3.38. Diffraction Diffraction is a scale phenomenon that can only be described by use of waves. Consider a plane wave front of wavelength incident on an aperture of width a. How does the wave change when it propagates through the aperture? We can understand the changes by using Huygens’ principle. There are two cases to consider Aperture width a greater than wavelength Aperture width comparable to the wavelength.

3.38. Diffraction Aperture width a greater than wavelength. Here plane wave front of wavelength incident the aperture of width a. In the aperture we can set up many secondary sources. These produce secondary wavelets. In the centre of the aperture the leading edges of the wavelets produces a plane wave front. At the edges, the wavelets form a curved wavefront. Hence the wave front appears to propagate through the aperture as if it were a plane wave in the centre but is only disrupted at the edges.

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3.38. Diffraction Aperture width a comparable to wavelength. Here plane wave front of wavelength incident the aperture of width a. In the aperture we can set up many secondary sources. These produce secondary wavelets. In the centre of the aperture the leading edges of the wavelets produces a plane wave front. At the edges, the wavelets form a curved wavefront. Hence the wave front appears to propagate through the aperture as if it were a plane wave in the centre but is only disrupted at the edges.

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3.39 Interference Consider two light sources that produce harmonic waves of equal amplitude A but equal frequency f. A detector is placed a distance x 1 from source 1 and a distance x 2 from source 2. Source 1 Source 2 Detector, x How does the signal recorded by a detector vary as we vary the distances x 1 and x 2 ? Need to use the principle of linear superposition to obtain the answer. x2x2 x1x1

Let the wave function for source 1 be y 1 (x,t) = Asin(kx 1 -  t +  1 (t)) and for source 2 y 2 (x,t) = Asin(kx 2 -  t +  2 (t)) From principle of linear superposition we have y t (x,t) = y 1 (x,t) + y 2 (x,t) y t (x,t) = Asin(kx 1 -  t +  1 (t)) + Asin(kx 2 -  t +  2 (t)) Which becomes

Thus at x we have a harmonic wave with phase that depends on the average distance of the detector from the sources multiplied by a term that depends solely on the difference in distance between the two sources. Let us assume that the initial phase of each wave is zero. We will discuss the significance of the initial phase later. The displacement at the point x is given by Term that depends on separation of sources Harmonic wave at ave distance

The detector responds to the intensity of the signal. Hence we find If we use the relationship We find

When k(x 1 -x 2 ) = 2nπ (n = integer) the intensity is a maximum When k(x 1 -x 2 ) = (2n+1)π (n = integer) the intensity is zero Thus there is a periodic modulation of the intensity that depends on the difference in distance between the two sources and the detector. The term (x 1 -x 2 ) is known as the path difference.

For an intensity maximum k(x 1 -x 2 ) = 2nπ But k = 2π/ So (x 1 -x 2 ) = n Thus when the path difference is an integer number of wavelengths there is an intensity maximum For an intensity minimum k(x 1 -x 2 ) = (2n+1)π But k = 2π/ So (x 1 -x 2 ) = (n+1/2) Thus when the path difference is a half integer number of wavelengths there is an intensity minimum.

If we now include the initial phase for each of the waves how is the intensity pattern affected? We find that the displacement is given by The intensity becomes

We can understand the effect of the initial phase with reference to the cosine term in the intensity The difference between  1 (t) and  2 (t) remains constant for all time then the interference pattern remains constant. If the difference between  1 (t) and  2 (t) varies in a random fashion then we find that the value of the cosine term fluctuates randomly between +1 and -1. As a result the intensity pattern varies in an unpredictable fashion and the fringes are “washed out”.

If  1 (t) -  2 (t) remains constant for all time the sources are said to be coherent. Example of a coherent source - loud speakers connected to the same frequency source For coherent sources the intensity I(x,t) is given by If  1 (t) -  2 (t) varies randomly as a function of time the sources are said to be incoherent. Example of incoherent source - car headlights For incoherent sources the intensity I(x,t) is given by

3.40 Young’s Slits Experiment: Interference from two sources Two slits s 1 and s 2 are separated by a distance d. A screen is placed a distance L from the slits (L>>d). The n th bright interference fringe is seen a B a height yn from the centre of the screen. The slits are illuminated by a white light source located behind the pinhole p. This ensures that the slits are illuminated by coherent radiation.

3.40 Young’s Slits Experiment: Interference from two sources For the nth bright fringe the path difference between S 2 B and S 1 B must be an integer number of wavelengths. S 1 B = x 1 S 2 B = x 2 S 2 B - S 1 B = x 2 - x 1 x 2 - x 1 = r r = dsina y n = Ltana x 2 - x 1 = n r = n

3.40 Young’s Slits Experiment: Interference from two sources S 1 B = x 1 S 2 B = x 2 S 2 B - S 1 B = x 2 - x 1 x 2 - x 1 = r r = dsina y n = Ltana Now the angle a is small and so sina = tana. Hence r= y n d/L so y n d/L = n Thus y n = Ln /d The spacing between successive fringes is ∆ = L /d

3.41 Single slit diffraction Consider a slit of width a. A plane wave is incident on the slit. A set of secondary sources are set up in the slit. We find that at an angle b the source at 1 is  (  /2) out of phase with the source at 5. The source at 5 is  (  /2) out of phase with the source at 9. Also the source at 5 is at the centre of the slit. So when the condition asinb = m is satisfied destructive interference occurs.