Question

An object is placed at some distance in front of a concave mirror of focal length $10cm$. The image is formed on the same side, but $10cm$ farther from the mirror than the object is. Find out the magnification of the mirror.

• A. -1.618

Answer 1

The correct option is A -1.618
Given:
Focal length of the mirror, $f=10cm$
The image distance is 10 more than the object distance.
i.e. if $u$ is the object distance, then the image distance $v=u+10$.
Using the mirror formula, $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ along with the sign convention,
$\frac{-1}{10}=-\frac{1}{u}-\frac{1}{(u+10)}$
$⟹\frac{1}{10}=\frac{1}{u}+\frac{1}{(u+10)}$
$⟹\frac{1}{10}=\frac{2u+10}{u^2+10u}$
$⟹u^2+10u=20u+100$
$⟹u^2-10u-100=0$
On solving the quadratic equation, the roots are $\frac{10\pm 10\sqrt{5}}{2}=5\pm 5\sqrt{5}$.
The possible object distances are $u_1=5+5\sqrt{5}=16.18cm$ and $u_2=5-5\sqrt{5}=-6.18cm$
Since, the image distance is greater than the object distance, the object must be placed somewhere between the centre of curvature and the Focus, i.e. $R>u>f$, where $R$ is the radius of curvature.
$20>16.18>10$
So, $16.18cm$ is the correct object distance to be chosen.
$\therefore v=u+10=26.18cm$
Magnification of a concave mirror, $m=\frac{-v}{u}=\frac{-26.18}{16.18}=-1.618$