7 Interference and Diffraction

Transcription

1 Physics 12a Waves Lecture 17 Caltech, 11/29/18 7 Interference and Diffraction In our discussion of multiple reflection, we have already seen a situation where the total (reflected) wave is the sum of many components. The different components can have different phase shifts between them, resulting in different amplitude for the total wave. If all the components are in phase, then they interfere constructively, giving rise to a large total wave; if the components are out of phase with each other, they interfere destructively, giving rise to a small total wave. In general, this kind of phenomena is called interference. In this section, we are going to discuss how interference can lead to different interference patterns and also diffraction. 7.1 Cylindrical and spherical wave Recall that in 1D, to generate a wave that travels in the +x direction, we can drive the x = 0 point to oscillate as e iωt. The wave generated is then e i(ωt kx). Similarly, in 2D and 3D, to generate a plane wave that travels in the +x direction, we can drive the x = 0 line / plane to oscillate as e iωt and the wave generated would be e i(ωt kx), where k = ω/v. More generally, we can drive the x = 0 line / plane as e i(ωt kyy) / e i(ωt kyy kzz). The wave generated is still a plane wave, taking the form e i(ωt kxx kyy) / e i(ωt kxx kyy kzz), where k x = k 2 k 2 y / k x = k 2 k 2 y k 2 z, k = ω/v. t a particular t, if we collect all the points with the same phase, we get parallel planes with distance λ = 2π k between them. Such planes are called wavefronts in the plane wave. What if we drive only one point x = 0, y = 0, (z = 0) as e iωt? What kind of wave is generated? In 2D, if we drive x = 0, y = 0 as e iωt, a cylindrical wave is generated which takes the form ψ( x, t) = r e i(ωt kr) (1) where r = x 2 + y 2. The wavefront are circles with distance λ = 2π k between them. In 3D, if we drive x = 0, y = 0, z = 0 as e iωt, a spherical wave is generated which takes the form ψ( x, t) = r ei(ωt kr) (2) where r = x 2 + y 2 + z 2. The wavefronts are spheres with distance λ between them. 1 r and 1 r The scaling of the amplitude is to satisfy energy conservation: the energy flux (energy flow per unit time and unit area) is proportional to amplitude squared in a wave. The surface area across which energy flows scales as r in 2D and r 2 in 3D. Therefore, to ensure energy conservation, the amplitude scales as 1 r in 2D and 1 r in 3D. 1

2 7.2 Single slit and diffraction Consider a wall with a slit. Suppose we send a plane wave in the +x direction. It hits the wall at x = 0. If the slit is infinitely small, it forms a point source (in 2D) for wave in the region x > 0. The wave generated is, at x > 0, The intensity is uniform in all directions! ψ(x, y, t) = r e i(ωt kr) (3) This is the phenomena of diffraction. Light after passing through the slit does not travel purely in the +x direction any more. This is a strong manifestation of the wave nature of light. In reality, we do expect that the angular spread of the wave after passing through the slit to be finite. This is because every slit has a finite width L. What is the intensity distribution pattern for diffraction at a finite width slit? Imagine dividing the slit into N pieces. s N, each piece generates a cylindrical wave. The total wave is then a superposition of waves from each piece of the slit. ψ(x, y, t) = L 2 L 2 L dỹ 1 e i(ωt kr ỹ) rỹ Here (0, ỹ) is the location of the piece in the slit, (x, y) is where we are measuring the wave, rỹ = x 2 + (y ỹ) 2. This is a complicated integral, but let s consider the case of large x. When x is large, r >> y, ỹ, rỹ x 2 + y 2 2yỹ = r 1 2yỹ yỹ r 2 r(1 r 2 ) r ỹ y (5) x 1 The variation in amplitude due to the factor rỹ is not important, we can take it simple to be a 1 uniform factor r. The difference in phase due to krỹ, on the other hand, is very important and cannot be ignored. (4) With these simplifications, the integral reduces to ψ(x, y, t) = L 2 L 2 = L dỹ 1 r e i(ωt kr) e ikỹy/x L rl(iky/x) e i(ωt kr) e ikỹyx 2 = rlk e kl i(ωt kr) sin( y) y L 2 (6) 2

3 The time averaged intensity is then proportional to I sin2 ( kl y) y 2 (7) From this formula (and the plot), we see that y = 0 is an intensity maximum, I max k2 L 2 4x 2. The intensity becomes zero at regularly spaced points kl 2nπx y = nπ, (n 0) y = kl = nλ x, (n 0) (8) L This is the so-called diffraction pattern of a single slit. The central peak is twice as wide as that of the side peaks. t large y, the envelope of the diffraction patterns decays very quickly as 1. s y 2 L 0, the central peak becomes infinitely wide and we recover the point source limit. In this analysis of single slit diffraction pattern, we have used the Huygens principle, which states that every point on a wave front (the slit) is itself the source of cylindrical / spherical waves. The total wave is then the superposition of all the component waves. We are going to use this principle to analyze more complicated situation in the following. 7.3 Double slits Infinitely thin double slits Now consider two small slits on a plane x = 0, with a distance a between them. If we shine a plane wave in the x direction, what would we observe on a plane at x = x 0? 3

4 Each slit makes a cylindrical wave. t a point (x, y), the total wave is a superposition of the two e i(ωt kr1) + e i(ωt kr 2) r1 rr (9) We can ignore the difference in amplitude, but the difference in phase is important. If we make the approximation that r 1 = r + y a x 2, r 2 = r y a (10) x 2 Then the total wave becomes ( ) e i(ωt kr) e ik y a x 2 + e ik y a x 2 = 2 ( e i(ωt kr) cos k y a ) (11) r r x 2 The time averaged intensity is then proportional to ( ) ka I cos 2 y (12) In this interference pattern, the intensity reaches its peak at y = nλx a. This is easy to understand: at the peaks, the optical path difference of the two waves are a y x = nλ (13) resulting in a phase difference of n2π. The two waves interference constructively and we see a peak in the intensity. Half way between the peaks, where y = (2n+1)λx 2a, the two waves have an optical path difference of half wave length, resulting in a π phase difference. The two wave interference destructively and we see a zero in the intensity. ll peaks are regularly spaced, with the same width and the same height Finite width double slits What if each of the double slits has finite width L? For a slit with width L, the wave that emanates from each slit is not a simple cylindrical wave. Instead, it makes the diffraction pattern of the form y) ψ(x, y, t) = e i(ωt kr) sin ( kl rlk y (14) 4

5 When we have two slits, centered at ± a 2 respectively, they interfere as ψ + (x, y, t) + ψ (x, y, t) = r+ Lk ei(ωt kr +) sin ( kl (y a/2)) + y a/2 r Lk ei(ωt kr ) sin ( kl (y + a/2)) y + a/2 (15) When x >> L, a, we can approximate r + as r a y 2 x and r as r + a y 2 x. Similar to the previous case, we ignore the difference in amplitude of the two waves so that r+ Lk sin ( kl (y a/2)) y a/2 r Lk sin ( kl (y + a/2)) y + a/2 But we need to take into account the difference in phase factor so that sin ( kl (y)) rlk y ψ + (x, y, t) + ψ (x, y, t) sin ( kl (y)) ( ) ka e i(ωt kr) cos rlk y y The total intensity is then proportional to ( I sin2 kl (y)) ( ) ka y 2 cos 2 y (16) (17) (18) That is, the interference pattern is modulated by the diffraction pattern Many slits If there are more than 2 slits, what is the interference pattern like? First let s assume that the slits are infinitely narrow and ignore diffraction. When the path difference between different slits is integer multiples of λ, then the waves from different slits interfere constructively with each other, giving rise to a peak in the interference pattern. If we ignore the difference in the amplitude of the waves, then the intensity distribution on the screen is proportional to I ( e ikay/ + e ikay/ + e 3ikay/ + e 3ikay/ e (2N 1)ikay/ + e (2N 1)ikay/) 2 where a is the distance between neighboring slits and there are in total 2N slits. If we make a plot of the intensity vs. ya/xλ, we see that (19) 5

6 the (highest) peak locations are the same, but with more slits, the peaks become sharper. useful optical device is made by cutting a large number of very thin slits into an opaque plane. This is called the diffraction grating. n important function of diffraction grating is to separate light of different wavelength. In particular, the location of the second, third... maximums depend (is proportional to) on the wavelength λ. Therefore, if white light shines perpendicularly on the grating, the central peak is still white, but the second, third etc. peak would be rainbow like: different wave lengths have peak at different locations. Now if we take into consideration the fact that the slits have finite width, the interference pattern shown above is going to be modulated by the diffraction pattern of each individual slit. 6