Question

An observer whose least distance of distinct vision is $d$ , views his own face in a convex mirror of radius of curvature $r$ . The magnification produced can not exceed?

• A. $\frac{\text{r}}{\text{d}+\sqrt{{\text{r}}^2+{\text{d}}^2}}$
• B. $\frac{\text{r}}{\text{d}+\sqrt{{\text{d}}^2-{\text{r}}^2}}$
• C. $\frac{\text{r}}{\text{d}-\sqrt{\text{r}+\text{d}}}$
• D. $\frac{\text{r}}{\text{d}+\sqrt{\text{d}+\text{r}}}$

Answer 1

Answer: (A)

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 $\Rightarrow$ 2x² + 2(r - d)x - rd = 0
$\text{2}{\left(\text{x}\right)}^2+2\left(\text{r}-\text{d}\right)\text{x}-\text{r}\text{d}=0\Rightarrow \text{x}=\frac{-2\left(\text{r}-\text{d}\right)\pm \sqrt{\text{4}{\left(\text{r}-\text{d}\right)}^2+8\text{r}\text{d}}}{\text{4}}$
$\text{x}=\frac{2\left(\text{d}-\text{r}\right)\pm \sqrt{\text{4}\left({\left(\text{r}\right)}^2+{\left(\text{d}\right)}^2\right)}}{\text{4}}\Rightarrow \text{x}=\frac{\left(\text{d}-\text{r}\right)\pm \sqrt{\left({\left(\text{r}\right)}^2+{\left(\text{d}\right)}^2\right)}}{\text{2}}$
$2\text{x}=\left(\text{d}-\text{r}\right)\pm \sqrt{{\left(\text{r}\right)}^2+{\left(\text{d}\right)}^2}$
putting the value of x in equation (1)
$\text{m}=\frac{\text{r}}{\text{r}+[\left(\text{d}-\text{r}\right)\pm \sqrt{{\left(\text{r}\right)}^2+{\left(\text{d}\right)}^2}]}$
For m to be maximum, x should be minimum
${\text{m}}_{\text{min}}=\frac{\text{r}}{\text{d}+\sqrt{{\text{r}}^2+{\text{d}}^2}}$