^(2)`
Substituting this in Eq. (1) we get f= `c^(2)/(a+b)`