Reflection (Continuation)

Now, suppose that a beam of parallel rays between HA and LC is incident on a plane mirror, and imagine a plane wavefront AB which is normal to the rays, reaching the mirror surface, see figure 1 below. At this instant the point A acts as a centre of disturbance. Suppose we require the new wavefront at a time corresponding to the instant when the disturbance at B reaches C. The wavelet from A reaches the surface of a sphere of radius AD at this instant. When other points between AC on the mirror, such as P, are reached by the disturbances starting at AB, wavelets of smaller radius than AD such as PM are obtained at the instant we are considering. The new wavefront is the surface CMD that touches all the wavelets.

FIGURE 1 Credits: Freeman Ira

In the absence of the mirror, the plane wavefront AB would reach the position EC in the time considered. Thus AD=AE=BC, and PN=PM, where PN is perpendicular to EC. The triangles PMC, PNC are each 90°, and PN=PM.

So the angles marked q in the above figure are all equal.

Law of Reflection. We can now deduce the law of reflection for the angles of incidence and reflection. The incident wavefront AB and the reflected wavefront CD make equal angles q with the mirror AC. Since the incident and reflected rays such as HA and HD are normal (90°) to the wavefronts, these rays also make equal angles with AC. So the angles of incidence and reflection are equal.

At a considerable distance from a small source of light, a limited portion of the spherical waves will be practically plane. The rays, which are always perpendicular to the waves, will in this case be very parallel (Figure 2a). Sunlight is an example. If a parallel beam of light hits a plane mirror, the Law of Reflection tells us that the rays will also be parallel after reflection (Figure 2b). This is called regular reflection. By contrast, when such a beam of light strikes a rough or irregular surface diffuse reflection takes place (Figure 2c). At each point on the surface the angles of incidence and reflection are equal, but the various portions of the surface have different directions, and so do the reflected rays. As a result, a rough surface will be visible from almost any portion, while in order to receive light from a mirror your eye must be in this particular direction in which the effect of their surface texture on the light they diffusely reflect. A perfectly smooth, clean mirror would not be visible; what you would see would be the source of light rather than the mirror.

A highly polished silver surface reflects about 95 percent of light that falls on it perpendicularly. An ordinary mirror, consisting of a sheet of glass silvered on the back, reflects about 90 percent. The surfaces of a transparent substance, such as a sheet of glass, reflects some light even though they may not be silvered. You notice, for instance, that at night the interior of a lighted room in which you are sitting can be seen reflected in the windows. Only about 8 percent of light falling perpendicularly on a sheet of glass is reflected, half of this from the rear surface; but at large angles on incidence ('grazing' incidence) almost all of the incoming light is reflected at the front surface. This explains why the reflection of the sun in a lake is not extremely bright when the sun is overhead but is too dazzling to look at when the sun is low in the sky.