Physics - Geometrical Optics Question with Solution | TestHub

Magnification, ​$\text{m}=\frac{\text{f}}{\text{f}-\text{u}}\Rightarrow \text{m}=\frac{\text{f}}{\text{f}+\text{x}}$

$\text{m}=\frac{\text{r}/2}{\text{r}/2+\text{x}}\Rightarrow \text{m}=\frac{\text{r}}{\text{r}+2\text{x}}\text{-------}\left(\text{1}\right)$

from mirror formula

$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}\Rightarrow \frac{\text{1}}{(\text{d}-\text{x})}+\frac{1}{-\text{x}}=\frac{1}{\text{f}}\Rightarrow \frac{1}{\text{d}-\text{x}}-\frac{1}{\text{x}}=\frac{2}{\text{r}}$

$\frac{+\text{x}-(\text{d}-\text{x})}{\text{x}(\text{d}-\text{x})}=\frac{2}{\text{r}}\Rightarrow \left(2\text{x}-\text{d}\right)\text{r}=2\text{x}\left(\text{d}-\text{x}\right)$

2rx - rd = 2xd - 2x²$\mathrm{\Rightarrow }$2x²+ 2(r - d)x - rd = 0

$\text{2}{\text{x}}^2+2\left(\text{r}-\text{d}\right)\text{x}-\text{r}\text{d}=0\Rightarrow \text{x}=\frac{-2\left(\text{r}-\text{d}\right)\pm \sqrt{\text{4}{\left(\text{r}-\text{d}\right)}^2+8\text{r}\text{d}}}{\text{4}}$

$\text{x}=\frac{2\left(\text{d}-\text{r}\right)\pm \sqrt{\text{4}\left({\text{r}}^2+{\text{d}}^2\right)}}{\text{4}}\Rightarrow \text{x}=\frac{\left(\text{d}-\text{r}\right)\pm \sqrt{\left({\text{r}}^2+{\text{d}}^2\right)}}{\text{2}}$

$2\text{x}=\left(\text{d}-\text{r}\right)\pm \sqrt{{\text{r}}^2+{\text{d}}^2}$

putting the value of x in equation (1)

$\text{m}=\frac{\text{r}}{\text{r}+[\left(\text{d}-\text{r}\right)\pm \sqrt{{\text{r}}^2+{\text{d}}^2}]}$

For m to be maximum, x should be minimum

${\text{m}}_{\text{min}}=\frac{\text{r}}{\text{d}+\sqrt{{\text{r}}^2+{\text{d}}^2}}$