Calculating Intensity in Single Slit Diffraction Experiments

In summary,The principle of single-slit diffraction is used to calculate the intensity of light on a screen located very far away from the slit. It is assumed that the width of the slit consists of an infinite number of Huygens' sources, and each Huygens' source creates a wave front that spreads in all directions. The interference of all the electromagnetic radiation emanating from all the sources creates the diffraction pattern that we see on the screen. But what is confusing me is that the principle is only applied at the wave front that is present at the slit. For example, suppose that the single-slit has an infinitesimal width. Therefore, the slit itself would be a Huy
  • #1
amjad-sh
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In a single slit diffraction experiment, when we want to calculate the intensity of light on a screen located very far away from the slit, usually Huygens' principle is adopted as a model to perform the calculations.

It is assumed that the width of the slit consists of an infinite number of Huygens' sources, where each Huygens' source is a source of electromagnetic radiation that spreads in all directions. The interference of all the electromagnetic radiation emanating from all the sources creates the diffraction pattern that we see on the screen.

But what is confusing me is that the principle is only applied at the wave front that is present at the slit.

For example, suppose that the single-slit has an infinitesimal width. Therefore, the slit itself would be a Huygens' source. The Huygens' source will create a wave front and this initial wave front contains infinite number of Huygens' sources, and all of these infinite number of Huygens' sources will create their own wave front and so on.

I think in all textbooks, they don't take the electromagnetic field emanating from the other wave fronts ( the wave front that is not present at the slit) into account in summing up the electromagnetic radiation to get the intensity at a certain position. Shouldn't we take them into consideration to calculate the intensity? or is there a proof that the superposition of all the electromagnetic fields emanating from wave wave front 1 creates electromagnetic field at position p (on wave front 2 just after wave front 1) with an amplitude just equal to the amplitude of the electromagnetic field there but in the case when we don't suppose that wave front 1 has infinite number of Huygens'
sources, e.g., when the electromagnetic field at p is just a continuation to the electromagnetic field emanated from the single infinitesimal-width slit and directed towards position p?

See the attachment below for more clarification to my question.
 

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  • #2
Let us generate a single photon and send it toward a single slit in the experiment. The photon has probability to collide with walls and not exit from the slit. Say ##\alpha## the probability for the photon to exit from space of the slit. Then we have probability ##\alpha## to get a spot of photon on the screen. The probability to get a spot around some special position say (x,y) on the screen is not homogeneous but has a pattern. Let us describe it P(x,y) .
[tex]\int dx \int dy \alpha P(x,y) = \alpha [/tex] or
[tex]\int dx \int dy P(x,y) = 1 [/tex]
where P(x,y) dx dy gives probability to get a spot in area dxdy around (x,y) when surely we have one spot somewhere on the screen. Huygens principle gives us P(x,y) with the normalization above.
Let us repeat it one by one N photons we get not probability but visible dot pattern of N##\alpha##P(x,y) on the screen. We may also do it with N photon beams at a time but please take care that Huygens principle or superposition principle applies for each photon: a photon interferes with itself not with other photons.
 
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  • #3
amjad-sh said:
For example, suppose that the single-slit has an infinitesimal width. Therefore, the slit itself would be a Huygens' source. The Huygens' source will create a wave front and this initial wave front contains infinite number of Huygens' sources, and all of these infinite number of Huygens' sources will create their own wave front and so on.

Yes that is correct. But remember these are lines of a particular phase on the wavefront and these lines are typicaly curved. At the point that the curve is complicated then the construction is not a simplification. Nor do these sources necessarilly have the same intensity as the original or each other.
 
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  • #4
amjad-sh said:
It is assumed that the width of the slit consists of an infinite number of Huygens' sources, where each Huygens' source is a source of electromagnetic radiation that spreads in all directions.
This is not an assumption. This is the requirement for the experiment. The light must be from point source far away (i.e. a "plane wave") or the equivalent. If the light is different Huyghens requires a different source. You don't just assume.
 
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  • #5
anuttarasammyak said:
Let us generate a single photon and send it toward a single slit in the experiment. The photon has probability to collide with walls and not exit from the slit. Say ##\alpha## the probability for the photon to exit from space of the slit. Then we have probability ##\alpha## to get a spot of photon on the screen. The probability to get a spot around some special position say (x,y) on the screen is not homogeneous but has a pattern. Let us describe it P(x,y) .
[tex]\int dx \int dy \alpha P(x,y) = \alpha [/tex] or
[tex]\int dx \int dy P(x,y) = 1 [/tex]
where P(x,y) dx dy gives probability to get a spot in area dxdy around (x,y) when surely we have one spot somewhere on the screen. Huygens principle gives us P(x,y) with the normalization above.
Let us repeat it one by one N photons we get not probability but visible dot pattern of N##\alpha##P(x,y) on the screen. We may also do it with N photon beams at a time but please take care that Huygens principle or superposition principle applies for each photon: a photon interferes with itself not with other photons.
Thanks for your answer, but that's not what I was inquiring about in my post. You are describing the quantum mechanical point of view of diffraction of light, and my question is just related to Huygens' principle.
 
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  • #6
amjad-sh said:
You are describing the quantum mechanical point of view of diffraction of light, and my question is just related to Huygens' principle.
Huygens' principle comes from Path Integral in QM. So you have questions on it you can get deeper understanding by investigating Path Integral.

amjad-sh said:
I think in all textbooks, they don't take the electromagnetic field emanating from the other wave fronts ( the wave front that is not present at the slit) into account in summing up the electromagnetic radiation to get the intensity at a certain position. Shouldn't we take them into consideration to calculate the intensity?
Path should be integrated with any possible courses with equal weight. Cohesion and incohesion among paths should matter in integration to get the result, i.e. probability or denseness of a spot.
Through two slits, three slits, windows, backdoors or any other paths from light source to a spot should be counted although other paths than slits are negligible due to its small contribution after superposition in most cases.

Ref. http://www1.phys.vt.edu/~takeuchi/Tools/CSAAPT-Spring2019-Huygens&Feynman.pdf
 
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  • #7
hutchphd said:
At the point that the curve is complicated then the construction is not a simplification. Nor do these sources necessarilly have the same intensity as the original or each other.
Yes I know that the sources on a wave front do not necessarily have the same intensity, but my question is that Is the intensity at a specific point p on an arbitrary wave front k, in the case when the wave fronts (other than the initial wave front at the slit) contain Huygens' sources, the same to the intensity at that specific point p on an arbitrary wave front k in the case when the wave fronts don't contain any Huygens' sources?

I guess that they would be the same, because otherwise there would be a flaw in the way the principle is used to calculate the intensity of light on a screen located very far away from the slit. Because in most sources that I have read they state that only the wave front on the slit contains infinite number of Huygens' sources and ignore that the other wave fronts have also Huygens' sources. So, for their formulation to be correct, the two intensities must be the same.
 
  • #8
anuttarasammyak said:
Huygens' principle is derived from Path Integral in QM.
That seems interesting. Do you know any reference or a link in the web that covers this derivation rigorously? It would be the answer to my question.
 
  • #9
hutchphd said:
This is not an assumption. This is the requirement for the experiment. The light must be from point source far away (i.e. a "plane wave") or the equivalent. If the light is different Huyghens requires a different source. You don't just assume.
I think it is a requirement for the Huygens' model. Huygens' principle states that "all points of a wave front of light in a vacuum or transparent medium may be regarded as new sources of wavelets that expand in every direction at a rate depending on their velocities".
 
  • #10
amjad-sh said:
That seems interesting. Do you know any reference or a link in the web that covers this derivation rigorously? It would be the answer to my question.
I found
https://www.osapublishing.org/josaa/abstract.cfm?uri=josaa-4-9-1751
but not free. You may search much better than I.

Huygens principle tells wave sources anywhere.
Feynman Integral tells any path does contribution with equal weight.
Though the both are funny but I prefer the latter as simpler and more "natural" view.
 
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  • #12
anuttarasammyak said:
Huygens' principle comes from Path Integral in QM. So you have questions on it you can get deeper understanding by investigating Path Integral.
This is not well stated. Obviously the principle does not "come from" QM because Christian Huyghens lived (1622-1695), somewhat before the time of Feynman. It is in fact the beginning of formal Green's Function technique developed in the 1800's.
You should study path integrals, but the two are related because QM deals with waves.
@anuttarasammyak Please be more careful in your commentary.
 
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  • #13
Of course Huygens's principle comes not only from path integrals in quantum mechanics. Indeed at this level of the description of light there's no need for quantum (field) theory at all, but it comes from the very important concept of Green's functions.

So let's consider the most simple case of some "wave field" ##\Phi(t,\vec{x})## with some "source" ##J(t,\vec{x})##, and consider the corresponding inhomogeneous wave equation
$$\frac{1}{c^2} \partial_t^2 \Phi-\Delta \Phi=J.$$
To make life easier and to also specialize to the plane-wave case already mentioned above we consider fields with harmonic time dependence, and that's most easily done working with a complex potential, i.e., we write
$$\Phi(t,\vec{x})=\Phi_0(\vec{x}) \exp(-\mathrm{i} \omega t), \quad J(t,\vec{x})=J_0(\vec{x}) \exp(-\mathrm{i} \omega t).$$
For the case of real fields (as if you consider the electromagnetic field, where ##\Phi## may be some Cartesian component of the electric or magnetic field), just take the real part at the end of the calculation.

Plugging this into the wave equation, after cancelling the exponential factor, we get Helmholtz's equation,
$$-\frac{1}{c}^2 \omega^2 \Phi_0-\Delta \Phi_0=-(\Delta+k^2) \Phi_0=J_0, \quad k=\omega/c. \qquad (1)$$
Now it makes sense to solve this equation first for a very simple source, the ##\delta##-distribution,
$$\Delta G(\vec{x}-\vec{x}')=-\delta(\vec{x}-\vec{x}').$$
The idea is that, if you have solved this special case of a "point source" (though it's a pretty unphysical source described by a generalized function, Dirac's ##\delta## distribution), then you also can solve the general case with an integral,
$$\Phi_0(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' G(\vec{x}-\vec{x}') J_0(\vec{x}').$$
Indeed, acting with the "Helmholtz operator" ##-(\Delta+k^2)## on this and just moving it under the integral sign on the right-hand side you get (1).

To solve the equation, just set ##\vec{y}=\vec{x}-\vec{x}'##. Then you have
$$(\Delta+k^2) G(\vec{y})=-\delta^{(3)}(\vec{y}). \qquad (2)$$
Now for ##\vec{y} \neq 0## you can introduce spherical coordinates. With ##r=|\vec{y}|## then the equation reads
$$\frac{1}{r} \partial_r^2 [r G(r)]+k^2 G=0 \quad r \neq 0,$$
where we have used the rotational invariance of the problem around ##\vec{y}=0##.

Multiplying with ##r## we get for ##H(r)=r G(r)##
$$\partial_r^2 H=-k^2 H \; \rightarrow \; H = A \exp(\mathrm{i} k r) + B \exp(-\mathrm{i} k r)$$
or
$$G(r)=\frac{A}{r} \exp(\mathrm{i} k r) + \frac{B}{r} \exp(-\mathrm{i} k r).$$
Taken this together with the time-dependent factor ##\exp(-\mathrm{i} \omega t)## that means the general solution can be written as a superposition of an outgoing (the part ##\propto A##) and an incoming (the part ##\propto## B).

That's characteristic for the wave equations ("hyperbolic differential equations"), which are time-reversal invariant: You always get two linearly independent equations, one describing the time-reversed of the other. Here we want outgoing waves only, i.e., we have a source (for em. waves some accelerated charges), and we want to describe outgoing waves. This means of all possible Green's functions we choose the one with ##B=0##, i.e.,
$$G(r)=\frac{A}{r} \exp(\mathrm{i} k r).$$
The integration constant ##A## can then be found by integrating (2) over a very small sphere of radius ##\epsilon## around ##\vec{y}=0##. This gives
$$\int_{K_{\epsilon}} \mathrm{d}^3 y (\Delta + k^2 ) \frac{A}{r} \exp(\mathrm{i} k r)=-1.$$
The piece with ##k^2## cannot contribute anything in the limit ##\epsilon \rightarrow 0##, because the integral goes like
$$\int_{K_{\epsilon}} \mathrm{d}^3 y \frac{A}{r} \exp(\mathrm{i} k r) = \int_0^{\epsilon} \mathrm{d} r \int_0^{2 \pi} \mathrm{d} \varphi \int_0^{\pi} \mathrm{d} \vartheta r^2 \sin \vartheta \frac{A}{r} \exp(\mathrm{i} k r) = 4 \pi \int_0^{\epsilon} A r \exp(\mathrm{i} k r) \rightarrow 0 \quad \text{for} \quad \epsilon \rightarrow 0.$$
The part with the Laplace operator can be calculated with Gauß's integral theorem,
$$\int_{K_{\epsilon}} \mathrm{d}^3 y \Delta G(\vec{y})=\int_{\partial K_{\epsilon}} \mathrm{d}^2 \vec{f} \cdot \vec{\nabla} \cdot \vec{G}.$$
Now
$$\vec{\nabla} \cdot \vec{G}=\frac{A}{r^2} (\mathrm{i} k r -1) \vec{e}_r$$
Further
$$\mathrm{d}^2 \vec{f} = \mathrm{d} \vartheta \mathrm{d} \varphi \vec{e}_r \epsilon^2 \sin \vartheta.$$
Plugging this in above gives
$$\int_{K_{\epsilon}} \mathrm{d}^3 y \Delta G(\vec{y})=\int_0^{2 \pi} \mathrm{d} \varphi \int_0^{\pi} \mathrm{d} \vartheta A (\mathrm{i} k \epsilon-1) \exp(\mathrm{i} k \epsilon)=4 \pi A(\mathrm{i} k \epsilon-1) \exp(\mathrm{i} k \epsilon) \rightarrow -4 \pi A \stackrel{!}{=}-1 \; \rightarrow \; A=\frac{1}{4 \pi}.$$
So our Green's function is
$$G(\vec{y})=\frac{\exp(\mathrm{i} k |\vec{y}|)}{4 \pi |\vec{y}|}.$$
Plugging this into our solution we indeed get Huygens's principle,
$$\Phi_0(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\exp(\mathrm{i} k |\vec{x}-\vec{x}'|)}{4 \pi |\vec{x}-\vec{x}'|} J(\vec{x}'),$$
i.e., the wave field at ##\vec{x}## is a superposition of spherical waves going out of all points ##\vec{x}'##, where the source is not vanishing.

The application to the theory of diffraction is that one considers with Kirchhoff's approximate solution that the wave field in the region before the obstacle is undisturbed by the obstacle, i.e., it looks like the source as if the obstacle weren't there at all, and beyond the obstacle the field is given by the Green's function integrated over the openings (in your example the to slits).

In Fraunhofer approximation, where you observe the wave field very far away from the slits you can neglect the variation of the ##1/|\vec{x}-\vec{x}'|## factor and finally get that the wave field is just the Fourier transform of the openings (provided the light source before the obstacle is so far away that you can take it as a plane wave at the openings).

For more details, see, e.g., A. Sommerfeld, Lectures on Theoretical Physics, Vol. 4 (optics). There you also find a thorough discussion about the more precise theory of diffraction, and why the Kirchhoff theory is only an approximation, though a very good one for the usual diffraction experiments you demonstrate in the lecture hall :-).
 
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  • #14
vanhees71 said:
Of course Huygens's principle comes not only from path integrals in quantum mechanics. Indeed at this level of the description of light there's no need for quantum (field) theory at all, but it comes from the very important concept of Green's functions.

So let's consider the most simple case of some "wave field" ##\Phi(t,\vec{x})## with some "source" ##J(t,\vec{x})##, and consider the corresponding inhomogeneous wave equation
$$\frac{1}{c^2} \partial_t^2 \Phi-\Delta \Phi=J.$$
To make life easier and to also specialize to the plane-wave case already mentioned above we consider fields with harmonic time dependence, and that's most easily done working with a complex potential, i.e., we write
$$\Phi(t,\vec{x})=\Phi_0(\vec{x}) \exp(-\mathrm{i} \omega t), \quad J(t,\vec{x})=J_0(\vec{x}) \exp(-\mathrm{i} \omega t).$$
For the case of real fields (as if you consider the electromagnetic field, where ##\Phi## may be some Cartesian component of the electric or magnetic field), just take the real part at the end of the calculation.

Plugging this into the wave equation, after cancelling the exponential factor, we get Helmholtz's equation,
$$-\frac{1}{c}^2 \omega^2 \Phi_0-\Delta \Phi_0=-(\Delta+k^2) \Phi_0=J_0, \quad k=\omega/c. \qquad (1)$$
Now it makes sense to solve this equation first for a very simple source, the ##\delta##-distribution,
$$\Delta G(\vec{x}-\vec{x}')=-\delta(\vec{x}-\vec{x}').$$
The idea is that, if you have solved this special case of a "point source" (though it's a pretty unphysical source described by a generalized function, Dirac's ##\delta## distribution), then you also can solve the general case with an integral,
$$\Phi_0(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' G(\vec{x}-\vec{x}') J_0(\vec{x}').$$
Indeed, acting with the "Helmholtz operator" ##-(\Delta+k^2)## on this and just moving it under the integral sign on the right-hand side you get (1).

To solve the equation, just set ##\vec{y}=\vec{x}-\vec{x}'##. Then you have
$$(\Delta+k^2) G(\vec{y})=-\delta^{(3)}(\vec{y}). \qquad (2)$$
Now for ##\vec{y} \neq 0## you can introduce spherical coordinates. With ##r=|\vec{y}|## then the equation reads
$$\frac{1}{r} \partial_r^2 [r G(r)]+k^2 G=0 \quad r \neq 0,$$
where we have used the rotational invariance of the problem around ##\vec{y}=0##.

Multiplying with ##r## we get for ##H(r)=r G(r)##
$$\partial_r^2 H=-k^2 H \; \rightarrow \; H = A \exp(\mathrm{i} k r) + B \exp(-\mathrm{i} k r)$$
or
$$G(r)=\frac{A}{r} \exp(\mathrm{i} k r) + \frac{B}{r} \exp(-\mathrm{i} k r).$$
Taken this together with the time-dependent factor ##\exp(-\mathrm{i} \omega t)## that means the general solution can be written as a superposition of an outgoing (the part ##\propto A##) and an incoming (the part ##\propto## B).

That's characteristic for the wave equations ("hyperbolic differential equations"), which are time-reversal invariant: You always get two linearly independent equations, one describing the time-reversed of the other. Here we want outgoing waves only, i.e., we have a source (for em. waves some accelerated charges), and we want to describe outgoing waves. This means of all possible Green's functions we choose the one with ##B=0##, i.e.,
$$G(r)=\frac{A}{r} \exp(\mathrm{i} k r).$$
The integration constant ##A## can then be found by integrating (2) over a very small sphere of radius ##\epsilon## around ##\vec{y}=0##. This gives
$$\int_{K_{\epsilon}} \mathrm{d}^3 y (\Delta + k^2 ) \frac{A}{r} \exp(\mathrm{i} k r)=-1.$$
The piece with ##k^2## cannot contribute anything in the limit ##\epsilon \rightarrow 0##, because the integral goes like
$$\int_{K_{\epsilon}} \mathrm{d}^3 y \frac{A}{r} \exp(\mathrm{i} k r) = \int_0^{\epsilon} \mathrm{d} r \int_0^{2 \pi} \mathrm{d} \varphi \int_0^{\pi} \mathrm{d} \vartheta r^2 \sin \vartheta \frac{A}{r} \exp(\mathrm{i} k r) = 4 \pi \int_0^{\epsilon} A r \exp(\mathrm{i} k r) \rightarrow 0 \quad \text{for} \quad \epsilon \rightarrow 0.$$
The part with the Laplace operator can be calculated with Gauß's integral theorem,
$$\int_{K_{\epsilon}} \mathrm{d}^3 y \Delta G(\vec{y})=\int_{\partial K_{\epsilon}} \mathrm{d}^2 \vec{f} \cdot \vec{\nabla} \cdot \vec{G}.$$
Now
$$\vec{\nabla} \cdot \vec{G}=\frac{A}{r^2} (\mathrm{i} k r -1) \vec{e}_r$$
Further
$$\mathrm{d}^2 \vec{f} = \mathrm{d} \vartheta \mathrm{d} \varphi \vec{e}_r \epsilon^2 \sin \vartheta.$$
Plugging this in above gives
$$\int_{K_{\epsilon}} \mathrm{d}^3 y \Delta G(\vec{y})=\int_0^{2 \pi} \mathrm{d} \varphi \int_0^{\pi} \mathrm{d} \vartheta A (\mathrm{i} k \epsilon-1) \exp(\mathrm{i} k \epsilon)=4 \pi A(\mathrm{i} k \epsilon-1) \exp(\mathrm{i} k \epsilon) \rightarrow -4 \pi A \stackrel{!}{=}-1 \; \rightarrow \; A=\frac{1}{4 \pi}.$$
So our Green's function is
$$G(\vec{y})=\frac{\exp(\mathrm{i} k |\vec{y}|)}{4 \pi |\vec{y}|}.$$
Plugging this into our solution we indeed get Huygens's principle,
$$\Phi_0(\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 x' \frac{\exp(\mathrm{i} k |\vec{x}-\vec{x}'|)}{4 \pi |\vec{x}-\vec{x}'|} J(\vec{x}'),$$
i.e., the wave field at ##\vec{x}## is a superposition of spherical waves going out of all points ##\vec{x}'##, where the source is not vanishing.

The application to the theory of diffraction is that one considers with Kirchhoff's approximate solution that the wave field in the region before the obstacle is undisturbed by the obstacle, i.e., it looks like the source as if the obstacle weren't there at all, and beyond the obstacle the field is given by the Green's function integrated over the openings (in your example the to slits).

In Fraunhofer approximation, where you observe the wave field very far away from the slits you can neglect the variation of the ##1/|\vec{x}-\vec{x}'|## factor and finally get that the wave field is just the Fourier transform of the openings (provided the light source before the obstacle is so far away that you can take it as a plane wave at the openings).

For more details, see, e.g., A. Sommerfeld, Lectures on Theoretical Physics, Vol. 4 (optics). There you also find a thorough discussion about the more precise theory of diffraction, and why the Kirchhoff theory is only an approximation, though a very good one for the usual diffraction experiments you demonstrate in the lecture hall :-).
Waw! Thanks for this detailed comment, I will try to read through it thoroughly.
 
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  • #15
vanhees71 said:
Plugging this into our solution we indeed get Huygens's principle,
Is the integral that you had derived is the same as the Fresnel diffraction integral? because from this Wikipedia article (https://en.wikipedia.org/wiki/Fresnel_diffraction) they look similar.
The wave equation can be derived directly from Maxwell's equations. Can I conclude from this that Huygens' principle is a direct consequence of Maxwell's equations? or it can be only a consequence if we assume that the field has harmonic time dependence?
 
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  • #16
Yes, that's the same formula (before the approximations for "Fresnel diffraction" are made). For an arbitrary time-dependence you just can do a Fourier representation, using this formula for the harmonic time-dependence. What you then get is of course the usual solution with the retarded propagator.
 
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  • #17
vanhees71 said:
Plugging this into our solution we indeed get Huygens's principle,
I can notice that the integral formula is applied when the time t is fixed. I mean, for example, if we consider the wave front at the single slit, all the points on the wave front at the single slit will be regarded as sources and using the formula the amplitude ##\Phi_0(\vec{x})## of light wave of any phase can be calculated at any position away from the single slit at fixed time t. We can also consider another wave front different from that present at the slit ( the points at the wave front located at the single slit would not then be considered as sources though). In this case, all the points located at that wave front will be regarded as sources and using the formula we can also calculate the the amplitude ##\Phi_0(\vec{x})## of light waves located at any positions beyond that wave front. But, as @hutchphd mentioned in post #3 the wave front not located at the slit may be curved in a complicated way. Therefore, the sources will be located in a wave front of complicated surface. Therefore, the easier choice would be to consider the wave front at the slit because its curve is known and not complicated (linear).

Now in case my analysis above is true, I still have some uncertainties related to Huygens' principle.

1- Suppose that at time t the plane wave just reached the slit and haven't yet moved beyond the slit, and we want to calculate ##\Phi_0(\vec{x})## beyond the slit, can we still use the formula to calculate ##\Phi_0(\vec{x})##? I mean if we want to calculate ##\Phi_0(\vec{x})## beyond the slit at time t'>t, and provided that the waves emanating from the sources are propagating with the same speed c, some of the waves my reach position ##\vec{x}## at time t' but some would still can't reach this position at that time. So, how the formula would be interpreted in this case?
2-From this formula it seems that the waves are emanated in all directions from the source, but according to Huygens' principle the waves only emanate forward from the sources and not backwards. So, do we need to impose this assumption on the formula?
3- Does the wave fronts beyond the slit need to be spherical, or they can have any arbitrarily complicated shape?
 
  • #18
Have a look at A. Sommerfeld, Lectures on Theoretical Physics, vol. 4 (optics). There you get an explanation from the expert on diffraction theory. The exact theory is very complicated and analytically solvable only for very simple cases like the half-space. That's why one uses approximations, starting with Kirchhoff's theory of diffraction, which also neglects questions of polarization. It assumes that the field before the obstacles where the field is scattered is unaffacted by these obstacles entirely, and then you use the Green's function of the free Helmholtz equation to get the field behind the obstacles. That's the formula we've been discussing above, and then you make further approximations to simplify the integral, leading Fresnel or Fraunhofer diffraction formulae.
 
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Related to Calculating Intensity in Single Slit Diffraction Experiments

1. What is single slit diffraction?

Single slit diffraction is a phenomenon that occurs when a wave passes through a narrow opening or slit. The wave spreads out and creates a diffraction pattern on a screen placed behind the slit.

2. How is intensity calculated in single slit diffraction experiments?

The intensity of the diffraction pattern can be calculated using the equation I = (A^2)(sin^2θ)/(θ^2), where A is the amplitude of the wave and θ is the angle between the center of the diffraction pattern and the point on the screen where the intensity is being measured.

3. What factors affect the intensity in single slit diffraction experiments?

The intensity of the diffraction pattern is affected by the wavelength of the wave, the width of the slit, and the distance between the slit and the screen. The intensity also varies depending on the angle at which it is measured.

4. How does the width of the slit impact the intensity in single slit diffraction experiments?

The width of the slit has a direct impact on the intensity of the diffraction pattern. A wider slit will result in a narrower diffraction pattern with lower intensity, while a narrower slit will result in a wider diffraction pattern with higher intensity.

5. Can the intensity in single slit diffraction experiments be increased?

Yes, the intensity can be increased by increasing the amplitude of the wave, using a narrower slit, or by moving the screen closer to the slit. However, this will also result in a wider diffraction pattern.

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