Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
Point source of light, opaque screen with hole. photodiode problem
Reply to thread
Message
[QUOTE="Screwdriver, post: 5542717, member: 287001"] [h2]Homework Statement [/h2] 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?[h2]Homework Equations[/h2] Constructive interference: optical path difference is a multiple of the wavelength. Photocurrent is proportional to intensity. [h2]The Attempt at a Solution[/h2] (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. [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Point source of light, opaque screen with hole. photodiode problem
Back
Top