Question

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?

Short answer

1. Object distance from the mirror if the image formed is virtual and 3 times the size of the object is p=+8.0cm.
2. Object distance from the mirror if the image formed is real and 3 times the size of the object is 16cm
3. Object distance from the mirror if the image formed is real and 1/3 times the size

of the object is 48cm.

Step-by-step solution

Listing the given quantities

Radius of curvature r=24cm

Magnification for virtual image m=+3

Magnification for real image m = -3

Magnification for real image m= -1/3

Understanding the concepts of mirror equation

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:

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

Step 3: Calculation of the object distance from the mirror if the image formed is virtual and 3    times the size of the object

(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

Calculation of the object distance from the mirror if the image formed is real and 3 times the size of the object

(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.

Step 5: Calculation of the Object distance from the mirror if the image formed is real and 13 times the size of the object:

(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.