Question

Explain the concept of Interference of Light Waves for double slit problems to a grade 12 classmate who was absent from class. Use visual aids and examples.

Answer 1

We can analyze double-slit interference with the help of Figure 1, which depicts an apparatus analogous to Young’s. Light from a monochromatic source falls on a slit S0. The light emanating from S0 is incident on two other slits S1 and S2 that are equidistant from S0. A pattern of interference fringes on the screen is then produced by the light emanating from S1 and S2. All slits are assumed to be so narrow that they can be considered secondary point sources for Huygens’ wavelets (The Nature of Light). Slits S1 and S2 are a distance dapart (d≤1mm), and the distance between the screen and the slits is D(≈1m)​​​​​​

Fig 1:The double-slit interference experiment using monochromatic light and narrow slits. Fringes produced by interfering Huygens wavelets from slits S1 and S2 are observed on the screen.

Since S0 is assumed to be a point source of monochromatic light, the secondary Huygens wavelets leaving S1 and S2 always maintain a constant phase difference (zero in this case because S1 and S2 are equidistant from S0) and have the same frequency. The sources S1 and S2 are then said to be coherent. By coherent waves, we mean the waves are in phase or have a definite phase relationship. The term incoherent means the waves have random phase relationships, which would be the case if S1 and S2 were illuminated by two independent light sources, rather than a single source S0. Two independent light sources (which may be two separate areas within the same lamp or the Sun) would generally not emit their light in unison, that is, not coherently. Also, because S1and S2S2 are the same distance from S0, the amplitudes of the two Huygens wavelets are equal.

Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. In the following discussion, we illustrate the double-slit experiment with monochromatic light (single λ) to clarify the effect. Figure 2 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

Fig 2:The amplitudes of waves add. (a) Pure constructive interference is obtained when identical waves are in phase. (b) Pure destructive interference occurs when identical waves are exactly out of phase, or shifted by half a wavelength.

When light passes through narrow slits, the slits act as sources of coherent waves and light spreads out as semicircular waves, as shown in Figure 3(a). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

Figure 3:Double slits produce two coherent sources of waves that interfere. (a) Light spreads out (diffracts) from each slit, because the slits are narrow. These waves overlap and interfere constructively (bright lines) and destructively (dark regions). We can only see this if the light falls onto a screen and is scattered into our eyes. (b) When light that has passed through double slits falls on a screen, we see a pattern such as this.

To understand the double-slit interference pattern, consider how two waves travel from the slits to the screen (Figure 4). Each slit is a different distance from a given point on the screen. Thus, different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively. If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively. More generally, if the path length difference ΔΔl between the two waves is any half-integral number of wavelengths [(1 / 2)λ, (3 / 2)λ, (5 / 2)λ, etc.], then destructive interference occurs. Similarly, if the path length difference is any integral number of wavelengths (λ, 2λ, 3λ, etc.), then constructive interference occurs. These conditions can be expressed as equations:

Δl=mλ,form=0,±1,±2,±3…(constructive interference).... (1)

Δl=(m+12)λ,form=0,±1,±2,±3…(destructive interference). .... . (2)

Figure 4:Waves follow different paths from the slits to a common point P on a screen. Destructive interference occurs where one path is a half wavelength longer than the other—the waves start in phase but arrive out of phase. Constructive interference occurs where one path is a whole wavelength longer than the other—the waves start out and arrive in phase.

Figure 5(a) shows how to determine the path length difference Δlfor waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle θ between the path and a line from the slits to the screen [part (b)] is nearly the same for each path. In other words, r1and r2 are essentially parallel. The lengths of r1and r2 differ by Δl, as indicated by the two dashed lines in the figure. Simple trigonometry shows

Δl=dsinθ.... (3)

where d is the distance between the slits. Combining this result with Equation (1) , we obtain constructive interference for a double slit when the path length difference is an integral multiple of the wavelength, or

dsinθ=mλ,form=0,±1,±2,±3,…(constructive interference).... .(4)

Similarly, to obtain destructive interference for a double slit, the path length difference must be a half-integral multiple of the wavelength, or

dsinθ=(m+1/2)λ,form=0,±1,±2,±3,…(destructive interference)..... (5)

where λ is the wavelength of the light, d is the distance between slits, and θ is the angle from the original direction of the beam as discussed above. We call m the order of the interference. For example, m=4 is fourth-order interference.

Figure 5(a) To reach P, the light waves from S1 and S2 must travel different distances. (b) The path difference between the two rays is Δl.

The equations for double-slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes (Figure 6). The closer the slits are, the more the bright fringes spread apart. We can see this by examining the equation

dsinθ=mλ,form=0,±1,±2,±3… For fixed λ and m, the smaller d is, the larger θ must be, since sinθ=mλ/d. This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance d apart) is small. Small d gives large θ, hence, a large effect.

Referring back to part (a) of the figure, θ is typically small enough that sinθ≈tanθ≈Y(m) /D, where Y(m) is the distance from the central maximum to the mth bright fringe and D is the distance between the slit and the screen. Equation 4 may then be written as

dY(m)/D=mλ

or

Y(m) =mλD/d.... (6)

Figure 6: The interference pattern for a double slit has an intensity that falls off with angle. The image shows multiple bright and dark lines, or fringes, formed by light passing through a double slit.