When light hits an object, it does one of three things: it passes through it, it gets absorbed, or it gets reflected - that is, the light ray "bounces off" the object. There are two ways light can be reflected. One is diffuse reflection, which means the reflected photons bounce off the object in random directions (see Chapter 1). The other is specular reflection, which means all the light rays bounce of in a particular direction ( Figure 1).
When specular reflection occurs, the light bounces off the surface with exactly the same angle as it hit the surface, much like a ball bouncing off a hard floor. Specular reflection obeys a simple law: the angle of reflection is equal to the angle of incidence. The angles of reflection and incidence are usually measured from the surface normal (just like refraction), but for flat surfaces they can also be measured from the surface (see Figure 2).
The fact that the most specular reflective substances are metals is no coincidence. Metals have plenty of "free" electrons; that is, electrons that are free to move around in the metal. When photons of light are absorbed by a smooth metal surface, their energy causes the free electrons in the surface to move. The moving electrons generate an electrical field which creates new photons. These photons have the same velocity parallel to the metal surface as the original absorbed photons, but a reversed velocity perpendicular to it.
When we see a reflection in a mirror, what we’re really looking at is a virtual image. The virtual image is the same size as the object, and is as far behind the mirror as the object is in front of it. ( Figure 3)
The virtual image is the same distance from the mirror as the object. Figure 4(a) shows two rays from Figure 3 leaving a point P, and their virtual rays appearing to come from a point B. The angles of incidence and reflection have been drawn in where the two rays hit the mirror. From this, we have
Figure 4(b) shows the opposite angles \( \alpha^{\prime\prime} \) and \( \beta^{\prime\prime} \) , which are equal to \( \alpha^\prime \) and \( \beta^\prime \). From the above two equations, that means
Figure 4(c) shows the "supplementary" angles \( \gamma \) and \( \gamma^{\prime\prime} \). Since \( \gamma=180-\beta \) and \( \gamma^{\prime\prime}=180-\beta^{\prime\prime} \), we have
Finally, Figure 4(d) highlights two triangles. Both triangles have two angles the same ( \( \alpha, \gamma \) and \( \alpha^{\prime\prime}, \gamma^{\prime\prime} \) ). Also, both triangles share a common side (the part of the mirror between where the two rays hit), so that means both triangles are exactly the same size. If they are exactly the same size, that means the distance from P to the triangle's base is the same as the distance from B to the triangle's base. Thus P and B are equally far from the mirror.
Curved mirrors can be used to focus light in much the same way as a lens does. In fact, for some applications such as astronomical telescopes, curved mirrors are far better than lenses. Spherical curved mirrors have a surface which is a small slice of a sphere. There are two types of spherical mirrors: convex, where the centre of the mirror pushes outwards, and concave, where it is pushed inwards. Convex mirrors diverge light and create a virtual image. Concave mirrors converge light and usually create real images ( Figure 5).
Recall that the focal length of a lens is either
Similar definitions apply to mirrors: the focal length of a mirror is the distance from mirror to image when parallel light enters the mirror and is reflected to form the image, or the distance from mirror to object when light from the object is reflected from the mirror to leave it in parallel.
There is a very simple relationship between the focal length of a spherical mirror and the radius of the spherical mirror, which we shall derive with the help of Figure 6. Figure 6(a) shows a ray of light, parallel to the optic axis (which is the horizontal dotted line through \( C \)), striking a concave mirror at some point \( R \) and being reflected to the focal point \( F \) . All such parallel rays get reflected to the same place \( F \) because, as we will see below, the height of the ray above the optic axis is irrelevant. The centre of the spherical mirror surface is at \( C \).
We can show that the focal length of a spherical mirror is half the radius by the following steps (which use the figure above):
| Step | Statement | Reason |
|---|---|---|
| 1 | \( \tan(\alpha)\approx h/r \) | From Figure 6(b) |
| 2 | \( \alpha\approx h/r \) | \( \alpha \) is small, so use the
small angle approximation |
| 3 | \( \tan(2\alpha)\approx h/f \) | From Figure 6(f) |
| 4 | \( 2\alpha\approx h/f \) | Small angle approximation |
| 5 | \( 2(h/r)\approx h/f \) | Substitute \( \alpha \) from step (2) into step (4) |
| 6 | \( 2(1/r)\approx 1/f \) | Divide both sides by \( r \) |
| 7 | \( 2 = r/f \) | Multiply both sides by \( r \) |
| 8 | \( 2f = r \) | Multiply both sides by \( f \) |
| 9 | \( f = r/2 \) | Divide by 2 |
Thus, the focal length \( f \) of a spherical mirror is \( r/2 \), half the radius. So the power of a spherical mirror is \( F=1/f=2/r \).
What does the sign convention say about the sign of \( f \) ? The focal length \( f \) is measured from the mirror surface to the mirror’s centre, which is a leftward direction. The focal point of this mirror is created by light that has reflected off the mirror and is also travelling leftwards. So the focal length for the mirror in Figure6, measured from the mirror towards the focal point, is in the direction that the reflected light travels, and so it has to be positive. The upshot of this is that the mirror has a positive power \( F \), which makes sense since it converges light.
All of this is pretty involved, so it's easiest just to remember that concave (converging) mirrors have positive power.
The algebra for working out the focal length of a convex mirror is very similar to the above, and the power of a convex mirror is again \( f=r/2 \). However, the power of a convex mirror is negative, because it diverges light.
Curved mirrors create images in a similar way to lenses, by changing the vergence of the light hitting it. Unlike lenses, mirrors also reverse the direction of the light. This change in direction means that we can’t apply a single sign convention to the entire image calculation like we do when using lenses, because the light changes direction half way through.
There are equations for curved mirrors that look like the thin lens equation, but because of the sign convention they are tricky to use, and you are prone to making mistakes about whether the image is to the left or right of the mirror.
When calculating the position of an image formed by a mirror, then, the best approach is as follows:
Pretend the curved mirror is a lens with the same power, and use the thin lens equation \( V_{in}+F=V_{out} \).
Since the mirror reflects light, the image formed by the mirror is always on the opposite side of the image formed by the thin lens.
Don’t state the image distance as a signed number, because the sign convention is confusing for mirrors; instead, say whether the image is to the left or right of the mirror.
A few examples should clarify this advice:
An object is placed 75cm to the left of a concave mirror with power \( +5D \). Where does the image form?
The vergence of the light hitting the mirror is \( V_{in}=1/(-0.75)=-1.33\text{D} \). If we pretend the mirror is a thin lens, then the thin lens equation says that \( V_{out}=-1.333+5=+3.667\text{D} \). That means the image is \( 1/3.667 = 0.273\text{m} \) to the right of the lens.
Since the mirror reflects, the image actually forms \( 0.273\text{m} \) to the left of the mirror. This is in front of the mirror, and so it is a real image.
An object is placed 0.5m to the left of a convex mirror with power \( -4\text{D} \). Where does the image form?
The vergence of the light hitting the mirror is \( V_{in}=1/(-0.5)=-2\text{D} \). If we pretend the mirror is a thin lens, then the thin lens equation says that \( V_{out}=-2+-4 =-6\text{D} \). That means the image is \( 1/(-6) = -0.167\text{mm} \) to the left of the lens.
Since the mirror reflects, the image actually forms \( 0.167\text{m} \) to the right of the mirror. This is behind the mirror, and so it is a virtual image.