# Question #db044

Jan 19, 2017

Test the condition at the two limits given in the question, ${d}_{o} = f$ and ${d}_{o} = C$.
I show you how, below...

#### Explanation:

The equation for a spherical mirror is

$\frac{1}{{d}_{o}} + \frac{1}{{d}_{i}} = \frac{1}{f}$

which I will write as

$\frac{1}{{d}_{i}} = \frac{1}{f} - \frac{1}{{d}_{o}}$

If the object is located at $f$, then ${d}_{o} = f$, and the equation becomes

$\frac{1}{{d}_{i}} = \frac{1}{f} - \frac{1}{f} = 0$

This is only true if ${d}_{i}$ goes to infinity. So, one condition is demonstrated.

Next, recall that for a spherical mirror, $f$ is midway between $C$ and the vertex. So, an object placed at C has ${d}_{o} = 2 f$

The equation becomes

$\frac{1}{{d}_{i}} = \frac{1}{f} - \frac{1}{2 f} = \frac{1}{2 f}$

or ${d}_{i} = 2 f = C$, which is simply a way of stating that the image is at the centre of curvature. So, the second condition is shown.

It follows (I think!) that for any placement of the object between these limits, the image must be between the location of ${d}_{i}$ shown for these limits, namely between $f$ and $C$.

Jan 20, 2017

Assuming that the question is about spherical mirrors and has been posted under flat mirror by oversight only.

We know that Mirror formula is the relationship between object distance $u$, image distance $v$ and focal length $f$ of the mirror and is given as below*.

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

A. When Object is placed at $\text{C}$
Object distance $u = 2 f$

Inserting in equation (1) we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{2 f}$
$\implies \frac{1}{v} = \frac{1}{f} - \frac{1}{2 f}$
$\implies \frac{1}{v} = \frac{1}{2 f}$
$\implies v = 2 f$
Implies that image is formed at $\text{C}$

B. Object is placed at $\text{F}$
Object distance $u = f$

Inserting in equation (1) we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{f}$
$\implies \frac{1}{v} = \frac{1}{f} - \frac{1}{f} = 0$
$\implies v = \infty$
Implies that image is formed at $\infty$

C. Object is placed between $\text{C and F}$
Object distance $u = n f$, where $n$ lies between $1 \mathmr{and} 2$

Inserting in equation (1) we get
$\frac{1}{f} = \frac{1}{v} + \frac{1}{n f}$
$\implies \frac{1}{v} = \frac{1}{f} - \frac{1}{n f}$
$\implies \frac{1}{v} = \frac{n - 1}{n f}$
$\implies v = \frac{n f}{n - 1}$
Inserting any value of $n$ between $1 \mathmr{and} 2$, we see that image is formed between $\text{C} \mathmr{and} \infty$.

Note that for $n = 1$ we have case number A and for $n = 2$ we get case number B as discussed above.
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.

*Sign Convention: For spherical mirrors, normally object distance is taken as negative, as it is measured against direction of light.Image distance is positive for real images and is in the direction of travel of light. Focal length is positive accordingly for a converging mirror as it is measured in the direction of travel of light.
If sign convention is followed then Mirror Formula for a converging mirror becomes

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

and calculations are to be done appropriately.