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

Answer 1

Answer: (A)

Magnification, $m=\frac{f}{f-u}\Rightarrow m=\frac{f}{f+x}$
$m=\frac{r/2}{r/2+x}\Rightarrow m=\frac{r}{r+2x}$......(1)
from mirror formula
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{(d-x)}+\frac{1}{-x}=\frac{1}{f}\Rightarrow \frac{1}{d-x}-\frac{1}{x}=\frac{2}{r}$
$\frac{+x-(d-x)}{x(d-x)}=\frac{2}{r}\Rightarrow (2x-d)r=2x(d-x)$
$2rx-rd=2xd-2x^2\Rightarrow 2x^2+2(r-d)x-rd=0$
$2(x{)}^2+2(r-d)x-rd=0\Rightarrow x=\frac{-2(r-d)\pm \sqrt{4(r-d{)}^2+8rd}}{4}$
$x=\frac{2(d-r)\pm \sqrt{4\left((r{)}^2+(d{)}^2\right)}}{4}\Rightarrow x=\frac{(d-r)\pm \sqrt{\left((r{)}^2+(d{)}^2\right)}}{2}$
$2x=(d-r)\pm \sqrt{(r{)}^2+(d{)}^2}$
putting the value of $x$ in equation (1)
$m=\frac{r}{r+\left[(d-r)\pm \sqrt{(r{)}^2+(d{)}^2}\right]}$
For $m$ to be maximum, $x$ should be minimum
$m_{\min }=\frac{r}{d+\sqrt{r^2+d^2}}$