Physics Help Forum angles of superposition for two diffracted wavelengths

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 Aug 6th 2009, 03:01 AM #1 Senior Member   Join Date: Feb 2009 Location: Singapore Posts: 206 angles of superposition for two diffracted wavelengths two wavelengths are diffracted one 400 nm and the other 600 nm. they have one point of superposition at $\displaystyle 20^o$ now i need to find if the angle of the other if any point of superposition for the two wavelengths. the answer is supposed to be $\displaystyle 43.3^o$. i tried equating them: $\displaystyle \frac{n_1\lambda_1}{d}=\frac{n_2\lambda_2}{d}=sin \theta$ d=285000 lines/meter but i'm lost thanks
 Aug 6th 2009, 08:31 PM #2 Senior Member   Join Date: Apr 2008 Location: HK Posts: 886 You have mentioned the right equations. But you have to think carefully about the phrase" they have a point of superposition at $\displaystyle 20^o$". For the two light rays of different wavelengths to superimpose with each other, they must first meet each other. $\displaystyle \frac{n_1 \lambda_1}{d}=\frac{n_2 \lambda_2}{d}$ $\displaystyle n_1 \lambda_1=n_2 \lambda_2$ For n to be an interger, in fact, you are finding the L.C.M. of 400nm and 600nm which is 1200nm When you plug $\displaystyle n_1=3,\lambda_1=400nm, d=1/285000m^{-1}$, you find $\displaystyle \theta=20^o$ When you plug $\displaystyle n_2=2,\lambda_1=600nm, d=1/285000m^{-1}$, you find $\displaystyle \theta=20^o$ So, from these, the second time they superimpose with each other, it should be 2400nm, for the following work, you can handle it.

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