Timeline for Homework - Young's Double Slit Experiment

8 events

when toggle format	what		by	license	comment
Jan 14, 2016 at 15:29	vote	accept	Gummy bears
Jan 14, 2016 at 15:29	comment	added	Gummy bears		Aha... Now I see it... yes... Thanks for the help! And sorry for being so stupid :)
Jan 14, 2016 at 15:21	comment	added	Floris		Triangle made by $h$ and $D$, triangle made by $\frac{d}{2}$ and $\ell$, and triangle made by path difference $\Delta$ and the slit spacing $d$ all have the same angle $\theta$.
Jan 14, 2016 at 15:19	comment	added	Gummy bears		I'm not seeing it... Which two triangles are we talking about?
Jan 14, 2016 at 15:15	comment	added	Floris		If the light arriving at a distance $h$ below $O$ has zero phase difference, then the path length difference $\Delta$ at $O$ (for small angles $\theta$) is found from (similar triangles) $\frac{\Delta}{d} = \frac{h}{D} = \frac{d}{2\ell}$. In this, I assume that $\sin\theta = \tan\theta$ which is an OK assumption for small angles. Note that your expression simplifies to the same thing if you do an expansion (noting that $d\ll l$, write $\ell\left(\sqrt{1+\frac{d^2}{\ell^2}}-1\right)\approx \ell(1+\frac{d^2}{2\ell^2} - 1) = \frac{d^2}{2\ell}$ which is what I got the other way...
Jan 14, 2016 at 15:09	comment	added	Gummy bears		Well won't the path difference simply be $\sqrt{l^2 + d^2} - l$? But unfortunately, that's not the answer given... Or am I doing something wrong?
Jan 14, 2016 at 14:35	history	edited	Floris	CC BY-SA 3.0	added 176 characters in body
Jan 14, 2016 at 14:27	history	answered	Floris	CC BY-SA 3.0