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Screwdriver
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Homework Statement
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?
Homework Equations
Constructive interference: optical path difference is a multiple of the wavelength.
Photocurrent is proportional to intensity.
The Attempt at a Solution
(a) A direct ray travels a distance of [itex]D_1 = x + d [/itex] to the photodiode. A ray traveling 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.