Diffraction -- Calculate the interference lost orders due to 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.

 

[Charles Link]

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).

 

[RealKiller69]

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.

 

[Charles Link]

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.

 

[RealKiller69]

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.