DIFFRACTION AND INTERFERENCE

Transcription

1 DIFFRACTION AND INTERFERENCE We now turn to a consideration of what happens when two light waves interact with one another. We assume that the intensities are low enough that the disturbances add vectorially. The question then becomes whether the two waves are in phase or out of phase. There are basically four ways in which the waves can get out of phase. 1. They start out of phase. They reflect from surfaces 3. They travel different distances 4. They have different frequencies We will consider each of these in turn. WAVES START OUT OF PHASE In general waves coming from different places have random phase relations at the beginning. If they are both coming from the same laser they will begin in phase due to the operation of the laser. If they are coming from very nearby regions of a hot filament they will tend to start in phase. We will consider only cases in which the waves start in phase. WAVES REFLECT FROM SURFACES In general there may be a 18 o phase change upon reflection. This is similar to the effect we encountered with waves on a string, or with sound. Solving the reflection problem with boundary conditions, as noted in class or done in detail in the notes, leads to the conclusion that there will be a 18 o phase change if a wave reflects from a material with a higher index of refraction than that of the one in which it is travelling. For example, a wave moving in air which reflects from a water surface will have a phase change. Going the other way (water to air) it will not. DIFFERENT PATH LENGTHS We now come to the most interesting case. We consider a plane wave incident on a slit of width a as shown in the sketch below.

2 To make the geometry simple we suppose that the light pattern is observed on a screen a distance L from the slit, where L >> a, and h << L. In this case all the rays coming from the slit will be essentially parallel. Then the only path difference is that incurred at the start, as shown in the following sketch. We now consider the path difference between two neighboring pieces of the slit. It will be: We now make use of Huygen s Principle which states that each tiny area on a wave front can be thought of as emitting a spherical wave of amplitude proportional to its area. We can then find the resulting wave at the screen by integration: dy dasinkz t 1/ a z L h x

3 a 1/ a y dxsinkl h x t I~y where we have assumed linearly polarized light. Rather than do this somewhat intimidating integral we can make use of our complex number trick used for AC circuits and EM waves. We write the amplitude produced by a little element of the slit, dx, as: dy dx coskz t dx e 1 ikz t where z is the distance to the observation point. Then for the neighboring element, we have: i kzdx sin z it ikz ikdx sin ikdx sin dy dx e dx e e e dy1e But when you multiply complex numbers you multiply the magnitudes and add the angles. Thus each element, dx, contributes the same length, but has its angle increased by: kdxsin dxsin When we now add them up we get the following picture:

4 Since the dx, and hence the dσ are infinitesimal, the chords become the arc of a circle of radius R as shown. Then the magnitude of E at the observation point becomes: E Rsin But we know that the sum of all the little individual amplitudes must be the total amplitude reaching the slit: E E RR E sin E E sin I I E We now graph this result as I vs β. I sin asin I To get this figure we note that the zeros are at β any integral multiple of π except. At zero we note that sin(β) = β for β << 1. We find the maximums in the usual way: d sin sincos sin tan d 3

5 This gives β = π(1.43,.45, ) or roughly n n 3,5,7, Thus the maximums decrease rapidly with increasing order (1 st max, nd max, 3 rd max, etc). In fact their value is roughly: I 1 4 n 3,5, I n n Thus the result is that the light spreads out over an angle approximately given by: asin sin a Since λ ~ m, whereas most objects around us are of human dimensions (~ ft), we normally see light going in straight lines without significant diffraction. However, we can readily observe the effect if we choose, as seen in class. However there is one important effect of diffraction even when the objects are large a limit on the angular resolution of a lens or mirror. Consider two closely spaced objects at a distance R from the lens or mirror.

6 Each will produce a diffraction pattern at the lens as sketched below. The scale of the diagram is distorted in order to show what is happening. In practice h >> x and hence θ = h/r. The question is then how small θ can be and it still be possible to tell that there are two diffraction patterns rather than one. This of course depends on the sensitivity of the detector, but a reasonable estimate is that the minimum separation is when the first minimum of one corresponds with the central maximum of the other. Then we have: sin min 1 min a a As an example of this effect recall our discussion of the eye. We found that the size of the retinal cells limited the angular resolution to ~ rad. The size of the pupil is variable but ~ 3mm. Then for visible light we find: min ~ 1 rad 3 31 Hence both resolution limits are about the same. Of course they are. There would be no evolutionary advantage is having one much different than the other. As a second example consider trying to determine whether a given star system is single or binary. Suppose the system is on the other side of the galaxy and hence ~ 5 light-years away. Then with a telescope of diameter m we could resolve a separation of: d R R a a d m By comparison, the distance from earth to sun is m. Now consider trying to do it for a system in Andromeda which is about million light years away. Then:

7 d m 4 51 This is about half the distance to the nearest star to our sun. INTERFERENCE We now turn to a consideration of what happens when we have two or more slits. Suppose we have N slits of width a and separation d. Then the situation is as shown below. Again we suppose that L >> h so that the rays are parallel. We now proceed exactly as before with complex numbers. where E is the amplitude produced by a single slit. The total is then given by: N1 i i N1 i in n E E E e E e E e E e But this is just a geometric series in E E e 1 e i 1 We need the magnitude of E

8 in in in in e 1 e 1 11e e i i i i E E E E E e 1 e 1 11e e cosn 1cosN E E cos 1cos But cos cos sin 1sin Thus N sin N sin E E E sin sin where dsin But E sin sin sin N E sin where α is a constant. Then sin I A The intensity at the center of the pattern (θ = ) is Hence sin N sin N I I A AN A N

9 I 1 sin sin I N sin N Again it is useful to graph this result. Clearly it is the product of two factors the diffraction of one slit, and the interference of N slits. The interference pattern by itself would look like: I 1 sin N I N sin The numerator is when Nγ = nπ, n =,1,, or γ = nπ/n. The denominator is when γ=nπ, n =,1,,3, When both are we get I 1 N I N 1 Hence at γ = nπ we will get Suppose N = 3. Then the numerator is at,,,, 3 3 Hence there are zeros for I/I between the maximums at and π

10 Hence there must be a maximum between π/3 and π/3. In fact, it will be at γ = π/. The others will be at 3π/, 5π/ Thus the graph is I 91 9 In general there will be N- sub peaks with intensity 1/N the intensity of the main peaks. We saw all of these effects in the demonstrations in class. Note that as N becomes very large, the amplitude of the secondary maximums becomes very small and ultimately we simply get an series of equal intensity maximums. Of course this curve is now multiplied by the diffraction factor. The result is an envelope modulating the interference curve, as shown below.