Distance between each double-slit and the light band.

1. Oct 30, 2008

RProgrammer

1. The problem statement, all variables and given/known data

This #74 is on the SAT Practice Subject Test in Physics, which is also the Physics test administered in January 2003.

2. Relevant equations
I don't even know how to begin this problem.

3. The attempt at a solution
I know Young's double-slit equation: nλL=dx
But I don't know any equation for the 'x' and 'y' in the diagram, the distance from the actual slits to the band other than the Pythagorean theorem.
I tried solving with those, and I get a difference, but it's not one of the provided answers.
In my attempt I called the slits A (top) and B (bottom) (since 'x' is the fringe distance), and the distances (x and y in the diagram) Za and Zb.
The system of equations I tried to manipulate was:
Za2=L2 + (d/2 - x)2
Zb2=L2 + (d/2 + x)2
nλL=dx

2. Oct 30, 2008

alphysicist

That is the double slit equation for small angles; the original double slit equation is:

$$d \sin\theta= n \lambda$$

Look at how this equation was derived in your textbook; what does $d\sin\theta$ represent?

3. Oct 30, 2008

RProgrammer

I don't have a textbook, I'm just going over the questions I missed online.
I've gotten all of them except this one. If you could just point me in the right direction, I would be most grateful.

4. Oct 30, 2008

alphysicist

Look at the diagram at the top of this page (just ignore the lens in the middle of the diagram):

http://www.physics.hmc.edu/courses/Ph51/two_slit.html

Right next to the slits, you can see a part labeled $d\sin\theta$. If you consider that part, you can see that $d\sin\theta$ is how much more that one light ray has to travel than the other; it's that path length difference between the two paths.

For constructive interference (maximum intensity), the equation that describes it is the one in my last post: $d\sin\theta = n \lambda$. Putting both of these things together, do you see what the answer to the problem is now?

5. Oct 30, 2008

RProgrammer

Thank you, this helps a lot.