A concave mirror has a radius of curvature of 24cm. How far is an object from the mirror if the image formed is (a) virtual and 3.0 times the size of the object, (b) real and 3.0 times the size of the object, and (c) real and 1/3 the size of the object?

Radius of curvature r=24cm

Magnification for virtual image m=+3

Magnification for real image m = -3

Magnification for real image m= -1/3

We use the mirror equation to find the object distance from the mirror. Before that, we have to use the lateral magnification equation to find the image distance, and substituting this value in mirror equation, we can solve for the object distance.

Formula:

$$\begin{gathered} \frac{1}{p}+\frac{1}{i}=\frac{1}{f}\left(i\right) \\ m=-\frac{i}{p}\left(2\right) \\ f=\frac{r}{2}\left(3\right) \end{gathered}$$

(a)

The focal length f is related to the radius of curvature r of the mirror as

$f=\frac{1}{2}r$

Where r is positive for the concave mirror and negative for the convex mirror.

$$\begin{gathered} f=\frac{24cm}{2} \\ =12cm \end{gathered}$$

The mirror equation relates an object distance p, mirror’s focal length f and the image distance i as

role="math" localid="1663056728423" $\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$

The lateral magnification is

$m=-\frac{i}{p}$

where, absolute value of m is, m=3.0.

In this case, the image is virtual. Hence m will be positive.

So,

$$\begin{gathered} 3=-\frac{i}{p} \\ i=-3p \end{gathered}$$

Substituting i in (1), we get

role="math" localid="1663056871342" $$\begin{gathered} \frac{1}{p}+\frac{1}{-3p}=\frac{1}{f} \\ \frac{-2}{-3p}=\frac{1}{12cm} \\ p=+8.0cm \end{gathered}$$

Object distance from the mirror if the image formed is virtual and 3 times the size of the object is p = +8.0cm

(b)

The mirror equation relates an object distance p, mirror’s focal length f and the image distance i as

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$(2)

The lateral magnification is

$m=-\frac{i}{p}$

where, absolute value of m is, m=3.0.

In this case, the image is real. Hence m will be negative.

So,

$$\begin{gathered} -3=-\frac{i}{p} \\ i=3p \end{gathered}$$

Substituting i in (2), we get

$$\begin{gathered} \frac{1}{p}+\frac{1}{3p}=\frac{1}{f} \\ \frac{4}{3p}=\frac{1}{12cm} \\ p=16cm \end{gathered}$$

Object distance from the mirror if the image formed is real and 3 times the size of the object is 16cm.

(c)

The mirror equation relates an object distance p , mirror’s focal length f and the image distance i as

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$ (3)

The lateral magnification is

$m=-\frac{i}{p}$

where, absolute value of m is, m=1/3

In this case, the image is real. Hence m will be negative.

So,

$$\begin{gathered} -\frac{1}{3}=-\frac{i}{p} \\ i=\frac{p}{3} \end{gathered}$$

Substituting i in (3), we get

$$\begin{gathered} \frac{1}{p}+\frac{3}{p}=\frac{1}{f} \\ \frac{4}{p}=\frac{1}{12cm} \\ p=48cm \end{gathered}$$

Object distance from the mirror if the image formed is real and 1/3 times the size

Of the object is 48cm.