Question

A plane mirror is placed at a distance 50 cm from a concave mirror of focal length 16 cm. A short object is placed at a distance x cm from the plane mirror and face both the mirrors so its virtual image in the plane mirror coincides the real image in concave mirror. Find x.

Answer 1

$$\begin{gathered} f=-16cm,v=-\left(50+x\right) \\ u=-\left(50-x\right) \\ \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \frac{1}{-\left(50+x\right)}+\frac{1}{-\left(50-x\right)}=\frac{1}{-16} \\ \frac{1}{\left(50+x\right)}+\frac{1}{\left(50-x\right)}=\frac{1}{16} \\ \frac{50-x-50-x}{2500-x^2}=\frac{1}{16} \\ 2500-x^2=-32x \\ {x}^2-32x-2500=0 \\ x=\frac{32\pm \sqrt{1024+10000}}{2}=68.5 \\ Regards \end{gathered}$$