Use the mirror equation to deduce that:
- an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
- a convex mirror always produces a virtual image independent of the location of the object.
- the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
Exercise
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- For a concave mirror, the focal length (f) is negative.
When the object is placed on the left side of the mirror, the object distance (u)is negative.
$\therefore \ \text{u}<0$
For image distance v, we can write the lens formula as:
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}\dots(1)$
The object lies between f and 2f.
$\therefore \ 2\text{f}<\text{u}<\text{f}$ $(\therefore \ \text{u and f are negetive})$
$\frac{1}{2\text{f}}>\frac{1}{\text{u}}>\frac{1}{\text{f}}$
$-\frac{1}{2\text{f}}>-\frac{1}{\text{u}}>-\frac{1}{\text{f}}$
$\frac{1}{\text{f}}-\frac{1}{2\text{f}}<\frac{1}{\text{f}}-\frac{1}{\text{u}}<0\dots(2)$
Using equation (1), we get:
$\frac{1}{2\text{f}}<\frac{1}{\text{v}}<0$
$\therefore \ \frac{1}{\text{v}}$ is negative, i.e., v is negative.
$\frac{1}{2\text{f}}<\frac{1}{\text{v}}$
2f > v
-v > -2f
Therefore, the image lies beyond 2f.
- For a convex mirror, the focal length (f) is positive.
When the object is placed on the left side of the mirror, the object distance (u) is negative.
$\therefore \ \text{u}>0$
For image distance v, we have the mirror formula:
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$
Using equation (2), we can conclude that:
$\frac{1}{\text{v}}<0$
v > 0
Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, regardless of the object distance.
- For a convex mirror, the focal length (f) is positive.
When the object is placed on the left side of the mirror, the object distance (u) is negative,
$\therefore \ \text{u}>0$
For image distance v, we have the mirror formula:
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$
But we have u < 0
$\therefore \ \frac{1}{\text{v}}>\frac{1}{\text{f}}$
v < f
Hence, the image formed is diminished and is located between the focus (f) and the pole.
- For a concave mirror, the focal length (f) is negative
When the object is placed on the left side of the mirror, the object distance (u) is negative.
$\therefore \ \text{u}>0$
It is placed between the focus (f) and the pole.
$\therefore \ \text{f}>\text{u}>0$
$\frac{1}{\text{f}}<\frac{1}{\text{u}}<0$
$\frac{1}{\text{f}}-\frac{1}{\text{u}}<0$
For image distance v, we have the mirror formula:
$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\frac{1}{\text{v}}=\frac{1}{\text{f}}-\frac{1}{\text{u}}$
$\therefore \ \frac{1}{\text{v}}<0$
v > 0
The image is formed on the right side of the mirror. Hence, it is a virtual image. For u < 0 and v > 0, we can write:
$\frac{1}{\text{u}}>\frac{1}{\text{v}}$
v > u
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