Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
(a) For a concave mirror, the focal length (f) is negative.
∴f < 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴u < 0
For image distance v, we can write the lens formula as:
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ ...(1)
The object lies between f and 2f.
$\therefore 2 f
$\frac{1}{2 f}>\frac{1}{u}>\frac{1}{f}$
$-\frac{1}{2 f}<-\frac{1}{u}<-\frac{1}{f}$
$\frac{1}{f}-\frac{1}{2 f}<\frac{1}{f}-\frac{1}{u}<0$ ...(2)
Using equation (1), we get:
$\frac{1}{2 f}<\frac{1}{v}<0$
$\therefore \frac{1}{v}$ is negative, i.e., $v$ is negative.
$\frac{1}{2 f}<\frac{1}{v}$
$2 f>v$
$-v>-2 f$
Therefore, the image lies beyond 2f.
(b) For a convex mirror, the focal length (f) is positive.
∴ f > 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u < 0
For image distance v, we have the mirror formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
Using equation (2), we can conclude that:
$\frac{1}{v}<0$
$v>0$
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.
(c) For a convex mirror, the focal length (f) is positive.
∴f > 0
When the object is placed on the left side of the mirror, the object distance (u) is negative,
∴u < 0
For image distance v, we have the mirror formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
But we have $u<0$
$\therefore \frac{1}{v}>\frac{1}{f}$
$v
Hence, the image formed is diminished and is located between the focus (f) and the pole.
(d) For a concave mirror, the focal length (f) is negative.
∴f < 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴u < 0
It is placed between the focus (f) and the pole.
$\therefore f>u>0$
$\frac{1}{f}<\frac{1}{u}<0$
$\frac{1}{f}-\frac{1}{u}<0$
For image distance v, we have the mirror formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
$\therefore \frac{1}{y}<0$
$v>0$
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:
$\frac{1}{u}>\frac{1}{v}$
$v>u$
Magnification, $m=\frac{v}{u}>1$
Hence, the formed image is enlarged.