When an object is placed 20cm from a concave mirror, a real image magnified three times is formed. Find:
- The focal length of the mirror.
- Where must the object be placed to give a virtual image three times the height of the object?
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Given:u = -20cm, m = -3cm, for the real image
$\therefore \text{m}=-3=-\frac{\text{v}}{(-20)}$
$\Rightarrow\text{v}= -60\text{cm}$
We have
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{(-60)}+\frac{1}{(-20)}=\frac{1}{\text{f}}$
$\frac{1}{\text{f}}=-\frac{1}{60}-\frac{1}{20}=\frac{-1-3}{60}=\frac{1}{15}$
$\text{f}=-15\text{cm}$
$\therefore \text{m}=-3=-\frac{\text{v}}{(-20)}\Rightarrow \text{v}=-3\text{u}$
We have
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{\text{(3u)}}+\frac{1}{\text{u}}=\frac{1}{(-15)}$
$\Rightarrow \frac{-1+3}{\text{3u}}=-\frac{1}{15}$
$\Rightarrow \text{u}=-\frac{2\times15}{3}=-10\text{cm}$
So object should be placed 10cm from the concave mirror.
- We know that
$\therefore \text{m}=-3=-\frac{\text{v}}{(-20)}$
$\Rightarrow\text{v}= -60\text{cm}$
We have
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{(-60)}+\frac{1}{(-20)}=\frac{1}{\text{f}}$
$\frac{1}{\text{f}}=-\frac{1}{60}-\frac{1}{20}=\frac{-1-3}{60}=\frac{1}{15}$
$\text{f}=-15\text{cm}$
- For virtual image m = 3 , and f = 15cm
$\therefore \text{m}=-3=-\frac{\text{v}}{(-20)}\Rightarrow \text{v}=-3\text{u}$
We have
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{\text{(3u)}}+\frac{1}{\text{u}}=\frac{1}{(-15)}$
$\Rightarrow \frac{-1+3}{\text{3u}}=-\frac{1}{15}$
$\Rightarrow \text{u}=-\frac{2\times15}{3}=-10\text{cm}$
So object should be placed 10cm from the concave mirror.
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