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.

A) For a concave mirror, the focal length (f) is negative.
$\therefore f<0$
Object is placed on the left side of the mirror; therefore, the object distance (u) is negative.
$\therefore u<0$
For image distance 𝑣, we can write the mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}....\left(1\right)$

The object lies between f and 2f and since both f and 𝑢 are negative, we can write,
$2f<u<f$
$\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}$
$-\frac{1}{2f}<-\frac{1}{u}<-\frac{1}{f}$
$\frac{1}{f}-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}<\frac{1}{f}-\frac{1}{f}....\left(2\right)$
Using equation (1), we get
$\frac{1}{2f}<\frac{1}{v}<0$
We can see from above $\frac{1}{v}$ is negative, i.e. v is negative.
$\frac{1}{2f}<\frac{1}{v}$
$2f>v$
$-v>-2f$
Therefore, the image lies beyond 2f.
Final Answer: Image lies beyond 2f.

B) For a convex mirror, the focal length (f) is positive.
$\therefore f>0$
When the object is placed on the left side of the mirror, the object distance (u) is negative.
$\therefore u<0$
For image distance 𝑣, we can write the mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
Since f > 0 and u < 0, RHS is +ve
$\frac{1}{v}>0$
$v>0$
Thus, the image is formed on the back side of the mirror. Hence, a convex mirror always produces a virtual image.
Final Answer: Image is formed on the back side of mirror.

C) For a convex mirror, the focal length (f) is positive.
$\therefore f>0$
When the object is placed on the left side of the mirror, the object distance (u) is negative.
$\therefore u<0$
For image distance 𝑣, we can write the mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
Since f > 0 and u < 0,
$\frac{1}{v}>\frac{1}{f}\Rightarrow v<f$
And $\frac{1}{v}>\frac{1}{u}$
$\frac{v}{u}<1$
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.
$\therefore f<0$
When the object is placed on the left side of the mirror, the object distance (u) is negative.
$\therefore u<0$
Object is placed between the focus (f) and the pole u < f.
Mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
Since f < 0 , u < 0 and u < f,
$\frac{1}{v}>0\Rightarrow v>0$
The image is formed behind the mirror.
Hence, it is a virtual image.
Since $v>0$ and $u<0,\Rightarrow v>u\Rightarrow \frac{v}{u}>1$
Magnification, $m=\frac{v}{u}>1$
Hence, image formed will be enlarged.