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In a concave mirror experiment, an object is placed at a distance $x_{1}$ from the principal axis focus, the image is formed at a distance $' x_{2}'$ from the principal focus. The focal length of the mirror is

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$x_{1} x_{2}$

$\sqrt{x_{1} x_{2}}$

$\sqrt{\frac{x_{1}}{x_{2}}}$

$\frac{x_{1} + x_{2}}{2}$

answer is B.

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Detailed Solution

$u = - (f + x_{1})$

$v = - (f + x_{2})$

$f = - f$

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

$- \frac{1}{f} = \frac{- 1}{f + x_{1}} - \frac{1}{f + x_{2}}$

$- \frac{1}{f} = - (\frac{f + x_{2} + f + x_{1}}{(f + x_{1}) (f + x_{2})})$

$\frac{1}{f} = \frac{2 f + x_{2} + x_{1}}{f^{2} + f x^{2} + f x_{1} + x_{1} x_{2}}$

$f^{2} + f x_{2} + f x_{1} + x_{1} x_{2} = f (2 f + x_{2} + x_{1})$ $f^{2} + f x_{2} + f x_{1} + x_{1} x_{2} = 2 f^{2} + f x_{2} + f x$ $f^{2} = x_{1} x_{2}$

$f = \sqrt{x_{1} x_{2}}$

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