Question

An observer whose least distance of distinct vision is 'd', views his own face in a convex mirror or radius or curvature 'r'. The magnification produced cannot exceed $\frac{kr}{d+\sqrt{d^2+r^2}}$. Find $k$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A 1

solution:

$m=\frac{f}{f-u}\to m=\frac{f}{f+x}$

$m=\frac{r/2}{r/2+x}\to m=\frac{R}{r+2x}\dots \dots \dots \left(1\right)$

from mirror formula

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

$\frac{+x-(d-x)}{x(d-x)}=\frac{2}{r}\to (2x-d)r=2x(d-x)$

$2rx-rd=2xd-2x^2\to 2x^2+2(r-d)x-rd=0$

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

$x=-2(d-r)\pm \sqrt{\frac{4(r^2+d^2)}{4}}\to x=(d-r)\pm \frac{r^2+d^2}{2}$

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

Putting the value of x in equation (1)

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

hence the correct option: A