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.

a >

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.

a ~

3.39 InterferenceConsider two light sources that produce harmonic

waves of equal amplitude A but equal frequency f. A detector is placed a distance x1 from source 1 and a distance x2 from source 2.

Source 1

Source 2 Detector , x

How does the signal recorded by a detector vary as we vary the distances x1 and x2?

Need to use the principle of linear superposition to obtain the answer.

x2

x1

• Let the wave function for source 1 be • y1(x,t) = Asin(kx1-t + 1(t)) • and for source 2• y2(x,t) = Asin(kx2-t + 2(t)) • From principle of linear superposition we have• yt(x,t) = y1(x,t) + y2(x,t)• yt(x,t) = Asin(kx1-t + 1(t)) + Asin(kx2-t + 2(t))• Which becomes

€

yt x,t( ) =2Asink x 1 +x 2( )

2−t +

φ1 t( )+φ2 t( )

2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟X

cosk x 1 −x 2( )

2+φ1 t( )−φ2 t( )

2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

• 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

€

yt x,t( ) =2Asink x 1 +x 2( )

2−t

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟cos

k x 1 −x 2( )

2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

The detector responds to the intensity of the signal.

€

I x,t( ) =4A2 sin2k x 1 +x 2( )

2−t

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟cos2

k x 1 −x 2( )

2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

€

I (x,t ) ∝ yt

2

x,t( )

€

cos x( ) =2 cos2 x

2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟−1

Hence we find

If we use the relationship

We find

€

I x,t( ) =2A2 sin2k x 1 +x 2( )

2−t

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟1+cosk x 1 −x 2( )( )( )

When k(x1-x2) = 2nπ (n = integer) the intensity is a maximumWhen k(x1-x2) = (2n+1)π (n = integer) the intensity is zeroThus there is a periodic modulation of the intensity that depends on the difference in distance between the two sources and the detector.The term (x1-x2) is known as the path difference.

The term (x1-x2) is known as the path difference.

For an intensity maximum k(x1-x2) = 2nπBut k = 2π/So (x1-x2) = nThus when the path difference is an integer number of wavelengths there is an intensity maximum

For an intensity minimum k(x1-x2) = (2n+1)πBut k = 2π/So (x1-x2) = (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

€

yt x,t( ) =2Asink x 1 +x 2( )

2−t +

φ1 t( )+φ2 t( )

2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟X

cosk x 1 −x 2( )

2+φ1 t( )−φ2 t( )

2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

• The intensity becomes

€

I x,t( ) =2A2 sin2k x 1 +x 2( )

2−t +

φ1 t( )+φ2 t( )

2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟X

1+cosk x 1 −x 2( )+ φ1 t( )−φ2 t( )( )( )( )

• 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”.

€

1+cosk x 1 −x 2( )+ φ1 t( )−φ2 t( )( )( )( )

• 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

€

I x,t( ) =y1 x,t( )+y2 x,t( )2

€

I x,t( ) =I 1 x,t( )+ I 2 x,t( )

s1

s2

x1

x2dL

yn

aa

r

p

B

3.40 Young’s Slits Experiment: Interference from two sources

Two slits s1 and s2 are separated by a distance d. A screen is placed a distance L from the slits (L>>d).The nth 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.

s1

s2

x1

x2dL

yn

aa

r

p

B

3.40 Young’s Slits Experiment: Interference from two sources

For the nth bright fringe the path difference between S2B and S1B must be an integer number of wavelengths.

S1B = x1

S2B = x2

S2B - S1B = x2 - x1

x2 - x1 = rr = dsinayn = Ltana

x2 - x1 = nr = n

s1

s2

x1

x2dL

yn

aa

r

p

B

3.40 Young’s Slits Experiment: Interference from two sources

S1B = x1

S2B = x2

S2B - S1B = x2 - x1

x2 - x1 = rr = dsinayn = Ltana

Now the angle a is small and so sina = tana.Hence r= ynd/Lso ynd/L = nThus yn = Ln/dThe spacing between successive fringes is∆ = L/d

3.41 Single slit diffraction

ab

b

λ

1

5

9

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.