Interference by Wavefront Division

Transcription

1 nterference by Wavefront Division One of the seminal experiments in physics was conducted in 1801 by Thomas Young, an English physicist who cut a small hole in an opaque screen, set a second screen in front of it in which two small holes were cut about a mm apart, and placen observing screen in front of that one. The apparatus is illustrated below. The purpose of the first screen was to make certain that light waves entering the two pinholes came from a common source so that light of the same phase entered the two pinholes of the second screen. Each of these pinholes then acts as a source of spherical outgoing waves. These waves recombine on the observing screen producing a distribution of bright and dark regions called an interference pattern. The experiment is easier to carry out and to interpret if instead of pinholes one uses thin slits as shown. 13.1

2 Experiment 13.1 : nterference from Two Thin Slits Apparatus: Calibrated slits, He-Ne laser. Procedure: Mount the calibrated slit holder on a stannim a He-Ne laser at the slits. An interference pattern will be produced on a screen a few meters away. This pattern is large enough for quantitative information to be gathered. Describe this pattern qualitatively. Then tape a sheet of white paper to the screen and mark where the intensity minima are found. Use a ruler to measure the linear separation between points of minimum intensity and convert these into angular separations as measured from the slit screen. Next, change the slit-to-slit distance and repeat the above measurements. Do this for a total of at least three different slit spacings. s there a relationship between the angular widths of the bright regions and the slit spacings? f so, can it be made quantitative? Finally, repeat the above set of measurements using lasers of different wavelengths. Do this for a total of at least three different wavelengths. s there a relationship between the angular widths of the bright regions and the wavelengths of the light used to produce them? f so, can it be made quantitative? Question: Use your quantitative relation to predict the angular separation of the bright regions in a situation that has not been studied directly. Then go and test it. Question: Does your quantitative relation account for all the intensity minima that you see? f not, which ones are missed? Can you think of an explanation for these? Question: Unlike Thomas Young s experiment, we did not have a first screen lying between the incident light and the slit-screen. Why not? 13.

3 A simple explanation of the two slit interference pattern is given by considering the two waves emerging from the slits. These waves travel distances r 1 and r respectively to reach a point P on the observing screen. Let D denote the distance from the slit-screen to the observing screen, d the separation between the centers of the slits, and s the distance of P from the midplane of the two slits. Assume that the slits are very thin and that D >> d. Again assume for convenience that the field is polarized perpendicular to the plane of figure 3. The distance from slit #1 to P is (13.1) r 1 = [D + (s - d/) ] = [D + s - ds + d /4 ] Define (13.) r [D + s ] = distance of P from the midline of the slits Then (13.3) r 1 = [r - ds + d /4] = r [1 - d s/r + d /4r ] r [1 - ds/r + d /8r ] = r - ds/r + d /4r Similarly, (13.4) r = [D + (s + d/) ] r + ds/r + d /4r so that (13.5) r - r 1 ds/r = d sin Light from slit # takes longer to reach P than light from slit #1, the time difference being (13.6) t = (r - r 1 )/c (d/c) sin The phase difference between the light arriving at P from the two slits is (13.7) = t ( d/c) sin = kd sin = ( d/ ) sin t will be bright at P if the two waves are in phase, that is if = ±, ±4, ±6,..., or = m where m is an integer. The condition for brightness at P is therefore (13.8) d sin = ± m (m = 0, 1,, 3,...) Similarly, it will be dark at P if the phase difference is ±, ±3, ±5,..., or = (m + ½ ). The condition for darkness at P is (13.9) d sin = ±(m + ½) (m = 0, 1,, 3,...) The waves from the two slits are said to interfere constructively if destructively if = (m + ½). = m, to interfere 13.3

4 ntensity distribution in Two-Slit nterference Huygens principle gives us a way of computing the intensity distribution in the interference pattern. Referring to the previous section, the wave amplitude at P is given by (13.10) E(P,t) E(slit #1, t - r 1 /c) / r 1 + E(slit #, t - r /c) / r The electric fielt the position of the slits is given by E(t) = E 0 e - j t. Also, we assume that D is so large that 1/ r 1 1/ r, so that ω E( P, t) E e + E e j ( t r1 / c) jω ( t r / c) 0 0 jω ( t r1 / c) jω ( r r1 )/ c (13.11) = E0 e [ 1 + e ] jω ( t r1 / c) j k d sinθ = E0 e [ 1 + e ] The time-averaged intensity is therefore j k d sinθ (13.1) ( P) 1 + e 1+ cos( k d sin θ) i.e. 0 d sin (13.13) ( P) = + cos 1 π θ λ where 0 is the intensity at = 0. Problem 1: Show that (13.13) implies an intensity maximum whenever d sin = ±m where m = 0, 1,,... ann intensity zero wherever d sin = ±(m + ½). What does this imply about the angular width of the bright fringes? Problem : Graph the distribution given by equation (13.14) for = 63.8m, d = 0.4mm. Be sure to choose an appropriate scale. What About Diffraction? No optical experiment is free of the effects of diffraction. We have treated the slits as though they were indefinitely thin and therefore the source of outgoing cylindrical waves. This cannot be the whole story since the slits must have some finite width and these will cause diffraction effects to be present. Let the slits each have a width a and let the distance between the centers of the slits be d. Then in the usual Fraunhofer limit we have d + a jω( t r / c) 1 jω( t r / c) (13.14) E( P, t) e dy + e dy d + a where r 1 and r are both given by [D + (s - y) ] [D + s - sy] ½ r - y sin. Therefore, 13.4

5 (13.15) j k y sinθ y = + jωt e E( P, t) e + j k sinθ y = e j k j k y sinθ y = + sinθ y = j t j k d ka j k d ω sinθ sinθ ka e e sinc( sin θ) + e sinc( sin θ) jωt kd ka e cos( sin θ) sinc( sin θ) Define (13.16) u ½ k a sin and w ½ k d sin Then the time-averaged intensity is therefore (13.17) ( ) ( ) ( P) = cos w sinc u = cos 0 0 π d sinθ sinc λ πa sinθ λ Equation (13.17) shows that the intensity pattern consists of the product of a double slit pattern an single slit diffraction pattern. This is illustrated below for d = 10a, a =.01mm, = 500nm. The effect of diffraction is to modulate the double slit interference pattern by a diffraction envelope. The interference fringes appear at the same positions but their intensity decreases with angle away from the central maximum and the intensity vanishes at angles where the single slit diffraction pattern is zero. 13.5

6 Fringe Visibility Suppose that the two slits are not equally illuminated. What effect is this likely to have on the intensity distribution at the viewing screen? We can model this problem by again assuming very narrow slits and rewriting (13.11) as (13.0) ω E( P, t) E e + E e 1 { } ω = e E e + E e j ( t r1 / c) jω ( t r / c) 1 j t j k r j k r 1 Assuming E 1 and E are real, the intensity distribution is (13.1) j k ( r1 r ) ( P) E1 + E + E1 E Re{ e } = E + E + E E cos k d 1 1 ( sinθ) Problem: Verify that the above equation reduces to (13.13) when E 1 = E. The maximum intensity occurs when the value of the cosine in (13.1) is equal to one: (13.) max = E 1 + E + E 1 E The minimum intensity occurs when the cosine is equal to -1: (13.3) min = E 1 + E - E 1 E Write 1 = ½ 0 ce 1 and = ½ 0 ce so that max = ( 1 ) and min = ( 1 ) There are many other experiments that produce interference fringes. n any of these experiments, one defines the fringe visibility V as (13.4) V = max max + min min so that if the minimum intensity is zero, V will be equal to 1 (perfect visibility) while if the intensity is uniform, V will equal 0 (fringes not visible). n the particular case of two-slit interference with unequally illuminated slits we have (13.5) V = ( 1 ) ½ / ( 1 + ) Problem 3: Give a physical interpretation of the quantities 1 and. Problem 4: Show that the fringe visibility given by (13.5) attains its maximum value when 1 = and give an explanation for this. 13.6

7 13.7