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, $m =\frac{ f }{ f - u } \Rightarrow m =\frac{ f }{ f + x }$
$m =\frac{ r / 2}{ r / 2+ x } \Rightarrow m =\frac{ r }{ r +2 x }$......(1)
from mirror formula
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{(d-x)}+\frac{1}{-x}=\frac{1}{f} \Rightarrow \frac{1}{d-x}-\frac{1}{x}=\frac{2}{r}$
$\frac{+ x -( d - x )}{ x ( d - x )}=\frac{2}{ r } \Rightarrow(2 x - d ) r =2 x ( d - x )$
$2 r x-r d=2 x d-2 x^{2} \Rightarrow 2 x^{2}+2(r-d) x-r d=0$
$2( x )^{2}+2( r - d ) x - rd =0 \Rightarrow x =\frac{-2( r - d ) \pm \sqrt{4( r - d )^{2}+8 rd }}{4}$
$x =\frac{2( d - r ) \pm \sqrt{4\left(( r )^{2}+( d )^{2}\right)}}{4} \Rightarrow x =\frac{( d - r ) \pm \sqrt{\left(( r )^{2}+( d )^{2}\right)}}{2}$
$2 x =( d - r ) \pm \sqrt{( r )^{2}+( d )^{2}}$
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
$m _{\min }=\frac{ r }{ d +\sqrt{ r ^{2}+ d ^{2}}}$