Physics - Geometrical Optics Question with Solution | TestHub

Let's say ${\mathrm{H}}_1{\mathrm{H}}_2=\mathrm{D},{\mathrm{E}}_1{\mathrm{E}}_2=\mathrm{d}$ ${\mathrm{H}}_1{\mathrm{H}}_2=$ width of persons face. ${\mathrm{E}}_1{\mathrm{E}}_2=$ separation between the eyes. Let's say ${\mathrm{M}}_1{\mathrm{M}}_2=$ Width of mirror required. Then following will be the ray diagram and calculations.

$$\begin{array}{rlrlr}& {\mathrm{H}}_1{\mathrm{M}}_1^{\prime }=\frac{{\mathrm{H}}_1{\mathrm{E}}_2}{2}\&{\mathrm{H}}_2{\mathrm{M}}_2^{\prime }=\frac{{\mathrm{H}}_2{\mathrm{E}}_1}{2}& {\mathrm{H}}_1{\mathrm{E}}_2=\mathrm{D}-\frac{1}{2}(\mathrm{D}-\mathrm{d})=\frac{\mathrm{D}+\mathrm{d}}{2}={\mathrm{H}}_2{\mathrm{E}}_1& {\mathrm{M}}_1^{\prime }{\mathrm{M}}_2^{\prime }=\mathrm{D}-{\mathrm{H}}_1{\mathrm{M}}_1^{\prime }-{\mathrm{H}}_2{\mathrm{M}}_2^{\prime }& =\mathrm{D}-\left(\frac{\mathrm{D}+\mathrm{d}}{2}\right)=\frac{\mathrm{D}-\mathrm{d}}{2}\end{array}$$