Draw a ray diagram showing the image formation when a concave mirror produces a real, inverted and magnified image of an object and hence obtain the…

\[\frac{A^{\prime} B^{\prime}}{M P}=\frac{B^{\prime} F}{F P} \] or \(\displaystyle \frac{A^{\prime} B^{\prime}}{A B}=\frac{B^{\prime} F}{F P} \quad(\because M P=A B)\) In similar triangles \(\displaystyle \triangle A^{\prime} B^{\prime} P\) and \(\displaystyle \triangle A B P\) \[\frac{A^{\prime} B^{\prime}}{A B}=\frac{P B^{\prime}}{P B} \] from equation ($\displaystyle 1$) and \[\frac{B^{\prime} F}{F P}=\frac{P B^{\prime}}{P B} \] \(\displaystyle \frac{P B^{\prime}-P F}{F P}=\frac{P B^{\prime}}{P B}\) \[\frac{(-v)-(-f)}{(-f)}=\frac{(-v)}{(-u)} \] on solving we get \[\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \]