Physics - Geometrical Optics Question with Solution | TestHub

Given: ${\mathrm{f}}_0=50\mathrm{cm},{\mathrm{f}}_{\mathrm{e}}=5\mathrm{cm}$ $\mathrm{d}=25\mathrm{cm},{\mathrm{u}}_0=-200\mathrm{cm}$ Magnification $\mathrm{M}=$ ? As $\frac{1}{{\mathrm{v}}_0}-\frac{1}{{\mathrm{u}}_0}=\frac{1}{{\mathrm{f}}_0}$ $$\Rightarrow \frac{1}{{\mathrm{v}}_0}=\frac{1}{{\mathrm{f}}_0}+\frac{1}{{\mathrm{u}}_0}=\frac{1}{50}-\frac{1}{200}=\frac{4-1}{200}=\frac{3}{200}$$ or ${\mathrm{v}}_0=\frac{200}{3}\mathrm{cm}$ Now ${\mathrm{v}}_{\mathrm{e}}=\mathrm{d}=-25\mathrm{cm}$ From, $\frac{1}{v_e}-\frac{1}{u_e}=\frac{1}{f_e}$ $$\begin{array}{rl}& -\frac{1}{u_e}=\frac{1}{f_e}-\frac{1}{v_e}\\ & =\frac{1}{5}+\frac{1}{25}=\frac{6}{25}\end{array}$$ or, ${\mathrm{v}}_{\mathrm{e}}=\frac{-25}{6}\mathrm{cm}$ $$\begin{array}{rl}& \text{ Magnification }\mathrm{M}={\mathrm{M}}_0\times {\mathrm{M}}_{\mathrm{e}}\\ & =\frac{{\mathrm{v}}_0}{{\mathrm{u}}_0}\times \frac{{\mathrm{v}}_{\mathrm{e}}}{{\mathrm{u}}_{\mathrm{e}}}=\frac{-200/3}{200}\times \frac{-25}{-25/6}\\ & =-\frac{1}{3}\times 6=-2\end{array}$$