Question

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.]

Answer 1

a) 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

For image distance $v$ we can write

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ ...(1)

The object lies between $f$ and $2f$.

$2f<u<f$

$\frac{1}{2f}>\frac{1}{u}>\frac{1}{f}$

$\frac{-1}{2f}<\frac{-1}{u}<\frac{-1}{f}$

$\frac{1}{f}-\frac{1}{2f}<\frac{1}{f}-\frac{1}{u}$ .......$\left(2\right)$

from eq. 1

$\frac{1}{2f}<\frac{1}{v}<0$ ; $v$ is negative

$2f>v$

$-v>-2f$

Therefore, the image lies beyond $2f$.

b) 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

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 eq ($2$)

$\frac{1}{v}<0$

$v>0$

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

When the object is placed on the left side of the mirror, the object distance (u) is negative

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 $u<0$

therefore $\frac{1}{v}>\frac{1}{f}$

$v<f$

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

When the object is placed on the left side of the mirror, the object distance ($u$) is negative

It is placed between the focus (f) and the pole

therefore $f>u>0$

$\frac{1}{f}<\frac{1}{u}<0$

$\frac{1}{f}-\frac{1}{u}$

For image distance $v$, we have the mirror formula

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$\frac{1}{v}<0$

$v>0$

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:

$\frac{1}{u}>\frac{1}{v}$ so $v>u$

magnification $m=\frac{v}{u}>1$

Hence, the formed image is enlarged.