If an object is placed at A (OA f),where f | vedclass

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(C) Using the lens formula,we have $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Here,$u = -OA$ and $v = OB$,so $\frac{1}{f} = \frac{1}{OB} - \frac{1}{-OA

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$OC^2/AB$

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(C) Using the lens formula,we have $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Here,$u = -OA$ and $v = OB$,so $\frac{1}{f} = \frac{1}{OB} - \frac{1}{-OA} = \frac{1}{OB} + \frac{1}{OA} = \frac{OA + OB}{OA \cdot OB}$.Thus,$f = \frac{(OA)(OB)}{OA + OB} = \frac{(OA)(OB)}{AB} \dots(i)$.In the right-angled triangle $	riangle BCA$,$OC$ is the altitude to the hypotenuse $AB$. By the geometric mean theorem in right triangles,$OC^2 = (OA)(OB)$.Substituting this into equation $(i)$,we get $f = \frac{OC^2}{AB}$.

If an object is placed at $A$ $(OA > f)$,where $f$ is the focal length of the lens,the image is found to be formed at $B$. $A$ perpendicular is erected at $O$ and $C$ is chosen on it such that the angle $\angle BCA$ is a right angle. Then the value of $f$ will be

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