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?
sin[tex]\theta[/tex] = 1.22[tex]\lambda[/tex]/D
The Attempt at a Solution
sin[tex]\theta[/tex] = opp/hyp = x/r
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.