A problem on double slit experiment The setup of double slit experiment is shown, find the minimum value of d for which O is the brightest point.
I've tried to solve this using one method, but the answer don't match. The solution manual suggests that i use Pythagoras theorem and then finally binomial expansion to determine the path difference. I want to know where i'm wrong.
The answer should be: ((lambda)*D)^(1/2)


 A: I think I know what's the problem. First of all, the $d\sin\theta$ is only meant to be an approximation, but that's not the important thing to know. The most important thing is that it is an approximation that works only when the screen is so far away that the two paths are nearly parallel. 
In this situation, the problem explicitly says that the two paths have a non-negligible angle $\theta$ to each other. Therefore, you can't invoke the $d\sin\theta$ by design of the problem. 
You'd have to use Samuel Weir's suggestion and look at the path length difference directly using the Pythagorean Theorem. 
The bottom path has length $2D$ and the top path has length $2\sqrt{d^{2}+D^{2}}$. The path difference is 
$$2\sqrt{d^{2}+D^{2}}-2D = 2D\left( \sqrt{1+\tfrac{d^{2}}{D^{2}}}-1 \right) \approx 2D\left( \left(1+\tfrac{1}{2}\tfrac{d^{2}}{D^{2}}\right) -1 \right) = \frac{d^{2}}{D}. $$
As you already realized, you want the path difference to be $d^{2}/D = \lambda$ so then $d = \sqrt{\lambda D}$.

You can see the issues a little more clearly if you actually look at an accurate representation of the setup and compare the two paths with a circle/compass.

If you look closely, you'll see that the blue line and the diagonal black line do not form a right angle, so assuming a path difference of $d\sin\theta$ wouldn't work.
However, like I already said, the real problem is that the two paths are not near parallel. That is what prevents us from invoking the $d\sin\theta$ approximation.
