# Diffraction: Intensity in the double slit situation

1. Oct 3, 2012

### assed

Hello. I have been studying interference and diffraction and one doubt has appeared. When you consider the double slit experiment forgeting the effects of diffraction you get the following equation for intensity

$I^{}=4I_{0}cos^{2}(\frac{πdsin(θ)}{λ})$

where d is the distance between the slits.
For the single slit diffraction we get

$I^{}=I_{0}(\frac{sin(x)}{x})^{2}$
$x^{}=(\frac{asinθπ}{λ})$

where a is the width of the slit.

Then for the double-slit case considering diffraction we get

$I^{}=4I_{0}cos^{2}(\frac{πdsin(θ)}{λ})(\frac{sin(x )}{x})^{2}$

My doubt raises when i consider the two limit cases:
1.For a/λ going to 0 the expression becomes that of the interference-only case.
2.But when we consider d=0(the distance between the centers of the slits) the expression obtained is

$I^{}=4I_{0}(\frac{sin(x)}{x})^{2}$

which is different from that of the single slit case although doing d=0 we are turning two slits of width a in one slit of width a.

My thoughts trying to solve this problem have considered that maybe I am taking the limit case wrong (although I haven't found where) or some expression for the intensity is wrong.

2. Oct 3, 2012

### Simon Bridge

... if d is the center-to-center distance between slits of width a, then this is the same as a single slit width d+a with a barrier of width d-a in the middle.

Even in the case where we let (d--->a) that is just the limit that the barrier gets very small ... there may still be a barrier in the math; just like the limiting case for a-->0 is not the same as for a solid wall with no slits.

3. Oct 4, 2012

### Philip Wood

By merging the positions of the two slits you haven't turned either of them off. You have two superimposed slits. They are giving out waves in phase. The resultant amplitude at any point is twice that due to either slit by itself, so the intensity is 4 times that due to either by itself.