The Mirror Equation and Magnification

Real and Virtual Images: What Are We Looking For?

When light rays leave a point on an object and bounce off a mirror, they may come together again at another point. If the reflected rays actually converge and physically meet at a location, that location is a real image of the original point. You could place a screen there and see the image projected on it.

Sometimes, though, the reflected rays spread apart after leaving the mirror. They never actually meet. But if you look at the mirror and trace those diverging rays backwards, they appear to come from a point behind the mirror. That point is a virtual image. It looks like the light is coming from there, but no light actually passes through that point.

In both cases, the mirror creates a point-to-point match: every point on the object has a corresponding point on the image, established through reflection.

Four Convenient Rays for Drawing Ray Diagrams

In principle, you could trace any two rays from a point on the object, find where they meet after reflection, and locate the image. In practice, four particular rays are especially easy to draw because their paths follow simple rules:

• Ray parallel to the principal axis — After reflection, this ray passes through the focus $F$ (for a concave mirror) or appears to come from $F$ (for a convex mirror).
• Ray through the centre of curvature — Since this ray hits the mirror along the normal (the radius is always perpendicular to the surface), it bounces straight back along the same path.
• Ray through the focus — After reflection, this ray travels parallel to the principal axis. (For a convex mirror, the ray is directed toward $F$ behind the mirror.)
• Ray hitting the pole at any angle — This ray reflects obeying the law of reflection, with the principal axis acting as the normal at the pole.

Pick any two of these four rays from a point on the object, draw their reflected paths, and where they intersect is the image of that point.

Fig 9.5: Ray diagram for image formation by a concave mirror

Figure 9.5 shows this in action. An object $AB$ is placed beyond $C$ in front of a concave mirror. Three rays from point $A$ are drawn, and all three meet at a single point $A'$ after reflection. The image $A'B'$ forms between $C$ and $F$, and it is real and inverted.

Keep in mind that an infinite number of rays leave point $A$ in all directions, and every single one of them, after reflecting off the concave mirror, passes through $A'$. The three rays shown are simply the easiest ones to draw.

Deriving the Mirror Equation

Now let us connect three key quantities: the object distance ($u$), the image distance ($v$), and the focal length ($f$). The relationship between them is called the mirror equation.

Step 1: Finding Similar Triangles

Look at the ray diagram in Fig. 9.5. There are two useful pairs of similar triangles here.

First pair: triangles $A'B'F$ and $MPF$

$M$ is the point where the ray parallel to the axis hits the mirror surface, and $P$ is the pole. For paraxial rays (rays close to the axis), the segment $MP$ is effectively a straight line perpendicular to the principal axis. These two triangles share the angle at $F$ and both have a right angle, so they are similar. This gives:

$\frac{B'A'}{PM} = \frac{B'F}{FP} \qquad \text{(9.4)}$

Since $PM = AB$ (both are equal to the object height, because $M$ is at the same height as $A$ for a ray travelling parallel to the axis):

$\frac{B'A'}{BA} = \frac{B'F}{FP}$

Second pair: triangles $A'B'P$ and $ABP$

The angle $\angle APB$ equals $\angle A'PB'$ (these are the angles the incident and reflected rays make at the pole). Both triangles also contain a right angle. So they are similar, giving:

$\frac{B'A'}{BA} = \frac{B'P}{BP} \qquad \text{(9.5)}$

Step 2: Combining the Two Results

Both equations (9.4) and (9.5) have the same left side, $B'A'/BA$. Setting the right sides equal:

$\frac{B'F}{FP} = \frac{B'P}{BP}$

Now, $B'F = B'P - FP$ (the distance from $B'$ to $F$ equals the distance from $B'$ to $P$ minus the distance from $F$ to $P$). Substituting:

$\frac{B'P - FP}{FP} = \frac{B'P}{BP} \qquad \text{(9.6)}$

Step 3: Applying the Sign Convention

Up to this point, all the distances are just magnitudes (positive numbers). Now we bring in the Cartesian sign convention. In the setup of Fig. 9.5, light travels from the object toward the mirror, and that direction is taken as positive. To reach the object $AB$, the image $A'B'$, and the focus $F$ from the pole $P$, you travel in the direction opposite to the incident light. So all three distances pick up a negative sign:

$B'P = -v, \quad FP = -f, \quad BP = -u$

Step 4: Substituting and Simplifying

Put these signed values into Equation (9.6):

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

Simplify the numerator on the left: $-v + f$

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

Multiply both sides by $-1$:

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

Expand the left side:

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

Rearrange:

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

Now divide every term by $v$:

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

Rewriting in the standard order:

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \qquad \text{(9.7)}$

This is the mirror equation. It links the object distance, image distance, and focal length in a single neat expression.

Linear Magnification

Knowing where the image forms is only half the picture. You also want to know how big the image is compared to the object. The quantity that captures this is the linear magnification $m$, defined as:

$m = \frac{h'}{h} \qquad \text{(9.8)}$

where $h'$ is the height of the image and $h$ is the height of the object. Both are measured from the principal axis, with heights above the axis taken as positive and heights below taken as negative.

Deriving the Formula

From the similar triangles $A'B'P$ and $ABP$, we already know:

$\frac{B'A'}{BA} = \frac{B'P}{BP}$

Now apply the sign convention. $B'A' = -h'$ (the image is inverted, so it is below the axis, giving a negative height) and $BA = h$ (the object is above the axis). Similarly, $B'P = -v$ and $BP = -u$. Substituting:

$\frac{-h'}{h} = \frac{-v}{-u}$

$\frac{-h'}{h} = \frac{v}{u}$

Multiply both sides by $-1$:

$\frac{h'}{h} = -\frac{v}{u}$

Therefore:

$m = \frac{h'}{h} = -\frac{v}{u} \qquad \text{(9.9)}$

Reading the Sign of Magnification

The sign of $m$ tells you about the orientation of the image:

Sign of $m$	Meaning
$m < 0$	Image is inverted (flipped upside down relative to the object)
$m > 0$	Image is erect (same orientation as the object)

The magnitude $|m|$ tells you about size:

Magnitude of $m$	Meaning
$	m
$	m
$	m

Universality of the Mirror Equation and Magnification Formula

The derivation above was carried out for a specific case: a real, inverted image formed by a concave mirror with the object placed beyond $C$. However, when the sign convention is applied properly, Equations (9.7) and (9.9) work for every case, whether the mirror is concave or convex and whether the image is real or virtual.

Fig 9.6: Image formation by (a) a concave mirror with the object between P and F, and (b) a convex mirror

Figure 9.6 shows two other situations:

• (a) Concave mirror, object between $P$ and $F$: The reflected rays spread apart after leaving the mirror and never actually meet. Tracing them backwards, they appear to come from behind the mirror, giving a virtual, erect, and magnified image.
• (b) Convex mirror: No matter where the object is placed, the reflected rays always diverge. The image is always virtual, erect, and diminished, located between $P$ and $F$ behind the mirror.

In both cases, plugging the correctly signed values of $u$ and $f$ into the mirror equation gives the correct value and sign for $v$, and the magnification formula gives the correct size and orientation.

Solved Examples

Example 9.1: What If Half the Mirror Is Blocked?

Problem: Suppose the lower half of a concave mirror’s reflecting surface is covered with opaque (non-reflective) material. What effect does this have on the image of an object placed in front of the mirror?

Solution:

At first glance, you might expect to see only half the image. But that is not what happens.

Every point on the object sends out rays in all directions. Many of these rays hit different parts of the mirror. Even with the lower half blocked, the upper half still reflects rays from every point on the object, both the top and the bottom of the object.

So the complete image still forms. The laws of reflection hold at every point on the remaining exposed surface, and those reflected rays still converge (or appear to diverge from) the same image points.

The only difference is the brightness: since the reflecting area has been cut in half, the number of rays reaching the image is also halved. The image has the same shape and position, but its intensity is reduced to half.

Example 9.2: A Phone Along the Principal Axis

Problem: A mobile phone lies along the principal axis of a concave mirror, as shown in Fig. 9.7. Show the formation of its image and explain why the magnification is not uniform. Does the distortion depend on the phone’s location?

Fig 9.7: Image of a phone placed along the principal axis of a concave mirror

Solution:

The key insight is that the magnification formula $m = -v/u$ depends on both $u$ and $v$, and these are linked through the mirror equation. When the object is not flat (perpendicular to the axis) but instead stretched along the axis, different parts of the object sit at different distances from the mirror.

The end of the phone closer to the mirror has one value of $u$, while the end farther away has a different value. Each part maps to a different image distance $v$ and therefore a different magnification. Parts closer to the focus get magnified more than parts farther from it.

This unequal stretching distorts the image: the phone’s image does not look like a proportionally scaled copy of the phone. Yes, the degree of distortion depends on where exactly the phone is placed. If the phone straddles a region where $v$ changes rapidly with $u$ (near the focus), the distortion is severe. If it sits far from the mirror where $v$ changes slowly, the distortion is milder.

Example 9.3: Numerical Problem with a Concave Mirror

Problem: An object is placed at (i) 10 cm and (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case.

Solution:

First, find the focal length. Since $f = R/2$ and the mirror is concave (focus in front of the mirror):

$f = -\frac{15}{2} = -7.5 \text{ cm}$

Case (i): Object at 10 cm ($u = -10$ cm)

Apply the mirror equation:

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

$\frac{1}{v} + \frac{1}{-10} = \frac{1}{-7.5}$

$\frac{1}{v} = \frac{1}{-7.5} - \frac{1}{-10} = -\frac{1}{7.5} + \frac{1}{10}$

Find a common denominator (75):

$\frac{1}{v} = -\frac{10}{75} + \frac{7.5}{75} = \frac{-10 + 7.5}{75} = \frac{-2.5}{75}$

$v = \frac{75}{-2.5} = -30 \text{ cm}$

The negative sign tells us the image is on the same side as the object (in front of the mirror), so it is a real image, located 30 cm from the mirror.

Now find the magnification:

$m = -\frac{v}{u} = -\frac{-30}{-10} = -\frac{30}{10} = -3$

The magnitude is 3, meaning the image is three times the size of the object. The negative sign means it is inverted. So the image is real, magnified, and inverted.

Case (ii): Object at 5 cm ($u = -5$ cm)

$\frac{1}{v} + \frac{1}{-5} = \frac{1}{-7.5}$

$\frac{1}{v} = \frac{1}{-7.5} + \frac{1}{5} = -\frac{1}{7.5} + \frac{1}{5}$

Common denominator (37.5):

$\frac{1}{v} = -\frac{5}{37.5} + \frac{7.5}{37.5} = \frac{-5 + 7.5}{37.5} = \frac{2.5}{37.5}$

$v = \frac{37.5}{2.5} = +15 \text{ cm}$

The positive sign means the image is on the opposite side of the mirror from the object, that is, behind the mirror. This makes it a virtual image, 15 cm behind the mirror.

Magnification:

$m = -\frac{v}{u} = -\frac{15}{-5} = +3$

The magnitude is again 3, so the image is three times the object’s size. The positive sign means it is erect. So the image is virtual, magnified, and erect.

Notice the contrast: moving the object from 10 cm to 5 cm (crossing the focal point at 7.5 cm) completely changed the nature of the image, from real-inverted to virtual-erect.

Example 9.4: Jogger Approaching a Convex Side-View Mirror

Problem: You are sitting in a parked car and notice a jogger approaching in the side-view mirror, which has $R = 2$ m. The jogger runs at a steady 5 m/s. How fast does the jogger’s image appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m, and (d) 9 m away?

Solution:

The side-view mirror is convex, so $f = +1$ m (positive, since the focus is behind a convex mirror).

From the mirror equation, solve for $v$:

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

Since the jogger runs at 5 m/s, after each second the jogger is 5 m closer. We calculate the image position at two instants (1 second apart) and find the shift.

(a) Jogger at 39 m ($u = -39$ m):

$v = \frac{1 \times (-39)}{-39 - 1} = \frac{-39}{-40} = \frac{39}{40} \text{ m}$

After 1 second, the jogger is at 34 m ($u = -34$ m):

$v = \frac{1 \times (-34)}{-34 - 1} = \frac{-34}{-35} = \frac{34}{35} \text{ m}$

Shift in image position over 1 second:

$\Delta v = \frac{39}{40} - \frac{34}{35} = \frac{39 \times 35 - 34 \times 40}{40 \times 35} = \frac{1365 - 1360}{1400} = \frac{5}{1400} = \frac{1}{280} \text{ m}$

Average image speed $\approx \frac{1}{280}$ m/s.

(b) Jogger at 29 m: Using the same approach:

Image speed $\approx \frac{1}{150}$ m/s.

(c) Jogger at 19 m:

Image speed $\approx \frac{1}{60}$ m/s.

(d) Jogger at 9 m:

Image speed $\approx \frac{1}{10}$ m/s.

Pattern: Even though the jogger moves at a constant 5 m/s, the image speed increases dramatically as the jogger gets closer. At 39 m, the image barely crawls. At 9 m, the image moves roughly 28 times faster than at 39 m.

This is something you can observe in everyday driving. Objects in a convex side-view mirror seem to suddenly “rush” toward you as they get close, even if they are approaching at a steady speed. It is the reason these mirrors carry the warning: “Objects in mirror are closer than they appear.”