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This question already has an answer here:

I've come across a question that I don't know how to tackle:

An alternative way of observing Fraunhofer diffraction uses lenses to provide appropriate conditions. Sketch an optical configuration for observing Fraunhofer diffraction using a point source of light and two lenses.

A parallel beam of light is incident normally on a diffraction grating which consists of a regular array of 200 narrow slits per mm, each slit being 1 µm wide. Light emerging from the grating is focussed by a lens of focal length 300 mm onto a screen. The lens and screen are centred on a line perpendicular to the centre of the grating, and are parallel to the grating. It is intended to observe the 3rd diffraction order with light at a wavelength of around 460 nm. Where on the screen does this diffraction order appear?

I've done the first part of the question (sketching the optical configuration, with the aperture between two lenses, and the screen and source at a distance f away from each other lenses).

Everything I've learnt about lenses is used to magnify images, and I have no idea how to deal with a ray coming in at an angle. Surely you need to know at what distance the aperture is from the lens?

Any help/hints would be greatly appreciated.

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marked as duplicate by DavePhD, Brandon Enright, Jim, Kyle Kanos, Valter Moretti Apr 21 '14 at 20:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ en.wikipedia.org/wiki/… may be useful. $\endgroup$ – Qianyi Guo Apr 19 '14 at 13:58
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    $\begingroup$ possible duplicate of Diffraction by a lens $\endgroup$ – Stefan Bischof Apr 19 '14 at 15:07
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    $\begingroup$ I find it odd that you're in a situation where somebody is asking you about Fraunhofer diffraction when your total knowledge of lenses is that they "magnify images". Are you taking a class that you haven't taken prerequisites for? $\endgroup$ – Colin K Apr 19 '14 at 16:52
  • $\begingroup$ Sorry, I should clarify. All we've learnt is the thin lens equation, its derivation and some applications. I don't think we've covered how to apply it in this situation (or at least I don't know how). $\endgroup$ – user138901 Apr 19 '14 at 20:00
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The key to this kind of problem is (i) to think of a lens as a Fourier transformer and (ii) use the principle of Linear superposition.

Take a look at my drawing below (it's one I drew to train people in the use of infinity conjugate optics, so don't worry about the "tested objective"). The key point here is that a point source on the focal plane of a lens transforms into a plane wave (or an approximation thereto, limited by the system's aperture) tilted at an angle $\theta$ to the optical axis given by $\tan\theta= \frac{r}{f}$, where $r$ is the transverse distance of the point source from the optical axis. You can derive this from a simple ray diagram: draw a ray from the point source through the optical centre and you've got it. The ray, in the wave world, represents the wavevector, whose direction is the direction of propagation of a plane wave.

Fourier Transformer Lens

So, we have, roughly, making a paraxial approximation

$$\delta(\vec{x} - \vec{x}_0)\,\leftrightarrow\, \exp\left(i\,\frac{k}{f}\,\vec{X}\cdot \vec{x}_0\right)\tag{1}$$ where I write on the left the field distribution on the focal plane and on the right the output field distribution over the transverse plane through the optical centre of the equivalent thin lens, or at the system aperture. So now you simply use linear superposition to calculate the field distribution on the output transverse plane when the input field is $g(x,\,y)$ on the focal plane $\mathscr{F}$ or, re-writing it in terms of an inner product:

$$g(\vec{x}) = \int_\mathscr{F} \delta(\vec{x}^\prime - \vec{x})\, g(\vec{x}^\prime)\,\mathrm{d}^2 x^\prime\tag{2}$$

by linear superposition of the "basic response" in (1) as weighted in (2), the transverse distribution at the system output must be:

$$G(\vec{X}) = \int_\mathscr{F} \exp\left(i\,\frac{k}{f}\,\vec{X}\cdot \vec{x}\right)\,g(\vec{x})\,\mathrm{d}^2 x$$

where $\vec{X}$ is the transverse position in the output plane. This is, of course, the Fraunhofer diffraction integral.

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