Physics - Geometrical Optics Question with Solution | TestHub

$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow \text{let}{\text{v}}_1=\text{distance of near end of image}$

$\frac{1}{{\text{v}}_1}+\frac{1}{-\text{u}}=\frac{1}{-\text{f}}\Rightarrow \frac{1}{{\text{v}}_1}=\frac{1}{\text{u}}+\frac{1}{-\text{f}}\Rightarrow \frac{1}{{\text{v}}_1}=\frac{\text{f}-\text{u}}{\text{u}\text{f}}$

${\text{v}}_1=\frac{\text{u}\text{f}}{\text{f}-\text{u}}$

$\Rightarrow \frac{1}{{\text{v}}_2}+\frac{1}{\infty }=\frac{1}{-\text{f}}\Rightarrow {\text{v}}_2=-\text{f}$(${\mathrm{v}}_2=$the distance of last end of image )

Length of Image$\Rightarrow \left|{\text{v}}_1-{\text{v}}_{\text{2}}\right|=\frac{\text{u}\text{f}}{\text{f}-\text{u}}-\left(-\text{f}\right)\Rightarrow \left|{\text{v}}_1-{\text{v}}_2\right|=\frac{\text{u}\text{f}-\text{u}\text{f}+{\text{f}}^2}{\text{u}-\text{f}}$

$\left|{\text{v}}_1-{\text{v}}_{\text{2}}\right|=\frac{{\text{f}}^2}{\text{u}-\text{f}}$