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An object is placed at A(OA gt f). Here,...

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

Text Solution

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