# Effect of slits and a lens

We have a screen with two slits (Young style) separated by a distance $d$, one of them receives a planar wave of $600nm$, the other receives a planar wave of $400nm$. Behind the slits there's a screen we observe.

Only with this, as the waves have different wavelengths, I guess there can't be any interference, we will only see the difraction pattern, the two functions of the form $\sin^2(x)/x^2$, with the principal maximums separated a distance $d$. Am I right here?

Now, we put a convex lens behind the slits so the screen in which we observe is in the focal plane. Ok, I have three configurations:

1.- In the first one, the system is configured such that the slits are far away from the lens. Here, we can approximate the wave that arrives as a planar wave, and therefore the lens will perform the Fourier transform in the focal plane of the screen. The diffraction of the slits also performs the Fourier transform, so this configuration should lead to having only two bars of light in the screen, centered in the focus. Am I right?

2.- The slits are in the focal plane on the lens, such that the lens is in the middle of slits-screen. Here, the same thing should happen, right? As the light comes from the focal plane, the lens must do the fourier transform with no extra things, and we should get the two bars, again both of them in the same line (center of the screen). Am I right here?

3.- The last one, I can't see... the lens is just behind the slits so the distance between slits and lens is $\approx 0$. My guess here is that the two centers of the intesity distributions $\sin^2(x)/x^2$ will go to the center of the screen (the focus) because they go perpendicular to the lens, but the rest of the pattern will just be compressed a little. Again the wave don't interfere, so the intensities just sum up, and the resulting will be:

$$I=I_1\frac{\sin^2(\alpha x)}{(\alpha x)^2}+I_2\frac{\sin^2(\beta x')}{(\beta x')^2}$$

Being $\alpha$ and $\beta$ the some factor of compression due to the lens, that could actually be a function of $x$. Am I right here? Am I completely wrong? What would happen?

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## 1 Answer

Only with this, as the waves have different wavelengths, I guess there can't be any interference, we will only see the difraction pattern, the two functions of the form sin2(x)/x2, with the principal maximums separated a distance d. Am I right here?

Sort of. The diffraction pattern is visible "at infinity", which is in fact your case #3. I'll explain there.

1.- In the first one, the system is configured such that the slits are far away from the lens. Here, we can approximate the wave that arrives as a planar wave, and therefore the lens will perform the Fourier transform in the focal plane of the screen. The diffraction of the slits also performs the Fourier transform, so this configuration should lead to having only two bars of light in the screen, centered in the focus. Am I right?

Sort of. Your lens has a limited diameter, so if you place it far from the slits, it will capture only the central portion of the diffraction pattern, i.e. the top of the sinx/x function. In other words, you will loose the fringe pattern and reconstruct slits without fringes.

2.- The slits are in the focal plane on the lens, such that the lens is in the middle of slits-screen. Here, the same thing should happen, right? As the light comes from the focal plane, the lens must do the fourier transform with no extra things, and we should get the two bars, again both of them in the same line (center of the screen). Am I right here?

This will be somewhat different because you do the image of the slits at infinity, i.e. a blurry image at close distances. Depending on how close is the lens, you may only be taking the "no-fringe" portion of the diffraction pattern.

3.- The last one, I can't see... the lens is just behind the slits so the distance between slits and lens is 0.

That is exactly the typical school case. The diffraction pattern is, before the lens, located at infinity, or in other words, the fringes are defined as angles and not position ($\sin\theta/\theta$). The role of the lens is to bring these fringes at a finite distance (the focal length). You probably learned that a lens makes an image, initially located at infinity, located at the focal length. That's the same with the diffraction fringes.

Now, as the two slits are both close to the lens, you won't do an image of them. That means that you should not see separated slit images, but instead, you should see two superimposed diffraction patterns, centered at the same point.

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Thank you for your answer, specially the third case. So, the lens still compresses the difraction pattern of the $\sin^2 x/x^2$, doesn't it? i understand they're centered at the same point, but is there a way to calculate quantitatively the position of the fringes in the focal plane of the lens? –  MyUserIsThis Jun 12 '13 at 13:28
The formula $\sin^2x/x^2$ (or also written $\mathrm{sinc}^2x$) is already in the focal plane of the lens. Without a lens, the fringe pattern is calculated in terms of angles: $\mathrm{sinc}^2((\pi d/\lambda)\sin\theta)$. The effect of the lens can be calculated easily as $x=f\sin\theta$ from geometrical optics. You then obtain your initial formula $\mathrm{sinc}^2(\pi xd/(f\lambda))$. –  fffred Jun 12 '13 at 23:46
Ok, thank you, that: $x=f\sin \theta$ was what I needed, I was using $x$ as a generic variable, meaning angular coordinates: $\mathrm{sinc} ^2(kpd/2)$. Thanks for your help. –  MyUserIsThis Jun 12 '13 at 23:52
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