Use the mirror equation to deduce that:

Solution:

(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.