Problem

Let us consider two slits \(\mbox{F}_1\) and \(\mbox{F}_2\) made on an opaque screen that are infinitely narrow, vertical, and spaced by \(a=2~\mbox{mm}\). They are illuminated by a point source \(\mbox{S}\), monochromatic with a wavelength of \(\lambda_0 = 550~\mbox{nm}\). The source slit \(\mbox{S}\) is equidistant to \(\mbox{F}_1\) and \(\mbox{F}_2\). The interference pattern is observed on a screen \(\mbox{E}\), which is parallel to the slits and \(d_0= 1~\mbox{m}\) away from them.

Question

Calculate the illumination on screen \(\mbox{E}\). Determine the interfringe distance and the contrast.

Solution

The set-up of the model and the notations used are shown in figure 25.

Let us take the case of a two wave interferometer with perfectly coherent light, since the source is punctual and monochromatic, the distribution of illumination is:

I = I 0 2 ( 1 + cos ( 2 π δ λ ) ) I = {I_0}over{2} left( 1+ "cos" left ( {2 %pi %delta} over {%lambda} right ) right)

In the case of Young's slits, the interfringe distance is expressed as:

i = λ d 0 a = 550 × 10 − 9 × 1 2 × 10 − 3 = 0.275 mm i= {%lambda d_0} over a = {550 times 10^{-9} times 1} over {2 times 10^{-3}} = 0.275 "mm"

Since the source is perfectly coherent, the maximum contrast is equal to 1.

Question

How does the preceding result change if \(\mbox{S}\) is replaced by a vertical, infinitely narrow slit \(\mbox{F}_0\) that is parallel to \(\mbox{F}_1\) and \(\mbox{F}_2\) ?

Solution

If the source \(\mbox{S}\) is replaced by an infinitely narrow vertical slit, the different source points are incoherent amongst themselves and the light is partially spatially coherent. The optical path difference for a point \(\mbox{M}\) on the screen is given by the relation in paragraph D.3 of this course (Evaluation of the Optical Path Difference):

δ ( S ' ) = δ 0 + 1 d 1 ( u 1 ⃗ − u 2 ⃗ ) . d S ⃗ %delta (S') = %delta_0 + {1} over {d_1} left(vec{u_1}-vec{u_2} right). d vec{S}

Here the vectors \(\overrightarrow{u_1} - \overrightarrow{u_2} = \overrightarrow{\mbox{F}_2 \mbox{F}_1}\) and \(d\overrightarrow{S}\) are orthogonal (figure 26).

Consequently \((\overrightarrow{u_1} - \overrightarrow{u_2}) \times d\overrightarrow{S} = 0\) and the degree of coherence is equal to 1, the contrast does not diminish.

Question

What can be observed if we introduce a constant phase shift \(\varphi_0\) on one of the trajectories? At \(\mbox{F}_1\) we interpose a very thin glass plate with parallel faces, of index \(n=1.5\). When this plate is placed on screen \(\mbox{E}\), all of the fringes translate by \(t= 2.9~\mbox{mm}\).

  1. What is the direction of this translation?

  2. How many fringes are affected by this?

  3. Deduce the thickness \(e\) of the glass plate.

Solution

A given fringe corresponds to a given optical path difference \(\delta\) If we interpose a glass plate in the trajectory of \(\mbox{F}_1\) (1) the optical path increases because the index of the plate is superior to the index of the air. In order to compensate for this difference, the path length of the trajectory (2) must increase so that the value of the optical path difference does not change.

  1. Therefore, the fringes pass through the side where the plate has been inserted.

  2. The \(2.9~\mbox{mm}\) shift is caused by the passing of the fringes:

    \(k=\dfrac{t}{i}=\dfrac{2.9}{0.275}=10.54\) because a spatial period corresponds to an interfringe distance.

  3. When the glass plate is interposed, the optical path difference varies because on a thickness of \(e\) the air index 1 is replaced by the index \(n\) of the glass. The variation is thus: \(\Delta \delta = (n-1) e\)

    The passage from one fringe to the next of the same nature (dark or bright) corresponds to a variation in the optical path difference of \(\lambda\), from which: \(e = \dfrac{k \lambda }{n-1} = \dfrac{10.54 \times 0.550}{1.5 -1} = 11.6~ \mbox{μm}\)

Question

We replace the plate with an absorbing screen with an optical density of \(2\) (\(\mbox{DO} = 2\)). Determine the new contrast.

Solution

Both slits do not have the same intensity, so we must split the current relation:

I = I 1 + I 2 + 2 I 1 I 2 cos ( 2 π δ λ ) I = I_1+ I_2 + 2 sqrt{I_1 I_2} "cos" left ( {2 %pi %delta} over {%lambda} right )

The relation between both intensities verifies:

DO = − log 10 ( I 1 I 2 ) "DO" = -log_{10} left ({I_1} over {I_2} right)

from which:

I 2 = 100 I 1 {I_2} = 100 I_1

i.e.:

I = I 1 [ 101 + 20 cos ( 2 π δ λ ) ] I = I_1 left [ 101 + 20 "cos" left ( {2 %pi %delta} over {%lambda} right ) right ]

which produces a contrast of:

C = I max − I min I max + I min = 121 − 81 121 + 81 = 0.2 C = {I_{"max"}-I_{"min"}} over {I_{"max"}+I_{"min"}} = {121-81} over {121+81} = 0.2

Question

We remove the density filter and now use a spatially incoherent source.

Describe the phenomenon that is observed if the source is polychromatic (white light).

Solution

The case of a white light source is the same as the one described in the case study in paragraph II.B "Case of white light"

Question

The source is \(d_1 = 100~\mbox{mm}\) away from \(\mbox{F}_1\) and \(\mbox{F}_2\) and has a width of \(e= 6.8~\mbox{μm}\). What happens to the light at point \(\mbox{M}\) on screen \(\mbox{E}\)? How does the visibility vary with \(e\) ? What is the maximum width for which the decrease in contrast remains below \(10\%\) ?

Solution

The sources still have the same intensity, but the source is no longer spatially coherent because it has a width of \(e\). The set-up and notations used are represented in 27.

The interference fringes have a weaker contrast. The intensity distribution on the screen is given by:

I = I 0 2 ( 1 + ℜ [ L ~ ( σ 0 α , σ 0 β ) L ~ ( 0 ,0 ) ] cos ( 2 π σ 0 δ 0 ) ) I = {I_0 over 2} left(1+ {Re left [ {tilde{L} left(σ_0 %alpha,σ_0 %beta right) } over {tilde{L} left(0,0 right)} right ]} cos left( 2 %pi σ_0 %delta_0 right)right)

\(\widetilde{L}\) represents the Fourier transform of the source's brightness function.

The source's spectral distribution of luminance can be described by a rectangle function with a width of \(e\) :

L ( X , Y ) = Π e ( X ) L (X,Y) =%PI_{e} left(X right )

for which the Fourier transform is:

L ~ ( σ 0 α ,0 ) = e sinc ( e σ 0 α ) tilde{L} left(σ_0 %alpha,0 right) = e` "sinc" left ( e σ_0 %alpha right )

where \(\sigma_0\) is the wave number of the source and \(\alpha = a/d_1\) is the angle from which the slits are seen from the source center.

Consequently, the degree of spatial coherence is equal to:

γ c , s = ℜ [ L ~ ( σ 0 α ,0 ) L ~ ( 0 ,0 ) ] = sinc ( e a λ 0 d 1 ) = sinc ( 2 × 6.8 × 10 − 3 0.555 × 10 − 3 × 100 ) = 0.9 %gamma_{c,s} = {Re left [ {tilde{L} left(σ_0 %alpha,0 right) } over {tilde{L} left(0,0 right)} right ]} ="sinc" left ({ e a} over {%lambda_0 d_1} right ) ="sinc" left ({ 2 times 6.8 times 10^{-3}} over {0.555 times 10^{-3} times 100} right ) = 0.9

Question

We use as source a grid made up of slits that are parallel to \(\mbox{F}_1\) et \(\mbox{F}_2\). Determine the grid spacing and the width of the slits for which the contrast remains of the preceding value.

Solution

First, let's consider a single source slit that has been shifted on the plane \((X , Y)\) by a quantity \(p\) following the \(\mbox{X}\) axis. For a point \(\mbox{M}\) on the screen the optical path difference varies by \(\Delta \delta = a p /d_1\). The calculation is analogous to the one in the Case Study. If this optical path difference is equal to \(k \lambda\) then the figure of interference is identical to the preceding one. The figure is translated downwards by the equivalent of one interfringe distance. A bright fringe (respectively dark) has replaced the following bright fringe (respectively dark).

Consequently, the single source slit can be replaced by a lattice:

p = λ d 1 a = 0.550 × 100 2 = 27.5 μ m p = {%lambda d_1} over a = {0.550 times 100} over 2 = 27.5 %mu "m"

Question

The source is once again a slit reduced to \(\mbox{F}_0\) but the source of monochromatic light is replaced by a sodium vapor source, which has a spectrum essentially made up of two radiations with a wavelength of \(\lambda_1 = 589.3~\mbox{nm}\) and \(\lambda_2 = 589.9~\mbox{nm}\).

Find the energy spectral density function \(S_S (\sigma)\) in terms of the average wave number \(\sigma = (\sigma_1 + \sigma_2 ) /2\) and the deviation of the doublet \(\Delta \sigma = \sigma_1 - \sigma_2\).

Find the illumination normalized for screen \(\mbox{E}\).

Solution

The source is now spatially coherent, but partially temporally incoherent.

As seen in paragraph C.3.b ("Examples" in Temporal Coherence), the energy spectral density is expressed as:

S S ( σ ) = I 0 δ ( σ − σ 0 − Δ σ 2 ) + I 0 δ ( σ − σ 0 + Δ σ 2 ) {S}_S left(σ right) = I_0 bold{%delta}left (σ-σ_0-{{%DELTA σ}over{2}} right )+I_0 bold{%delta}left (σ-σ_0+{{%DELTA σ}over{2}} right )

and we have (see paragraph C.3.b "Examples"):

I ( δ ) = I 0 2 [ 1 + cos ( π Δ σ δ ) cos ( 2 π σ 0 δ ) ] I left(%delta right) = { I_0} over 2 left [ 1+ cos left (%pi %DELTA σ %delta right ) cos left( 2 %pi σ_0 %delta right) right ]

Question

Carefully represent the previous function.

Solution

Graph of the intensity distribution on the screen: the interferogram has beats and a similar structure to the one shown on figure 28.

Question

What is the value of the interfringe?

Solution

The interferogram of the doublet is also written as:

I ( δ ) = I 0 2 [ 1 + cos ( π Δ σ δ ) cos ( 2 π δ λ 0 ) ] I left(%delta right) = { I_0} over 2 left [ 1+ cos left (%pi %DELTA σ %delta right ) cos left( 2 %pi %delta over %lambda_0 right) right ]

The inferfringe distance is thus given by:

i = λ 0 d 0 a i = {%lambda_0 d_0} over a

with \(\lambda_0 = 1/ \sigma_0 = 2 \lambda_1 \lambda_2 / (\lambda_1 + \lambda_2) = 589.3~\mbox{nm}\).

Therefore, we have:

i = 0.5893 × 10 − 3 × 1000 2 = 0.294 mm i = {0.5893 times 10^{-3} times 1000 } over 2 = 0.294 "mm"

Question

For which values of the optical path difference is there contrast cancellation?

Solution

The contrast function \(\gamma_{c,~t) = cos( \pi \Delta \sigma \delta)\) cancels out when the cosine function is equal to \(\pi /2 + k \pi\), i.e.:

δ = 1 2 Δ σ + k Δ σ %delta = 1 over {2 %DELTA σ } + k over {%DELTA σ }

Question

A source emits a spectral line with an energy spectral density function that can by described by a gaussian shape:

S S ( σ ) = I 0 exp ( − κ ( σ − σ 0 ) 2 Δ σ 2 ) {S}_S left(σ right)= {I_0} exp left ( -%kappa{left (σ-σ_0 right )^2} over {%DELTA σ}^2 right )

where \(\kappa = -4 \ln 2\), \(\sigma_0\) is the average wave number. The line is centered on the wavelength where \(\Delta \sigma\) is the width at mid-height \(\lambda_0 = 546.1~\mbox{nm}\). The source has a spectral width of \(\Delta \lambda = 50~\mbox{nm}\).

What is the value of \(\Delta \sigma\) ?

Solution

Same procedure as the previous excercise.

The value of \(\Delta \sigma\) is given by:

Δ σ = Δ λ λ 2 = 50 × 10 − 9 ( 546.1 × 10 − 9 ) 2 = 1.676 × 10 3 cm − 1 {%DELTA σ } = {%DELTA %lambda } over {%lambda^2} = {50 times 10^{-9}} over {(546.1 times 10^{-9})^2} =1.676 times 10^{3} "cm"^{-1}

Question

Find the normalized illumination for screen \(\mbox{E}\).

Calculate the coherence length of the source.

Solution

The energy spectral density is:

S S ( σ ) = I 0 exp ( − κ ( σ − σ 0 ) 2 Δ σ 2 ) {S}_S left(σ right)= {I_0} exp left ( -%kappa{left (σ-σ_0 right )^2} over {%DELTA σ}^2 right )

The spectral shape is thus:

S 0 ( σ ) = I 0 exp ( − κ σ 2 Δ σ 2 ) S_0 left(σ right) = {I_0} exp left ( -%kappa{σ^2} over {%DELTA σ}^2 right )

The Fourier transform of \(\mbox{S}_0 (\sigma )\) is:

S 0 ~ ( δ ) = π Δ σ 2 κ I 0 exp ( − ( π Δ σ δ κ ) 2 ) tilde{S}_0 left(%delta right) = {sqrt{%pi} %DELTA σ^2} over {sqrt%kappa} {I_0} exp left ( -left ({{%pi %DELTA σ %delta} over {sqrt{%kappa}}}right )^2 right )

The degree of coherence is:

γ c , t ( δ ) = ℜ [ S 0 ~ ( δ ) S 0 ~ ( 0 ) ] = exp ( − 1 κ π 2 Δ σ 2 δ 2 ) %gamma_{c,t} left(%delta right) = {Re left [ {tilde{S}_0 left(%delta right) } over {tilde{S}_0 left(0 right)} right ]} =exp left ( -{ 1 over {%kappa}} {{%pi^2 %DELTA σ^2 %delta^2}} right )

The interference signal is expressed as:

I ( δ ) I 0 = 1 2 [ 1 + exp ( − 1 κ π 2 Δ σ 2 δ 2 ) cos ( 2 π σ 0 δ ) ] {I left(%delta right)} over {I_0} = 1 over 2 left [ 1 + exp left ( -{ 1 over {%kappa}} {{%pi^2 %DELTA σ^2 %delta^2}} right ) cos left( 2 %pi σ_0 %delta right) right ]

The coherence length of the source can be calculated by using the expression:

l c = ∫ − ∞ + ∞ | γ c , t ( δ ) | 2 d δ l_c = int_{-infinity} ^{+infinity} lline %gamma_{c,t} left(%delta right) rline^2 d %delta

i.e.:

l c = ∫ − ∞ + ∞ exp ( − 1 κ π 2 Δ σ 2 δ 2 ) d δ l_c = int_{-infinity} ^{+infinity} exp left ( -{ 1 over {%kappa}} {{%pi^2 %DELTA σ^2 %delta^2}}right ) d %delta

Taking into account that:

∫ − ∞ + ∞ exp ( − x 2 ) dx = π int_{-infinity} ^{+infinity} exp left(- x^2 right) dx = sqrt %pi

the result is:

l c = 0.664 Δ σ = 0.664 λ 2 Δ λ l_c = {0.664}over{%DELTA σ} = 0.664{%lambda^2}over{%DELTA %lambda}

When this value is applied it produces:

l c = 0.664 1.6766 × 10 3 = 3.96 × 10 − 4 cm = 3.96 μ m l_c = {0.664}over{1.6766 times 10^{3}} = 3.96 times 10^{-4} "cm" = 3.96 %mu "m"

Question

Carefully graph the normalized illumination.

Solution

The standard illumination has an average value of \(1/2\). It is shown on figure 29.

Question

What is the value of the interfringe distance?

Solution

The cosine function always has the same period 1 / σ 0 = λ 0 1/σ_0 = %lambda_0 , the geometry does not change so the period on the screen and the interfringe remain the same.

Question

Re-do the previous questions for the case of a two-wavelength sodium source, taking into account the physical nature of each line: the fact that each line is described by a gaussian shape. Assume Δ λ 0 ≈ 0.002 nm %DELTA %lambda_0 approx 0.002 "nm" as the width of each line, which corresponds to 300K of inhomogeneous broadening due to the Doppler effect, and Δ λ = 0.6 nm %DELTA %lambda = 0.6 "nm" for the distance between the two lines.

What are the values of Δ σ 0 %DELTA σ_0 and Δ σ %DELTA σ ?

Solution

The value of Δ σ 0 {%DELTA σ_0 } is given by:

Δ σ 0 = Δ λ 0 λ 2 = 0.005 × 10 − 7 ( 589.3 × 10 − 7 ) 2 = 0.057 cm − 1 {%DELTA σ_0 } = {%DELTA %lambda_0 } over {%lambda^2} = {0.005 times 10^{-7}} over {(589.3 times 10^{-7})^2} =0.057 "cm"^{-1}

and the value of Δ σ {%DELTA σ} is given by:

Δ σ = Δ λ λ 2 = 0.6 × 10 − 7 ( 589.3 × 10 − 7 ) 2 = 17.277 cm − 1 {%DELTA σ} = {%DELTA %lambda } over {%lambda^2} = {0.6 times 10^{-7}} over {(589.3 times 10^{-7})^2} =17.277 "cm"^{-1}

Question

Recalculate the normalized illumination considering that each sodium doublet can be described by the Gaussian shape.

Graph the spectrum of the source and the obtained interferogram.

Solution

When each one of the sodium doublets can be described by a Gaussian function, the caseis similar to the real situation. The energy spectral density function is written as:

S S ( σ ) = I 0 exp ( − κ ( σ − σ 0 ) 2 Δ σ 0 2 ) ∗ [ δ ( σ − ( σ 0 + Δ σ 2 ) ) + δ ( σ − ( σ 0 − Δ σ 2 ) ) ] {S}_S left(σ right)= {I_0} exp left ( -%kappa {left (σ-σ_0 right )^2} over {%DELTA σ_0}^2 right ) * left [ bold{%delta} left(σ- left(σ_0+{{%DELTA σ} over 2} right) right) +bold{%delta} left(σ- left(σ_0-{{%DELTA σ} over 2} right) right) right ]

The procedure is the same as in the previous exercise. We calculate the Fourier transform of the spectral energy density, which is equal to the product of the Fourier transforms of the two convoluted functions.

These Fourier transforms have been previously calculated.

The two Dirac distributions convoluted to the gaussian function are written as:

f ( σ ) = δ ( σ − σ 0 − Δ σ 2 ) + δ ( σ − σ 0 + Δ σ 2 ) f left(σ right) = bold{%delta} left(σ- σ_0-{{%DELTA σ} over 2} right) +bold{%delta} left(σ- σ_0+{{%DELTA σ} over 2} right)

and their Fourier transform is:

f ~ ( δ ) = cos ( π Δ σ δ ) tilde{f} left(%delta right) = cos left (%pi %DELTA σ %delta right )

From this, we can deduce the expression of the normalized interferogram:

I ( δ ) I 0 = 1 2 [ 1 + exp ( − 1 κ π 2 Δ σ 0 2 δ 2 ) cos ( π Δ σ δ ) cos ( 2 π σ 0 δ ) ] {I left(%delta right)} over {I_0} = 1 over 2 left [ 1 + exp left ( -{1 over %kappa }{{%pi^2 %DELTA σ^2_0 %delta^2} } right ) cos left (%pi %DELTA σ %delta right ) cos left( 2 %pi σ_0 %delta right) right ]

The intensity as a function of the optical path difference is represented in figure 30.