question_answer

The distances of an object and its real image, measured from the focus of a concave mirror, are a and b respectively. Show that \[{{f}^{2}}=ab\].

Answer:

                Here \[u=-(f+a),u=-(f+b),f=-f\]                 As           \[\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\]     \[\therefore \] \[f=\frac{uv}{u+v}\]                 or            \[-f=\frac{[-(f+a)\times [-(f+b)]}{-(f+a)-f(a+b)}\] \[=\frac{{{f}^{2}}+af+bf+ab}{-(2f+a+b)}\]                 or \[2{{f}^{2}}+af+bf={{f}^{2}}+af+bf+ab\] or \[{{f}^{2}}=ab\]