question_answer

An object is placed exactly midway between a concave mirror of radius of curvature 40 cm and a convex mirror of radius of curvature 30 cm.

The mirrors face each other and are 50 cm apart. Determine the nature and position of image formed by successive reflections first at the concave mirror and then at the convex mirror.

Answer:

        Consider the formation of image of an object AB as shown below.             (i) For concave mirror, \[{{u}_{1}}=-25cm,\] \[{{f}_{1}}=-20cm.\] From mirror formula, \[\frac{1}{{{v}_{1}}}=\frac{1}{{{f}_{1}}}-\frac{1}{{{u}_{1}}}=-\frac{1}{20}+\frac{1}{25}=\frac{-1}{100}\] \[\therefore \]      \[{{v}_{1}}=-100cm\] As, \[{{v}_{1}}\] is negative, the image A?B? is real and is formed in front of concave mirror such that \[{{p}_{1}}B'=100cm.\] (ii) For convex mirror, the image A?B? acts as virtual object. \[\therefore \]      \[{{u}_{2}}=+(100-50)=50cm,\] \[{{F}_{2}}=+50cm.\] Hence,  \[\frac{1}{{{v}_{2}}}=\frac{1}{{{f}_{2}}}-\frac{1}{{{u}_{2}}}=\frac{1}{15}-\frac{1}{50}=\frac{7}{150}\] \[\Rightarrow \]   \[{{v}_{2}}=+21.43cm\] As, \[{{v}_{2}}\] is positive, the final image A?B? is virtual and is formed behind the convex mirror such that             \[{{P}_{2}}B''=21.43\,cm.\]