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Diffraction from a circular aperture

  • Thread starter Brewer
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  • #1
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Homework Statement


A ruby laser ([tex]\lambda[/tex] = 694.3nm) is aimed in free space at a target 10,000km away. If the laser beam is initially 14cm in diameter, how large will its diameter be when it hits the target taking account of diffraction at the circular aperture of the laser?


Homework Equations


I think:
sin[tex]\theta[/tex] = 1.22[tex]\lambda[/tex]/D


The Attempt at a Solution


sin[tex]\theta[/tex] = opp/hyp = x/r

therefore:
sin[tex]\theta[/tex] = 1.22[tex]\lambda[/tex]/D = x/r

===> r= xD/(1.22[tex]\lambda[/tex]), which after plugging in the numbers gives r = 1.65 * 10^12 m.

Then by a bit of Pythagoras it can be found that y(the diameter) = 1.65 * 10^12m.

Now I'm not sure about this. I couldn't find an equation in my notes or text book that seemed to explain this entirely. However the equation given says that 85% of the light is in the first circular maximum, so I thought it would be a valid assumption to assume that this would be the diameter of the image formed. I also have trouble believing my answer. It seems a little large, and I don't know if I've approached it the correct way either.

Any comments and/or hints would be appreciated.

Brewer
 
Last edited:

Answers and Replies

  • #2
hage567
Homework Helper
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I think your approach is basically right, but I'm confused about your equation for "r". What is x? I think you can use your relevant equation to find the angle of the beam as it leaves the aperture. After that, use trig to find out what the size is on the screen.

I get a different answer than you (I'm not saying mine is right), but I'm having trouble understanding your solution. Could you show more detail in you work?
 
  • #3
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I'll scan in my work and post it in a minute or two. Thanks for reading this by the way.
 

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