# Interference viewed in the focal plane

I'm doing a problem where we have plane monochromatic waves incident on finite double slits (each of width a, separation d), and the transmitted light is observed in the focal plane of a lens of focal length f. The position of the lens isn't specified.

So I have worked out that the intensity distribution (ignoring the lens at the moment) in the Fraunhofer regime would be
I(θ)=I(0)cos2(kdsinθ/2)sin2(kasinθ/2)/(kasinθ/2)2

However I'm finding it hard to understand how the lens fits into the problem exactly. If the lens was right after the screen, I would write sinθ≈tanθ=x/f where x is the position on the screen, and then rewrite my intensity in terms of x instead of θ - this works because the length between slits and screen is clearly f. However, if the lens was anywhere else, which it could well be, the distance between slits and screen wouldn't simply be f anymore. I don't see how the above could hold anymore, and so how would I rewrite the intensity in terms of f now?. So the problem is how does this work out when the lens is in any position

Thanks for any help!

Note, I would have this same issue if the aperture was simply two infinitesimal double slits with intensity pattern
I(θ)=I(0)cos2(kdsinθ/2)
so feel free to use this if necessary to simplify things.

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Drakkith
Staff Emeritus
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Be patient. This isn't exactly a question we get often, so it may take some time for someone to answer.

Be patient. This isn't exactly a question we get often, so it may take some time for someone to answer.

I appreciate that, it's just it's probably more likely to slip down the list and be forgotten about...

Andy Resnick
I'm doing a problem where we have plane monochromatic waves incident on finite double slits (each of width a, separation d), and the transmitted light is observed in the focal plane of a lens of focal length f. The position of the lens isn't specified.

It's just a scaling factor that 'converts' the angular extent of the diffraction pattern into a linear extent. Say the lens was placed a distance 'f' from the slits; the interference pattern would be located a distance 'f' from the lens and the linear extent of the angular diffraction pattern given by x'= λf/x.

http://ocw.mit.edu/courses/mechanic...fer-functions-transforms/MIT2_71S09_lec18.pdf

It's just a scaling factor that 'converts' the angular extent of the diffraction pattern into a linear extent. Say the lens was placed a distance 'f' from the slits; the interference pattern would be located a distance 'f' from the lens and the linear extent of the angular diffraction pattern given by x'= λf/x.

http://ocw.mit.edu/courses/mechanic...fer-functions-transforms/MIT2_71S09_lec18.pdf

Did you mean 'x' when you defined the first 'f'?

Even so, I still can't quite understand where you obtained x'=λf/x from. I assume you are using the diagram on slide 15/16 with z=x and x''=x', but it's not clear exactly what to do? After this, how would you insert this into the intensity profile exactly (i.e how would you replace θ)?

Andy Resnick
Even so, I still can't quite understand where you obtained x'=λf/x from. I assume you are using the diagram on slide 15/16 with z=x and x''=x', but it's not clear exactly what to do? After this, how would you insert this into the intensity profile exactly (i.e how would you replace θ)?

Using text makes the expressions a little clumsy- let me try again.

In the case for an object placed a distance 'z' in front of a lens with focal length f, the amplitude and phase of the light at image position (xf, yf) in the rear focal plane of the lens is given by the amplitude and phase of the object spectrum at frequencies (xf/λf, yf/λf) with an overall phase factor that goes as 1/λf (1-z/f). Alternatively, the image spectrum (fx, fy) = (xf/λf, yf/λf).

Now, you have a 1-D intensity pattern: I(0)cos2(kdsinθ/2)sinc2(kasinθ/2) so the situation is somewhat simpler, but in any case you have the spectrum specified in terms of an angular (θ) frequency (call it fx), so in order to figure out 'how big' the pattern is at your image plane simply substitute in the scaling rule above.

How's that?

Using text makes the expressions a little clumsy- let me try again.

In the case for an object placed a distance 'z' in front of a lens with focal length f, the amplitude and phase of the light at image position (xf, yf) in the rear focal plane of the lens is given by the amplitude and phase of the object spectrum at frequencies (xf/λf, yf/λf) with an overall phase factor that goes as 1/λf (1-z/f). Alternatively, the image spectrum (fx, fy) = (xf/λf, yf/λf).

Now, you have a 1-D intensity pattern: I(0)cos2(kdsinθ/2)sinc2(kasinθ/2) so the situation is somewhat simpler, but in any case you have the spectrum specified in terms of an angular (θ) frequency (call it fx), so in order to figure out 'how big' the pattern is at your image plane simply substitute in the scaling rule above.

How's that?

Hmm, θ in my intensity is just the angle subtended at the screen from the centre of the aperture, as here (just obviously we have a lens between aperture and screen):
http://www.citycollegiate.com/young1.gif

Andy Resnick
Hmm, θ in my intensity is just the angle subtended at the screen from the centre of the aperture, as here (just obviously we have a lens between aperture and screen):
http://www.citycollegiate.com/young1.gif

Right- adding a lens simply alters the relationship between 'y' and 'θ'.

Right- adding a lens simply alters the relationship between 'y' and 'θ'.

Without the lens, I can write sinθ=y/L and so the intensity distribution becomes
I(θ)=I(0)cos2(kdy/2L)sin2(kay/2L)/(kay/2L)2.

If the lens is a distance D from the slits, and f from the screen, so that L=D+f, I have sinθ=y/D+f, however I am not given what D is so how can I get rid of it?

Andy Resnick
That's what I wrote up in post #7- what you are calling 'D', I wrote as 'z'.

That's what I wrote up in post #7- what you are calling 'D', I wrote as 'z'.

I'm not quite sure what you mean by 'object spectrum at frequencies (xf/λf, yf/λf) with an overall phase factor that goes as 1/λf (1-z/f). Alternatively, the image spectrum (fx, fy) = (xf/λf, yf/λf)'. How did you get those expressions and what does 'image spectrum' mean?

Apart from this, I'm guessing it's impossible to eliminate z - the position of the lens does matter?

Andy Resnick