# Double-slit diffraction with UNEQUAL source amplitude

1. Apr 8, 2010

### hbal9604@usyd

In the following I am always talking about fraunhoffer diffraction

suppose you have two slits in a doulbe slit diffraction experiment which are both exposed to the same plane wave (say the same laser) as the original source of radiation. Usually, we expect that both slits will act as EQUAL sources of radiation to the extent that they produce a diffraction pattern on a distant screen. the Irradiance of this diffraction pattern is given by

I = I(0)(sin(B)/B)^2*cos(A)^2

where B = 1/2bksin(theta)
A = 1/2aksin(theta)
where b = slit width
a = centre to centre slit speration
theta = anglular position on the screen

BUT now suppose we change the situation so that the second slit has a film over it that prevents complete transmission. say only a fraction c of the radiation is transmitted through c. Then we have to rework the derivation of the double slit diffraction.

QUESTION:

1) I have come accross the following equation for the situation above with the second slit transmitting only a fraction c of the radiation that it would normally transmit (and which the first slit DOES transmit)

I = I(0)/K*(sin(B)/B)^2*(1+c^2+2c[cos(2A)])

where K is a constant.

HOW DO YOU DERIVE THIS EQUATION?????????????

I have been trying to derive this equation from integrating of the apperture as usual, but including the factor c in the second integral (ie multiplying it), but I simply cannot work it out. Can you either refer me to somewhere where this derivation might be set out? or set it out for me? I have tried and tried for hours, and I get something kind of close, but never manage to get it in this form? please help!!!

2) also, just a question about the sort of diffraction effect that would result from this situation: should you expect the intensity pattern on the screen to be symetrical still? I guess I could thing about the mathematics and make a conclusion, but could you explain physically?

3)If this situation were changed so that we were in teh Fresnel (near-slit) regime? would it be symmetric of asymmetric? again, could you explain physically? I'm not too familiar with Fresnel diffraction (of course I know what it is, but I haven't actually done any Fresnel analysis ever) so could you explain physically as best you can?

Thanks a lot.

2. Apr 8, 2010

### Andy Resnick

The far-field diffraction pattern is the Fourier Transform of the aperture. Your aperture can be written as:

rect(x/b)# (d(x+a)+d(x-a)) -c* rect(x/b)#d(x-a), where

rect(x/d) means a rectangular hole of width b, '#' means convolution, d(x+a) is a delta function displaced by 'a', and 1-c is the fractional transmission on the other hole.

The fourier transform of a rect function is a sinc, d(x+a)+d(x-a) is cos(ax), so the total diffraction pattern is:

sinc(bx)*cos(ax)-c*sinc(bx)*exp(i(x-a)), where I left out a lot of scale factors involving the wavelength.

Note- that's an amplitude. To get the intensity, square it. You still get the (sinc(bx)*cos(ax))^2, but there's some extra terms -2c*sinc^2(bx)cos(ax)+c^2sinc(bx)^2.

I seem to have different signs than you, but doing this more carefully should fix that...

But it is surprising that the final pattern is symmetric.... well, maybe not, since the diffraction pattern from a single displaced slit is identical to a centered slit (it's only a difference in phase).

Now in the near field, all bets are off. I have no desire to work that out- it would be a nightmare.

3. Apr 8, 2010

### hbal9604@usyd

hey thanks for the response and input,

but I'm afraid I don't really understand your advice. I have not studied fourier transforms as such, and have only a vague idea about their purpose specifically in the context of diffraction in 2nd year physics.

the only advice I have been given in this regard, is to use the fraunhoffer diffraction double integral to work such problems, that is

Electric Field at P due to Light Through Apperture = [double integral accross apperture, lying in the xy-plane] of e^-i(ux+vy) dxdy

where u = ksin(theta_x)
v = ksin(theta_y)

and since we are talking about two slits, we only need one of these variables (say the x). Then for each slit, (say of width b) and seperated by distance a, assuming slit 2 transmits only a fraction c of the radiation being transmitte though slit 1, then

Electric Field at P due to Light Through Slit 1 and Slit 2 =
A*[integral from -b/2 to b/2 of] e^-iux dx
+ Ac*[integral from a - b/2 to a + b/2 of] e^-iux dx

where A involves a heap of constants, and a complex number involving time variation which vanishes upon muliplying by the complex conjugate to obtain the E^2 expression for irradiance calucation.

the Ac in the second term represents the fact that the second slit transmits only a fraction c of the radiation that slit 1 transmits.

this is the only way I have learned to do these questions. Can you give me some more insight on how to proceed?

4. Apr 9, 2010

### Andy Resnick

I totally understand where you are coming from- the curse of text-based interweb forums is the inability to clearly write and communicate the relevant mathematics.

Ok- item #1: "Electric Field at P due to Light Through Apperture = [double integral accross apperture, lying in the xy-plane] of e^-i(ux+vy) dxdy"

That is the Fourier transform! The conjugate coordinates 'u' and 'v' can be understood as *angles*- the diffraction pattern has a constant angular size, and the size of the pattern on a screen depends on where the screen is.

Item#2- the odd functions 'rect', 'sinc', 'd', etc. etc. Those are functions that are very useful in optics, because they are very intuitive: the field amplitude is '1' inside the aperture and '0' outside of the aperture. A point source located a distance 'a' from the origin is d(x-a), a square aperture centered at the origin is rect (x/b,y/b), sinc(x) = sin(x)/x, etc.

I apologize for using them without any explanation. A good intro to those functions (and others, like 'circ', 'somb', 'comb'...) can be found in Gaskill's "Linear Systems,Fourier Transforms, and Optics". One additional point: the FT of two convolved functions (a # b) is the multiplication of the transformed functions (A * B).

Hopefully, you can see that using these odd functions and Fourier Transforms, I don't have to evaluate wacky integrals- that's the advantage.

In any case, you are on the right track (writing down the amplitude, squaring it to get the irradiance). Those integrals can be evaluated (you can evaluate int[-a, a] (exp(cx) dx), right?) directly.