Recent content by khaos89

@PeterO: Finishing what I have started: Min: \cos(\theta_1) = \frac{m}{3} which has the "first" solution when m=3, it gives \theta_1 = 0 ... which I think is wrong... :\ @ both: The problem says the light has been emitted with \lambda and then it incides on the...

Hey, thank you so much, I think u are right, but still, ahaha, maybe it's me but new problems here! Max: 2n\cos(\theta_1) = (2m+1) it becomes cos(\theta_1)=\frac{2m+1}{3} The first solution i found is when m=1, indeed cos(\theta_1)=1 so that \theta_1=0°, now, to find the...

Thanks a lot :)

Homework Statement We have a thin film of glass which has thickness t=\lambda and n=1.5 and light (\lambda) passing through it with an angle \theta_0. We have to find the minimum angle that allows us to see both constructive and destructive interference. Homework Equations Max...

Sorry, here we go with the pic: [PLAIN]http://img193.imageshack.us/img193/2271/schermata082455775alle1.png

Look at the picture below, I have to prove that the optical path length difference is \Delta=n(BC+CD)-BE=2nd\cos(r) [PLAIN]http://img200.imageshack.us/img200/2271/schermata082455775alle1.th.png The problem is just how to get 2nd\cos(r) I actually don't have any idea :\ I have...

Thank you for your answer, I think I am wrong because my book's solution is t=\frac{m\lambda}{2n}

Hello, I am trying to understand thin film (in air) interference but I have a problem: I know we have destructive interference when \delta=(2m+1)\pi. Now i can try to calculate the thickness of the film to get it, so since \delta =\frac{4nt\pi}{\lambda} - \pi where t is the thickness...