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The taillights of an automobile are $1.25\:\rm{m}$ apart. Assume the pupil of a person's eye has a diameter of $5\:\rm{mm}$ and the light has an average wavelength of $604\:\rm{mm}$. At night, on a long straight highway, how far away can the two taillight be resolved? Suppose you squint your eyes, forming a slit in which the limiting angle changes from $\theta=1.22 \lambda/D$ to $\theta=\lambda/D$. What is the new distance for resolution of the images?

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Don't you have to worry about how good the optics in the eye are? The eye performs a lot better than a pinhole camera with a pupil-sized hole, and I presume a lot worse than the best possible telescope with the same size aperture. I don't think there's enough information there to answer the question. – Peter Shor Apr 16 '12 at 14:49
The eye is not going to do a very good job at detecting light with a wavelength of 604mm. – user2963 Apr 16 '12 at 15:07
I amend the wavelength of 604mm to 604nm – Yojito Apr 16 '12 at 15:09
Hooray, we went metric here! – Pygmalion Apr 16 '12 at 15:17
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This is obviously a homework question, and the forum rules dictate I can explain the priciples involved but not give the answer.

I would guess the question expects you to treat the eye as optically perfect and work out the limit due to diffraction by the pupil. That's why you've been given the size of the pupil, because the size determines how much the light will be diffracted.

Googling for "diffraction pinhole" or something along those lines will find you lots of articles. From a quick look I found http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/cirapp2.html and I think this gives you all the info you need.

It's not obvious why squinting would affect the lateral resolution. Squinting will effectively make the pupil smaller in the vertical direction so you'll get greater vertical diffraction, but shouldn't affect the horizontal resolution.

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the problem doesn't specify the orientation of the observer's head or that of the car, maybe one of them is on it's side =) – tmac Apr 16 '12 at 15:36
Also, this may be useful: en.wikipedia.org/wiki/Angular_resolution – tmac Apr 16 '12 at 15:37
@tmac: good call, that Wikipedia article gives you the equation to use. In fact now I read the post properly, the question gives you that equation so it gives you the answer! – John Rennie Apr 16 '12 at 15:41

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