Determine resolution of use by modelling it as single slit

by twinklestar28
Tags: determine, modelling, resolution, single, slit
 P: 21 1. The problem statement, all variables and given/known data Model the pupil of the eye using a single slit. If the eye is able to notice changes in intensity of 20% show that it can distinguish two objects seperated by an angle of 0.00015 radians if =500nm and diameter=4mm. 2. Relevant equations I=Iosin^2((π/λ)asinθ)/ ((π/λ)asinθ)^2 sin θ=λ/a 3. The attempt at a solution I attempted to find a (width) from the intensity equation but it doesn't seem to work, so Im unsure of how to start the calculation, the question is worth 4 marks, any help would be appreciated :)
Mentor
P: 12,069
 Quote by twinklestar28 1. The problem statement, all variables and given/known data Model the pupil of the eye using a single slit. If the eye is able to notice changes in intensity of 20% show that it can distinguish two objects seperated by an angle of 0.00015 radians if =500nm and diameter=4mm. 2. Relevant equations I=Iosin^2((π/λ)asinθ)/ ((π/λ)asinθ)^2 sin θ=λ/a 3. The attempt at a solution I attempted to find a (width) from the intensity equation but it doesn't seem to work, so Im unsure of how to start the calculation, the question is worth 4 marks, any help would be appreciated :)
Looks like for the slit width you are to use the diameter of the pupil ("Model the pupil of the eye using a single slit")

You wrote I(θ) for a slit illuminated by a single source. You need to account for the fact that there are two sources, separated by an angle of 0.00015 radians.
 P: 21 O ok so the equation would be I=Iosin^2((π/λ)asinθ-∅)/ ((π/λ)asinθ-∅)^2 it gives a hint to consider the intensity halfway between the peaks of the two images so i did ((π/λ)asin(θ-∅)) = ∏/2 but where do i put the intensity change of 0.2% to get (θ-∅) = 0.00015?
Mentor
P: 12,069
Determine resolution of use by modelling it as single slit

 Quote by twinklestar28 O ok so the equation would be I=Iosin^2((π/λ)asinθ-∅)/ ((π/λ)asinθ-∅)^2
That's the intensity of a single source that is located an angle ∅ away from the central axis.

But we have two sources. Let's say one of them is an angle ∅ away from the central axis, and one of them is on the central axis. You need to consider the intensity due to each source, and combine them to get the intensity in the combined diffraction pattern.
 P: 2 Hello, I am fairly new here. I posted one new question but I forgot how to do that. Can anybody assist me by telling me how to post a new question?
 Mentor P: 12,069 See this thread: http://www.physicsforums.com/showthr...=1#post3977513

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