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Point source of light, opaque screen with hole. photodiode problem

  1. Aug 13, 2016 #1
    1. The problem statement, all variables and given/known data

    A point source of light (wavelength [itex] \lambda = 600 \, \text{nm} [/itex]) is located a distance [itex] x = 10\,\text{m} [/itex] away from an opaque screen with a small circular hole of radius [itex]b[/itex]. A very small photodiode is moved on an axis from very far away toward the screen. The first observation of maximum signal occurs when the photodiode is at a distance [itex] d = 10.2\,\text{m} [/itex] away from the screen.

    (a) Calculate the radius [itex]b[/itex] of the hole.
    (b) If the photodiode current is [itex]i_0 = 10\,\mu \text{A}[/itex], what will this current be if the screen is removed?


    2. Relevant equations

    Constructive interference: optical path difference is a multiple of the wavelength.

    Photocurrent is proportional to intensity.

    3. The attempt at a solution

    (a) A direct ray travels a distance of [itex]D_1 = x + d [/itex] to the photodiode. A ray travelling to the edge of the hole and then to the photodiode travels a distance of [itex]D_2 = \sqrt{x^2 + b^2} + \sqrt{d^2 + b^2}[/itex] which can be approximated as [itex] D_2 \approx x + d + (b^2/2)(1/x + 1/d) [/itex] since [itex]b[/itex] is very small. Therefore, the path difference is [itex] \Delta D = (b^2/2)(1/x + 1/d)[/itex]. This needs to be equal to an integer multiple of the wavelength, [itex]m \lambda [/itex], and in particular [itex] m = 1 [/itex] since we're talking about the first maximum after coming in from very far away. Therefore,

    [tex]
    b = \sqrt{\frac{2 \lambda x d}{x + d}}
    [/tex]

    (b) When the screen is present, the hole acts like a point source with intensity proportional to [itex]\pi b^2 / 4 \pi x^2 [/itex]. The intensity at the photodiode is therefore proportional to [itex](\pi b^2 / 4 \pi x^2)(1/4 \pi d^2) [/itex]. Without the screen, the intensity is simply proportional to [itex](1/4 \pi (x + d)^2) [/itex]. Therefore,

    [tex]
    \frac{i_1}{i_0} = \frac{1}{\pi b} \left(\frac{x}{x + d} \right)^2
    [/tex]

    My waves/optics knowledge is somewhat rusty and (b) is a complete guess.
     
  2. jcsd
  3. Aug 13, 2016 #2

    Simon Bridge

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    The argument seems reasonable given the assumptions ... ie. that m=1, and that the photodiode is approaching down the line through the point source and the center of the hole.

    You can increase your confidence levels by checking your assumptions.
    Since the detector is approaching from infinity, then the first max encountered will be the farthest from the screen.
    So how does d depend on m?

    For the second part. i1 and i2 are the intensities measured at the detector position right?
    Don't these also depend on the source intencities? ie. does all the light from the source pass through the hole? Is the hole really a point source (what path do light rays from the hole take to get to the detector)?
     
  4. Aug 13, 2016 #3
    Sorry, I don't follow: [itex]m[/itex] is just [itex]1[/itex].

    Yes.

    Sure, but the intensity of the source is an overall constant which divides out.

    I assume the source is isotropic, so no.

    I guess not really, but my idea is that it's so small that it's close enough - otherwise I'm not sure what to do.
     
  5. Aug 13, 2016 #4

    Simon Bridge

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    That was the assumption, yes. You are assuming that the first maximum encountered as the detector moves in from infinity, te one that is farthest away from the screen, is going to be the one that corresponds to m=1. What basis do you have to believe this? ie is the value of d when you assume m=2 closer to, or farther away from, the screen than the result assuming m=1?

    That would be true for the same source at two different distances.
    Aren't you comparing two different sources though (the point source and the hole)?
    How did you account for the possibility that these sources may not have the same strength?

    Sure - a point source usually means spheriical wavefronts ... so what does that mean about the amount of light coming from the hole compared with that coming from the point source?

    If the distance to the detector is large compared with the diameter of the source, then it is a good approximation to model the source as a point source. The trouble with this is that a point source does not exhibit interference maxima. Doesn't the existance of interference maxima suggest that the hole may not be well modelled as a point source?

    Your model for finding the interference maxima says that light from the edge of the hole must arrive at the detector in phase with light from the center of the hole. You will need a way to work out the net flux arriving at the detector. It is possible for an interference effect to result in a brighter image than could be produced without. See Feynman's example of interference off a plane mirror.

    I don't know what level this problem is supposed to be tackled at though.

    Don't get me wrong: I have not worked through the problem. I am responding this way because you do not seem confident that you have the correct answer/approach. One way to gain confidence is to identify your assumptions and check that they make sense. This, I encourage you to do.
     
  6. Aug 14, 2016 #5
    Closer. The path difference is small when the detector is far away and then increases as [itex]d[/itex] decreases. We'll "get to" [itex]m = 1 [/itex] first.

    The intensity of a point source goes like [itex]1/4 \pi r^2 [/itex], so if the intensity of the source is [itex]I_0[/itex], the intensity at the hole should be [itex]I_0/4 \pi x^2[/itex]. Only a fraction [itex] \pi b^2 [/itex] gets through the hole, so the intensity of the hole should be [itex]I_0 \pi b^2 /4 \pi x^2[/itex] (I accounted for this). The amount of light coming from the hole is (as you would expect) much less than the amount of light coming from the source.

    Well, in "real life" a point source diffracting through a circular aperture makes an Airy disk. The photodiode (according to the question) has no physical extent, so you're basically just getting the exact middle of the central maximum. Does the central maximum of an Airy disk itself experience maxima and minima as the distance of the screen moves? Yes. Furthermore, the intensity should still go like [itex]1/4 \pi r^2 [/itex]. This question is from an old exam so it should be solvable in something like 10 minutes.
     
  7. Aug 14, 2016 #6

    Simon Bridge

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    See how adding checking can lead to increased confidence?

    A student sitting an exam can be expected to remember things like worked examples made during the course - this gives them a shortcut not available to someone approaching from outside the course. When the exam is quite old, there is an additional complication that what students were expected to know as a matter of rote may have changed. I had trouble with a high-school level geometry exam once: it took minutes to work through some problems the students of the day would have done in seconds because they would have spent time memorizing which equations and formulae to use while I had to derive them.

    In this case I don't know what the course was so maybe students are expected to make some approximation that would not occur to us? However, I think you have good reason to be confident in your results: your reasoning is sound and the assumptions supported.
    I don't think anyone can, scientifically, say more.
     
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