Question 214: Ray Optics

Question

A concave mirror of focal length 10 cm forms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is
$\_\_\_\_$ cm.

Accepted Answer

For a mirror, magnification is given as $\mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}$, where v is the image distance and u is the object distance.
From the mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ :
$$ \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{u f} \Rightarrow v=\frac{u f}{u-f} $$
Substituting v into the magnification formula:
$$ \mathrm{m}=-\frac{\mathrm{f}}{\mathrm{u}-\mathrm{f}}=\frac{\mathrm{f}}{\mathrm{f}-\mathrm{u}} $$
A concave mirror ( $\mathrm{f}=-10 \mathrm{~cm}$ ) produces a magnification $|\mathrm{m}|=2$ in two cases:
Case 1 :
Real Image ( $\mathrm{m}=-2$ )
$$ -2=-\frac{10}{-10-\mathrm{u}_1} $$
$$\Rightarrow $$ $$ 20+2 \mathrm{u}_1=-10 \Rightarrow 2 \mathrm{u}_1=-30 \Rightarrow \mathrm{u}_1=-15 \mathrm{~cm} $$
Case 2 :
Virtual Image ( $\mathrm{m}=+2$ )
$$ 2=-\frac{10}{-10-\mathrm{u}_2} $$
$$\Rightarrow $$ $$ -20-2 \mathrm{u}_2=-10 \Rightarrow-2 \mathrm{u}_2=10 \Rightarrow \mathrm{u}_2=-5 \mathrm{~cm} $$
The distance between the two object positions is:
$$ \Delta \mathrm{u}=\left|\mathrm{u}_1-\mathrm{u}_2\right|=|-15-(-5)|=|-10|=10 \mathrm{~cm} $$
Therefore, the distance between the two positions is 10 cm .

Ray Optics

A concave mirror of focal length 10 cm forms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is \_\_\_\_ cm.

Integer / Numeric Answer Type