Revision Notes & Important Formulas on Wave Optics
Young’s double-slit experiment
A schematic diagram of the apparatus that Young used is shown in Figure 37.1a. Light is incident on a first barrier in which there is a slit S0 . The waves emerging from this slit arrive at a second barrier that contains two parallel slits S1 and S2 . These two slits serve as a pair of coherent light sources because waves emerging from them originate from the same wave front and therefore maintain a constant phase relationship. The light from S1 and S2 produces on a viewing screen a visible pattern of bright and dark parallel bands called fringes (Fig. 37.1b). When the light from S1 and that from S2 both arrive at a point on the screen such that constructive interference occurs at that location, a bright fringe appears. When the light from the two slits combines destructively at any location on the screen, a dark fringe results.Figure 37.1 (a) Schematic diagram of Young’s double-slit experiment. Slits S1 and S2 behave as coherent sources of light waves that produce an interference pattern on the viewing screen (drawing not to scale). (b) An enlargement of the center of a fringe pattern formed on the viewing screen with many slits could look like this.
We can describe Young’s double-slit experiment quantitatively with the help of Figure 37.4.
Figure 37.4 (a) Geometric construction for describing Young’s double-slit experiment (not to scale). (b) When we assume that r1 is parallel to r2 , the path difference between the two rays is r2 – r1 = d sin θ. For this approximation to be valid, it is essential that L >> d .
The viewing screen is located a perpendicular distance L from the doubleslitted barrier. S1 and S2 are separated by a distance d, and the source is monochromatic. To reach any arbitrary point P, a wave from the lower slit travels farther than a wave from the upper slit by a distance d sin θ. This distance is called the path difference δ (lowercase Greek delta). If we assume that r1 and r2 are parallel, which is approximately true because L is much greater than d, then δ is given by
The value of δ determines whether the two waves are in phase when they arrive at point P. If δ is either zero or some integer multiple of the wavelength, then the two waves are in phase at point P and constructive interference results. Therefore, the condition for bright fringes, or constructive interference, at point P is
The number m is called the order number. The central bright fringe at θ = 0 (m = 0) is called the zeroth-order maximum. The first maximum on either side, where m = ±1, is called the first-order maximum, and so forth.
When δ is an odd multiple of λ/2, the two waves arriving at point P are 180° out of phase and give rise to destructive interference. Therefore, the condition for dark fringes, or destructive interference, at point P is
It is useful to obtain expressions for the positions of the bright and dark fringes measured vertically from O to P. In addition to our assumption that L >> d, we assume that d >> λ. These can be valid assumptions because in practice L is often of the order of 1 m, d a fraction of a millimeter, and λ a fraction of a micrometer for visible light. Under these conditions, θ is small; thus, we can use the approximation sin θ ≈ tan θ. Then, from triangle OPQ in Figure 37.4, we see that
Solving Equation 37.2 for sin θ and substituting the result into Equation 37.4, we see that the positions of the bright fringes measured from O are given by the expression
Using Equations 37.3 and 37.4, we find that the dark fringes are located at
Distribution of light intensity of double-slit interference pattern
Calculate the distribution of light intensity associated with the double-slit interference pattern.
Suppose that the two slits represent coherent sources of sinusoidal waves such that the two waves from the slits have the same angular frequency ω and a constant phase difference φ. The total magnitude of the electric field at point P on the screen in Figure 37.5 is the vector superposition of the two waves. Assuming that the two waves have the same amplitude E0 , we can write the magnitude of the electric field at point P due to each wave separately as
Although the waves are in phase at the slits, their phase difference φ at point P depends on the path difference δ = r2 – r1 = d sin θ. Because a path difference of λ (constructive interference) corresponds to a phase difference of 2 rad, we obtain the ratio
This equation tells us precisely how the phase difference φ depends on the angle θ in Figure 37.4.
Using the superposition principle and Equation 37.7, we can obtain the magnitude of the resultant electric field at point P :
To simplify this expression, we use the trigonometric identity
Taking A = ωt + φ and B = ωt, we can write Equation 37.9 in the form
This result indicates that the electric field at point P has the same frequency ω as the light at the slits, but that the amplitude of the field is multiplied by the factor 2 cos(φ/2). To check the consistency of this result, note that if φ = 0, 2π, 4π, . . . , then the electric field at point P is 2E0 , corresponding to the condition for constructive interference. These values of φ are consistent with Equation 37.2 for constructive interference. Likewise, if φ = π, 3π, 5π, . . . , then the magnitude of the electric field at point P is zero; this is consistent with Equation 37.3 for destructive interference.
Finally, to obtain an expression for the light intensity at point P, the intensity of a wave is proportional to the square of the resultant electric field magnitude at that point (Eq. 34.20). Using Equation 37.10, we can therefore express the light intensity at point P as
Most light-detecting instruments measure time-averaged light intensity, and the time-averaged value of sin2(ωt + φ/2) over one cycle is ½ . Therefore, we can write the average light intensity at point P as
where Imax is the maximum intensity on the screen and the expression represents the time average. Substituting the value for φ given by Equation 37.8 into this expression, we find that
Alternatively, because sin θ ≈ y/L for small values of θ in Figure 37.4, we can write Equation 37.12 in the form
Figure 37.6 Light intensity versus d sin θ for a double-slit interference pattern when the screen is far from the slits (L >> d).
Constructive interference, which produces light intensity maxima, occurs when the quantity πdy/λL is an integral multiple of π, corresponding to y = (λL/d )m. This is consistent with Equation 37.5.
A plot of light intensity versus d sin θ is given in Figure 37.6. Note that the interference pattern consists of equally spaced fringes of equal intensity. Remember, however, that this result is valid only if the slit-to-screen distance L is much greater than the slit separation, and only for small values of θ.
We have seen that the interference phenomena arising from two sources depend on the relative phase of the waves at a given point. Furthermore, the phase difference at a given point depends on the path difference between the two waves. The resultant light intensity at a point is proportional to the square of the resultant electric field at that point. That is, the light intensity is proportional to (E1 + E2)2. It would be incorrect to calculate the light intensity by adding the intensities of the individual waves. This procedure would give E12 + E22, which of course is not the same as (E1 + E2)2. Note, however, that (E1+ E2)2 has the same average value as E12 + E22 when the time average is taken over all values of the phase difference between E1 and E2 . Hence, the law of conservation of energy is not violated.
Phasor diagrams of waves
Consider a sinusoidal wave whose electric field component is given by
where E0 is the wave amplitude and ω is the angular frequency. This wave can be represented graphically by a phasor of magnitude E0 rotating about the origin counterclockwise with an angular frequency ω, as shown in Figure 37.7a. Note that the phasor makes an angle ωt with the horizontal axis. The projection of the phasor on the vertical axis represents E1 , the magnitude of the wave disturbance at some time t . Hence, as the phasor rotates in a circle, the projection E1 oscillates along the vertical axis about the origin.
Figure 37.7 (a) Phasor diagram for the wave disturbance E1 = E0 sin ωt. The phasor is a vector of length E0 rotating counterclockwise. (b) Phasor diagram for the wave E1 = E0 sin(ωt + φ). (c) The disturbance ER is the resultant phasor formed from the phasors of parts (a) and (b).
Now consider a second sinusoidal wave whose electric field component is given by
This wave has the same amplitude and frequency as E1 , but its phase is φ with respect to E1 . The phasor representing E2 is shown in Figure 37.7b. We can obtain the resultant wave, which is the sum of E1 and E2 , graphically by redrawing the phasors as shown in Figure 37.7c, in which the tail of the second phasor is placed at the tip of the first. As with vector addition, the resultant phasor ER runs from the tail of the first phasor to the tip of the second. Furthermore, ER rotates along with the two individual phasors at the same angular frequency ω. The projection of ER along the vertical axis equals the sum of the projections of the two other phasors: Ep = E1 + E2.
It is convenient to construct the phasors at t = 0 as in Figure 37.8. From the geometry of one of the right triangles, we see that Because the sum of the two opposite interior angles equals the exterior angle φ, we see that α = φ/2; thus, Hence, the projection of the phasor ER along the vertical axis at any time t is This is consistent with the result obtained algebraically, Equation 37.10. The resultant phasor has an amplitude 2E0 cos(φ/2) and makes an angle φ/2 with the first phasor. Furthermore, the average light intensity at point P, which varies as Ep2, is proportional to cos2(φ/2), as described in Equation 37.11.
We can now describe how to obtain the resultant of several waves that have the same frequency:
- Represent the waves by phasors, as shown in Figure 37.9, remembering to maintain the proper phase relationship between one phasor and the next.
- The resultant phasor ER is the vector sum of the individual phasors. At each instant, the projection of ER along the vertical axis represents the time variation of the resultant wave. The phase angle α of the resultant wave is the angle between ER and the first phasor. From Figure 37.9, drawn for four phasors, we see that the phasor of the resultant wave is given by the expression EP = ER sin(ωt + α).
Phasor diagrams for two coherent sources
As an example of the phasor method, consider the interference pattern produced by two coherent sources. Figure 37.10 represents the phasor diagrams for various values of the phase difference φ and the corresponding values of the path difference δ, which are obtained from Equation 37.8. The light intensity at a point is a maximum when ER is a maximum; this occurs at φ = 0, 2π, 4π, . . . . The light intensity at some point is zero when ER is zero; this occurs at φ = π, 3π, 5π, . . . . These results are in complete agreement with the analytical procedure described in the preceding topic. Figure 37.10 Phasor diagrams for a double-slit interference pattern. The resultant phasor ER is a maximum when φ = 0, 2π, 4π, . . . . and is zero when φ = π, 3π, 5π, . . . .
Three-slit interference pattern
Using phasor diagrams, let us analyze the interference pattern caused by three equally spaced slits. We can express the electric field components at a point P on the screen caused by waves from the individual slits as
where φ is the phase difference between waves from adjacent slits. We can obtain the resultant magnitude of the electric field at point P from the phasor diagram in Figure 37.11.
The phasor diagrams for various values of φ are shown in Figure 37.12. Note that the resultant magnitude of the electric field at P has a maximum value of 3E0 , a condition that occurs when φ = 0, ±2π, ±4π, . . . . These points are called primary maxima. Such primary maxima occur whenever the three phasors are aligned as shown in Figure 37.12a. We also find secondary maxima of amplitude E0occurring between the primary maxima at points where φ = ±π, ±3π, . . . . For these points, the wave from one slit exactly cancels that from another slit (Fig. 37.12d). This means that only light from the third slit contributes to the resultant, which consequently has a total amplitude of E0 . Total destructive interference occurs whenever the three phasors form a closed triangle, as shown in Figure 37.12c. These points where ER = 0 correspond to φ = ±2π/3, ±4π/3, . . . . You should be able to construct other phasor diagrams for values of φ greater than π.
Figure 37.12 Phasor diagrams for three equally spaced slits at various values of φ. Note from (a) that there are primary maxima of amplitude 3E0 and from (d) that there are secondary maxima of amplitude E0 .
Figure 37.13 shows multiple-slit interference patterns for a number of configurations. For three slits, note that the primary maxima are nine times more intense than the secondary maxima as measured by the height of the curve. This is because the intensity varies as ER2. For N slits, the intensity of the primary maxima is N2 times greater than that due to a single slit. As the number of slits increases, the primary maxima increase in intensity and become narrower, while the secondary maxima decrease in intensity relative to the primary maxima. Figure 37.13 also shows that as the number of slits increases, the number of secondary maxima also increases. In fact, the number of secondary maxima is always N – 2, where N is the number of slits. Figure 37.13 Multiple-slit interference patterns. As N, the number of slits, is increased, the primary maxima (the tallest peaks in each graph) become narrower but remain fixed in position, and the number of secondary maxima increases.
Phase change produced by reflection
an electromagnetic wave undergoes a phase change of 180° upon reflection from a medium that has a higher index of refraction than the one in which the wave is traveling.
It is useful to draw an analogy between reflected light waves and the reflections of a transverse wave pulse on a stretched string. The reflected pulse on a string undergoes a phase change of 180° when reflected from the boundary of a denser medium, but no phase change occurs when the pulse is reflected from the boundary of a less dense medium. Similarly, an electromagnetic wave undergoes a 180° phase change when reflected from a boundary leading to an optically denser medium, but no phase change occurs when the wave is reflected from a boundary leading to a less dense medium. These rules, summarized in Figure 37.15, can be deduced from Maxwell’s equations, but the treatment is beyond the scope of this text.
Interference in thin films
- A wave traveling from a medium of index of refraction n1 toward a medium of index of refraction n2undergoes a 180° phase change upon reflection when n2 > n1 and undergoes no phase change if n2 < n1
- The wavelength of light λn in a medium whose refraction index is n is
where λ is the wavelength of the light in free space.
Let us apply these rules to the film of Figure 37.16, where nfilm > nair . Reflected ray 1, which is reflected from the upper surface (A), undergoes a phase change of 180° with respect to the incident wave. Reflected ray 2, which is reflected from the lower film surface (B), undergoes no phase change because it is reflected from a medium (air) that has a lower index of refraction. Therefore, ray 1 is 180° out of phase with ray 2, which is equivalent to a path difference of λn/2.
However, we must also consider that ray 2 travels an extra distance 2t before the waves recombine in the air above surface A. If 2t = λn/2, then rays 1 and 2 recombine in phase, and the result is constructive interference. In general, the condition for constructive interference in such situations is
This condition takes into account two factors: (1) the difference in path length for the two rays (the term mλn) and (2) the 180° phase change upon reflection (the term λn/2). Because λn = λ/n, we can write Equation 37.15 as
If the extra distance 2t traveled by ray 2 corresponds to a multiple of λn , then the two waves combine out of phase, and the result is destructive interference. The general equation for destructive interference is
The foregoing conditions for constructive and destructive interference are valid when the medium above the top surface of the film is the same as the medium below the bottom surface. The medium surrounding the film may have a refractive index less than or greater than that of the film. In either case, the rays reflected from the two surfaces are out of phase by 180°. If the film is placed between two different media, one with n < nfilm and the other with n > nfilm , then the conditions for constructive and destructive interference are reversed. In this case, either there is a phase change of 180° for both ray 1 reflecting from surface A and ray 2 reflecting from surface B, or there is no phase change for either ray; hence, the net change in relative phase due to the reflections is zero.
In Figure 37.17, where does the oil film thickness vary the least?
Another method for observing interference in light waves is to place a planoconvex lens on top of a flat glass surface, as shown in Figure 37.18a. With this arrangement, the air film between the glass surfaces varies in thickness from zero at the point of contact to some value t at point P. If the radius of curvature R of the lens is much greater than the distance r, and if the system is viewed from above using light of a single wavelength λ, a pattern of light and dark rings is observed, as shown in Figure 37.18b. These circular fringes, discovered by Newton, are called Newton’s rings. Figure 37.18 (a) The combination of rays reflected from the flat plate and the curved lens surface gives rise to an interference pattern known as Newton’s rings. (b) Photograph of Newton’s rings.
The interference effect is due to the combination of ray 1, reflected from the flat plate, with ray 2, reflected from the curved surface of the lens. Ray 1 undergoes a phase change of 180° upon reflection (because it is reflected from a medium of higher refractive index), whereas ray 2 undergoes no phase change (because it is reflected from a medium of lower refractive index). Hence, the conditions for constructive and destructive interference are given by Equations 37.16 and 37.17, respectively, with n = 1 because the film is air.
The contact point at O is dark, as seen in Figure 37.18b, because ray 1 undergoes a 180° phase change upon external reflection (from the flat surface); in contrast, ray 2 undergoes no phase change upon internal reflection (from the curved surface).
Using the geometry shown in Figure 37.18a, we can obtain expressions for the radii of the bright and dark bands in terms of the radius of curvature R and wavelength λ. For example, the dark rings have radii given by the expression r ≈ √ (m λR/n). We can obtain the wavelength of the light causing the interference pattern by measuring the radii of the rings, provided R is known. Conversely, we can use a known wavelength to obtain R.
A schematic diagram of the interferometer is shown in Figure 37.22. A ray of light from a monochromatic source is split into two rays by mirror M, which is inclined at 45° to the incident light beam. Mirror M, called a beam splitter, transmits half the light incident on it and reflects the rest. One ray is reflected from M vertically upward toward mirror M1 , and the second ray is transmitted horizontally through M toward mirror M2 . Hence, the two rays travel separate paths L1 and L2 . After reflecting from M1 and M2 , the two rays eventually recombine at M to produce an interference pattern, which can be viewed through a telescope. The glass plate P, equal in thickness to mirror M, is placed in the path of the horizontal ray to ensure that the two returning rays travel the same thickness of glass.
The interference condition for the two rays is determined by their path length differences. When the two rays are viewed as shown, the image of M2 produced by the mirror M is at M2’, which is nearly parallel to M1 . (Because M1 and M2 are not exactly perpendicular to each other, the image M2’ is at a slight angle to M1 .) Hence, the space between M2’ and M1 is the equivalent of a wedge-shaped air film. The effective thickness of the air film is varied by moving mirror M1 parallel to itself with a finely threaded screw adjustment. Under these conditions, the interference pattern is a series of bright and dark parallel fringes as described in Example 37.5. As M1 is moved, the fringe pattern shifts. For example, if a dark fringe appears in the field of view (corresponding to destructive interference) and M1is then moved a distance λ/4 toward M, the path difference changes by λ/2 (twice the separation between M1 and M2’). What was a dark fringe now becomes a bright fringe. As M1 is moved an additional distance λ/4 toward M, the bright fringe becomes a dark fringe. Thus, the fringe pattern shifts by one-half fringe each time M1 is moved a distance λ/4. The wavelength of light is then measured by counting the number of fringe shifts for a given displacement of M1 . If the wavelength is accurately known (as with a laser beam), mirror displacements can be measured to within a fraction of the wavelength.