An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of object is $d_{1}$ from C and the distance of the image formed is $d_{2}$ from C, then the radius of curvature of this mirror is

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

$\frac{d_{1} d_{2}}{d_{1} - d_{2}}$

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

$\frac{2 d_{1} d_{2}}{d_{1} - d_{2}}$

answer is A.

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Detailed Solution

Object distance
$\begin{array}{l} O b j e c t d i s \tan c e μ = - (R + d_{1}) \\ Im a g e d i s \tan c e v = (R - d_{2}) \\ f o c a l l e n g t h o f m i r r o r = f = \frac{- R}{2} \\ m i r r o r f o r m u l a; - \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{u + v}{u v} \end{array}$