UQ Physics

Mirrors

Mirrors can be used to direct light and to manipulate the wavefront to give focusing and defocusing. This section looks at how images are formed using mirrors.<

Geometric Construction

When attempting any problem using mirrors, it is important to draw yourself a diagram showing the position of the object in relation to the mirror and estimating the position of the image. This is best achieved by drawing a ray diagram. A number of different types of rays can be drawn and these are shown in the diagram below. The image is formed where they intersect.

Image Location

The equations to determine the position of an image are derived here. To achieve this, an assumption known as the paraxial approximation is made. Rays incident on a mirror are assumed to do so at only a small angle to the normal. This gives sin θ ~ θ. The assumption is mostly good as objects are usually a much larger distance away from the mirror than the diameter of the mirror.

The following diagram shows a spherical mirror with an object O placed a distance u from the axis of the mirror. C represents the centre of curvature of the mirror and I is the position of the image. To determine the location of the image, a ray is considered that is incident on the mirror at the point Q where it is reflected at equal angle from the normal to the mirror (QC).

From the diagram:

In the triangles ΔOQC & ΔQCI

β = α + γ and γ = β + θ

Hence

γ = 2β - α (2)

From (1) and (2)

Defining the focal length f = r/2 gives the mirror equation:

For a concave mirror, f is positive while for a convex mirror, f is negative.

If the mirror causes the light to converge then a real image is formed. Alternatively, if the mirror causes the light rays to diverge then a virtual image is formed behind the mirror. It is called virtual as it is not possible to place a screen in any location and view the image, however it is possible to use your eye (or another optical instrument) to view the image. Real objects and images have positive distances while virtual objects and images have negative distances.

Vergences for Mirrors

To derive a mirror equation for vergences, the above mirror equation can be rewritten:

where n is the refractive index of the surrounding medium. This then takes the form

V_f = P +V_i

where V_f= ± n/v is the final vergence, V_i = ± n/u is the initial vergence and P = ± 2n/r is the power of the mirror. Powers also have the unit of dioptre.

A sign convention also exists - a vergence is positive for converging waves and negative for diverging waves. The power is positive when a surface causes convergence and negative when the surface causes divergence. Note however that when working with vergences we treat object and image distances as just positive numbers - there are no signs associated with these values. The sign of the vergence is determined from whether the light is converging or diverging.

Focal Point

Why do we call the distance f=r/2 the focal point?

Firstly, consider a point source placed at the focal point of a concave mirror. The initial vergence is then V_i = -1/f. The power of the surface is P = +1/f. The final vergence is thus zero. What does this mean? A wavefront with zero vergence is a plane wave. The mirror is said to have "collimated" the light . This point is thus called the first focal point.

Secondly consider, a collimated beam of light (ie a plane wave) incident on the mirror. The initial vergence is zero, the power of the mirror is, as before, P = +1/f. The final vergence is then also +1/f implying that a real image is formed a distance f from the mirror. The collimated light has been focused to a point which is naturally called the focal point. We call this the second focal point which for a mirror is at the same location as the first focal point.

Examples - Mirrors

Example 1. An object is placed 12 cm from a concave mirror with a radius of curvature of 6 cm in air. Use vergences to calculate the position of the image.

Step 1. Draw a diagram - remember that the focal point is at half the distance to the centre of curvature of the mirror. This diagram shows the parallel, central and radial rays. The image location is determined to be between the focal point and the centre of curvature - ie in the range 3 - 6 cm.

Step 2. Calculate the initial vergence: An object generates light rays that diverge. Therefore the initial vergence is negative and given by

V_i = ± n/u = -1/0.12 = -8.33 D

(The refractive index for air is approximately 1. Remember to convert all distances to metres!)

Step 3. Determine the power: A concave (hollowed) mirror causes light rays to converge. The power is thus positive and given by<

P = ± 2n/r = +2/0.06 = + 33.33 D

Step 4. Calculate the final vergence:

V_f = P +V_i = -8.33 + 33.33 D = +25 D

Step 5. Calculate the final distance. The power is positive so the rays are converging to a real image:

V_f = ± n/v => v = n/V_f = 1/25 = 0.04 m = 4 cm.

Step 6. Compare this with your diagram. Do the numbers make sense??

Why use Vergences?? For this simple example, the mirror equation could also be used. However for more complicated systems using lenses, interfaces and mirrors, the vergence approach is much easier and requires one to memorise fewer equations.

Example 2. (Try this yourself) An object is placed 10 cm from a convex mirror with a radius of curvature of 10 cm. Use vergences to find the position of the image.

Step 1. Draw a diagram like above except with the mirror curved in the other direction. The centre of curvature and the focal point are behind the mirror. Draw the ray that reflects back on itself (must be aimed at the centre of curvature), the equal angle ray (hits the mirror on axis), and the parallel ray (reflected away from the focus). Do these rays cross?? Extrapolate the rays in your diagram back to where they intersect - this is the position of the virtual image.

Step 2. Calculate the initial vergence. As before, the object generates light rays that diverge.

Step 3. Determine the power: A convex mirror causes light rays to diverge. The power is thus negative.

Step 4. Calculate the final vergence.

Step 5. Calculate the final distance. The vergence is negative so the rays are diverging thus forming only a virtual image.

Step 6. Check that diagram again.

You should have found that there was a virtual image formed 3.33 cm behind the mirror. If not, then check each step carefully and examine your diagram.