Single-slit diffraction experiment and the conservation of energy

[amjad-sh]

In a single-slit diffraction experiment, a monochromatic light of wavelength $\lambda$ is passed through one slit of finite width $D$ and a diffraction pattern is observed on screen.
For a screen located very far away from the slit, the intensity of light $I$ observed on the screen in units of its maximum value ${I_0}$ and in terms of $sin(\theta)$, where $\theta$ is the angle between the slit and a position on the screen, is represented in figure1 attached below.

If we let $D=4 \times 10^{-9} m$ and $\lambda= 1 \times 10^{-6}m$, $I(\theta)$ will have the form in figure 2.

If we immensely increase the width of the slit to become $D=4 \times 10^{-1}m$ and keep the wavelength $\lambda$ to be the same , $I(\theta)$ will have the form in figure3.

You can see that in case $D$ is very small ( $D= 4 \times 10^{-9}m$), the intensity on the screen will be maximum along the whole screen, but when we increase the width $D$ to reach $4 \times 10^{-1}m$, the intensity of light on the screen will just be maximum on the center of the screen, but will be null on the rest.

What is confusing me is that if we suppose that in the both cases the same amount of light is entering the slit, shouldn't the total energy on the the screen be the same for the two cases? By comparing the intensities in the two cases, it is clear that the total energy that would be detected on the screen for case $D=4 \times 10^{-9}m$ will be higher than that of the case when $D=4 \times 10^{-1}m$. Doesn't this violate the law of conservation of energy? since the same amount of light got emanated from the slit , so we expect to detect the same amount of total energy or the same total intensities in the both cases.

Note that $I(\theta)=I_0\Big(\dfrac{sin(\frac{\beta}{2})}{\frac{\beta}{2}}\Big)^2$, where $\beta=\dfrac{2\pi Dsin(\theta)}{\lambda}$

 

[anuttarasammyak]

Looking at three sketches you show, higher the peak, narrower it is. Total energy through the slit get larger for higher and broader peaks. In your speculation are these two features considered ?

 

[amjad-sh]

Looking at three sketches you show, higher the peak, narrower it is. Total energy through the slit get larger for higher and broader peaks. In your speculation are these two features considered ?

But the peak of the intensity for the two cases is $I_0$, which is the same. It is the same because, as I said in the post, I am assuming that the same amount of energy is entering through the slit and I also assumed that the wavelength of light is the same for the two cases. Off course, when I increase the width $D$ of the slit, the total intensity $I$ of light passing through the slit will decrease but the peak $I_0$ will still the same.
what is making me confused is that when $D$ is large the total intensity detected on the screen will be much lower than when $D$ is small. As the intensity is defined to be energy per unit area, and provided that we didn't change the dimensions of the screen in the both cases, shouldn't the total intensity be the same for the both cases, or otherwise energy would not be conserved?

 

[anuttarasammyak]

what is making me confused is that when D is large the total intensity detected on the screen will be much lower than when D is small. As the intensity is defined to be energy per unit area, and provided that we didn't change the dimensions of the screen in the both cases

#1 When you open a window of your room, wider you open it more sunshine comes into the room. Energy increase.
#2 Getting cloudy you open the window more so that energy coming into the room the same.

Which case is similar to "If we immensely increase the width of the slit" in OP ?

 

[amjad-sh]

#1 When you open a window of your room, wider you open it more sunshine comes into the room. Energy increase.
#2 Getting cloudy you open the window more so that energy coming into the room the same.

Which case is similar to "If we immensely increase the width of the slit" in OP ?

#2 is very similar to what I am referring in my OP. I want to assume that the same amount of light is entering the slit despite the increase in the width $D$ of the slit.

 

[anuttarasammyak]

Then
For a narrower slit :
high intensity light enters
high peak
narrower zero intensity to neighbor zero intensity length

For a wider slit :
low intensity light enters
low peak
wider zero intensity to neighbor zero intensity length

The areas the graphs you attached in OP makes, which correspond to energy, would be the same.

 

[amjad-sh]

Then
For a narrower slit :
high intensity light enters
high peak
narrower zero intensity to neighbor zero intensity length

For a wider slit :
low intensity light enters
low peak
wider zero intensity to neighbor zero intensity length

The areas the graphs you attached in OP makes, which correspond to energy, would be the same.

For whether the slit gets narrower or wider, I think that the the peak of the intensity detected on the screen will be the same.
You can model the slit as an infinite number of Huygens sources with infinitesimal widths. The light emanating from a Huygens source in the slit and directed towards a location on the screen making an angle $\theta$ with it has the following form:
$$E_{\theta}=\triangle E_0sin(wt+\phi)$$
where $\phi$ is the phase angle between its neighboring light wave and $\triangle E_0$ is the amplitude of the wave, which is infinitesimal.
For slit of width $D$, the amplitude of the light wave is $\triangle E_0$. Also suppose that all the other light waves emanating from all the other light sources has the same amplitude $\triangle E_0$. At the center of the screen all the light waves interfere in phase, so the the total amplitude on the screen will be $N\triangle E_0= E_0$, where $N$ is the number of Huygens sources in the slit which is infinite.
If we increase the width $D$ of the slit, $E_{\theta}$ will have the form:
$$E_{\theta}=\triangle E_0^{'}sin(wt+\phi)$$
Also here, at the center of the screen all the light waves interfere in phase, so the total amplitude at the screen will be $2N\triangle E_0^{'} =E_0$ (suppose here that we doubled the width $D$).
Therefore, in both cases, the amplitude of the resultant wave at the screen will be the same, and hence, the amplitude of the intensity will also be the same, as it is defined to be equal to $\sqrt{\dfrac{\epsilon_0}{\mu_0}}E_0^{2}$.

 

[anuttarasammyak]

For whether the slit gets narrower or wider, I think that the the peak of the intensity detected on the screen will be the same.

Yes for the case #1 of my post #4 but you choose #2. Intensity of sunlight is lessened by cloud. E goes downer. Peak goes downer.

$$E_1^2 D_1= E_2^2 D_2$$

Say $D_1/D_2=1/2$, $\ E_1/E_2=\sqrt{2}$

 

[amjad-sh]

Yes for the case #1 of my post #4 but you choose #2. Intensity of sunlight is lessened by cloud. E goes downer. Peak goes downer.

$$E_1^2 D_1= E_2^2 D_2$$

Say $D_1/D_2=1/2$, $\ E_1/E_2=\sqrt{2}$

In my latest reply I mentioned that the "peak of the intensity" on the screen stays the same not the intensity. The intensity of light entering the slit is different in the two cases. Moreover, the energy entering the slit in both cases are the same. In the first case the amplitude of the wave is $N\triangle E_0=E_0$ and in the second case it is $2N\triangle E_0^{'}=E_0$. So, my problem refers to #2 of your post #4.

 

[anuttarasammyak]

For whether the slit gets narrower or wider, I think that the the peak of the intensity detected on the screen will be the same.

I agree they share the same position wrt the center of the slits, though brightness are different.

 

[amjad-sh]

I agree they share the same position wrt the center of the slits, though brightness are different

What position? I didn't talk about any position in the statement that you quoted. By "will be the same" I meant the magnitude not the position.

 

[PeroK]

What position? I didn't talk about any position in the statement that you quoted. By "will be the same" I meant the magnitude not the position.

Here is a lecture that discusses the single slit experiment among other things. The experiment itself starts at about 39:00. I don't understand your analysis of the "total intensity". The amount of light that gets through to the screen decreases, of course, as the slit gets narrower, as more light impacts the barrier on either side of the slit and is lost to the experiment:

 

[sophiecentaur]

What is confusing me is that if we suppose that in the both cases the same amount of light is entering the slit, shouldn't the total energy on the the screen be the same for the two cases?

I have read through most of this thread but the whole thing pivots on this statement. If the "same amount of light is entering the slit" then the maximum intensity of the diffraction pattern will have to be the same. BUT this is not a realistic experiment to discuss. You would have to put a variable density filter in front of the slit in order to obtain that condition.

The usual experiment to discuss and to carry out would use uniform illumination of the slit and the total energy through the slit would be proportional to the width. The Fraunhofer Integral for the diffraction pattern is (half way down) in this link. The quantity "a" represents the amplitude of the incident plane wave on the slit. The width of the slit sets the limits for the integration.

Note: If you come across a result that you think is violating basic stuff like conservation of energy then it is pretty certainly that there is no actual violation. You have to follow any calculations carefully and not miss out any steps. Do that and you will get 'the right / accepted answer' (unless you are a potential Nobel Prizewinner ). If you can't / won't follow the Maths then you really just have to believe the standard result.

 

[vanhees71]

Of course to get energy conservation, you have to take into account the entire em. field, i.e., not only the part going through the slits but also the part being reflected and/or absorbed.

 

[amjad-sh]

I have read through most of this thread but the whole thing pivots on this statement. If the "same amount of light is entering the slit" then the maximum intensity of the diffraction pattern will have to be the same. BUT this is not a realistic experiment to discuss. You would have to put a variable density filter in front of the slit in order to obtain that condition.

The usual experiment to discuss and to carry out would use uniform illumination of the slit and the total energy through the slit would be proportional to the width. The Fraunhofer Integral for the diffraction pattern is (half way down) in this link. The quantity "a" represents the amplitude of the incident plane wave on the slit. The width of the slit sets the limits for the integration.

Note: If you come across a result that you think is violating basic stuff like conservation of energy then it is pretty certainly that there is no actual violation. You have to follow any calculations carefully and not miss out any steps. Do that and you will get 'the right / accepted answer' (unless you are a potential Nobel Prizewinner ). If you can't / won't follow the Maths then you really just have to believe the standard result.

I think your comment resolved the issue. Indeed, my whole issue was pivoting around my claim that the "same amount of light would enter the slit", which is a wrong claim or at least very hard to make it feasible. I was ignoring the fact that the source of light that I'm illuminating into the slit also encounters the obstacles beyond the slit. Therefore, the total energy through the slit will inevitably increase when we increase the width of the slit and decrease when we narrow it. Moreover, when the total energy through the slit increases, the maximum intensity of the diffraction pattern will increase, and will decrease when the total energy through the slit decreases. Thank you.

 

[amjad-sh]

Here is a lecture that discusses the single slit experiment among other things. The experiment itself starts at about 39:00. I don't understand your analysis of the "total intensity". The amount of light that gets through to the screen decreases, of course, as the slit gets narrower, as more light impacts the barrier on either side of the slit and is lost to the experiment:

By the total intensity I meant the intensity summed over the whole screen not only the intensity on a certain location in the screen.

 

[amjad-sh]

Of course to get energy conservation, you have to take into account the entire em. field, i.e., not only the part going through the slits but also the part being reflected and/or absorbed.

exactly.

 

[hutchphd]

Perhaps you will also be interested in the optical theorem:
https://en.wikipedia.org/wiki/Optical_theorem
This will be in most intermediate texts.

 

[sophiecentaur]

Moreover, when the total energy through the slit increases, the maximum intensity of the diffraction pattern will increase, and will decrease when the total energy through the slit decreases.

An interesting aside:
If you consider all the light that doesn't get through the slit, you can do the Fraunhofer Integral over the ranges [-∞ to -w/2] and [w/2to ∞] the result will be what you would get for a thin strip, inserted in a wide uniform beam. The shape of the shadow formed will be the same but upside down, every peak will be in the corresponding place as a min and vice versa. Now there's your Energy Conservation at work.

 

[tech99]

An interesting aside:
If you consider all the light that doesn't get through the slit, you can do the Fraunhofer Integral over the ranges [-∞ to -w/2] and [w/2to ∞] the result will be what you would get for a thin strip, inserted in a wide uniform beam. The shape of the shadow formed will be the same but upside down, every peak will be in the corresponding place as a min and vice versa. Now there's your Energy Conservation at work.

Babinet's Principle.

 

[sophiecentaur]

Babinet's Principle.

Ah yes. I remember the name now.