Diffraction -- Calculate the interference lost orders due to diffraction

• RealKiller69
In summary, the conversation discusses the calculation of diffraction for a slit system, where the diffraction term is calculated in the far field case using a weighting function and a phase factor. The conversation also mentions the existence of lost interference maximums, which occur when the diffraction integral gives zero for an integer value. The speaker also mentions that the problem would be more educational if the case of five squares in a column, which does have lost orders, was included. Through further discussion and calculations, it is determined that the larger slit composed of five smaller ones does not have any extra zeros, but only modulates the diffraction.
RealKiller69
Homework Statement
Two 1D grating grids are given with period 5a, every slit of them are composed with 5 smaller and square slits with area a**2 with different spatial distribution . I have to calculate interference lost orders due to diffraction for the line q=0.
Relevant Equations
derived formulas are written on the paper.
I have calculated the diffraction part for each one of the slits but I am not getting any lost order.

This took me a while to figure out what they are asking, but I think I have it: ## \\ ## In one dimension you need to calculate the diffraction term across the ##x ## direction for the far field case: ## I(\theta)=| \int\limits_{single \, slit}E_o \, t(x) e^{i \phi(x)} \, dx|^2 ## where ## t(x) ## is the weighting function for the different parts of the slit, and ## \phi(x)=\frac{2 \pi x \sin{\theta}}{\lambda}##. ## \\ ## (##E_o ## is an arbitrary constant).## \\ ## ## I(\theta) ## is likely to have some zeros at places where ## m \lambda=(5a) \sin{\theta} ## which are the ## \theta's ## for the interference peaks for integer ## m ##. If the diffraction integral gives zero for an integer ## m ##, this is a lost interference maximum. ## \\ ## Note: You should be able to designate any part of the slit as the origin ## x=0 ##. Choosing a different location for the origin will just introduce a factor ## e^{i \phi_o} ##, which will be converted to a unity factor when you take ## I(\theta)= |E(\theta)|^2 ##. ## \\ ## And I think I solved it correctly=I didn't try all the cases, but a slit that has 5 squares in a column gives me a bunch of lost orders (I'm not going to give you the complete answer), but I don't think any other configuration gives any lost orders. (I haven't checked my result very carefully=I'll leave that part to you).

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the weighted function is t(x,y)=cte and using unit intensity (I(0)=1) for both of the larger slits( composed of 5 smaller 1's) I am getting for the diffraction on the x-axis the following exp. for the 1st slit (that looks like a plus sign) Itotal_1=I(0)(3+2cos(kpa))^2 (k=2pi/lambda) , this term has no zeros, the smallest value it gets to is when cos(kpa)=-1, so there won't be any value of sin(theta) where an interference maxima overlaps with a zero of diffaction. Same fot the second slit.

I worked the 2nd case in detail. (I did not work the first case in detail). If the OP would show their calculations in detail for the second case, I could verify them. It would appear these two cases may have no lost orders, As I previously mentioned, they should include the case of 5 squares in a column, which does have lost orders. The problem would be much more educational if they did that.

thanks man, i´ve figured what i was doing it wrong. I was forgetting the differential term given by the smaller slits, i was expecting some type of special diffraction effect with added extra zeros for the line y=0 due to the larger slit composed by 5 smaller ones. In this case, it only modulates the diffraction changing its form.For the line y=0 or q=0 we have:
Idiifraction_small_slit(q=0)=I0*sin(kpa/2)/kpa/2 , p=sin(theta);k=2pi/lambda [this part has its zeros ], Idiff_large_slit=Idiffraction_small_slit*(3+2cos (kpa))^2[ This part corresponds to the whole system of 5 smaller slits and it dosent add any extra zeros, the smallet it can get is when cos(kpa)=-1 (if have used the difference in phases using the centre slit of the as the origin) ] , I_grating_grid_1d_[5a_period]=I_diff_large_slit*interference_factor. So in this The zeros of diffraction that overlaps with the maximas of interference are the same as only having grating_grid with square slits of area a^2 of period 5a.

1. What is diffraction and how does it affect interference patterns?

Diffraction is the bending of waves around obstacles or through small openings. In the context of interference, diffraction causes the interference pattern to become less distinct and results in the loss of certain interference orders.

2. How do you calculate the interference lost orders due to diffraction?

The interference lost orders due to diffraction can be calculated using the formula m = dsinθ/λ, where m is the order of interference, d is the distance between the diffracting object and the screen, θ is the angle of diffraction, and λ is the wavelength of the incident light.

3. What factors affect the amount of interference lost due to diffraction?

The amount of interference lost due to diffraction is affected by the size of the diffracting object, the distance between the object and the screen, and the wavelength of the incident light. Smaller diffracting objects, larger distances, and shorter wavelengths result in less interference lost.

4. Can diffraction be eliminated in interference patterns?

No, diffraction is a natural phenomenon that cannot be completely eliminated. However, its effects can be minimized by using smaller diffracting objects, increasing the distance between the object and the screen, and using shorter wavelengths of light.

5. How does diffraction relate to the Heisenberg uncertainty principle?

The Heisenberg uncertainty principle states that it is impossible to know both the position and momentum of a particle with absolute certainty. Diffraction can be seen as an example of this principle, as the diffracting object's position and momentum are both affected by the bending of waves around it, making it impossible to accurately determine both at the same time.

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