An object and it’s image by a mirror of focal length ‘f’ have distances from the focus in the ratio 1 : 4, the object distance is

$\mathrm{u}=\mathrm{f}+x$

$\begin{array}{r}x=\mathrm{u}-\mathrm{f}\\ \mathrm{v}=\mathrm{f}+4x\end{array}$

$4x=\mathrm{v}-\mathrm{f}$

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

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

$\mathrm{f}=\frac{(\mathrm{f}+x)(\mathrm{f}+4x)}{\mathrm{f}+x+\mathrm{f}+4x}$

$f=\frac{f^2+4xf+xf+4x^2}{2f+5x}$

$\mathrm{f}(2\mathrm{f}+5x)={\mathrm{f}}^2+4x\mathbf{f}+x\mathbf{f}+4x^2$

$2{\mathrm{f}}^2+5x\mathrm{f}={\mathrm{f}}^2+5x\mathrm{f}+4x^2$

$\begin{array}{r}2f^2-f^2=4x^2\\ f^2=4x^2\end{array}$

$x^2=\frac{{\mathrm{f}}^2}{4}$

$x=\frac{f}{2}$

$\therefore \mathrm{u}=\mathrm{f}+x$

$=f+\frac{f}{2}$

$\mathrm{u}=\frac{2\mathrm{f}+\mathrm{f}}{2}$

$\therefore \mathrm{u}=\frac{3\mathrm{f}}{2}$