Intensity - Double slit diffraction

[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

[itex]I^{}=4I_{0}cos^{2}(\frac{πdsin(θ)}{λ})[/itex]

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

[itex]I^{}=I_{0}(\frac{sin(x)}{x})^{2}[/itex]

[itex]x^{}=(\frac{asinθπ}{λ})[/itex]

where a is the width of the slit.

Then for the double-slit case considering diffraction we get

[itex]I^{}=4I_{0}cos^{2}(\frac{πdsin(θ)}{λ})(\frac{sin(x)}{x})^{2}[/itex]

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

[itex]I^{}=4I_{0}(\frac{sin(x)}{x})^{2}[/itex]

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.

Thanks in advance for your attention, expecting a good answer...

PS.:I also apologize in advance for any mistake in my english writing...Brazilian here.

 

[gtoguy97]

The expression for the intensity in the double slit case considering diffraction is correct, however you are not taking the limit case correctly. When the distance between the slits (d) is zero, the equation for intensity simplifies to:I^{}=4I_{0}(\frac{sin(x)}{x})^{2}where x = (\frac{a sinθπ}{λ}), a being the width of the slit. This is the same equation for intensity as the single slit diffraction case, since the two slits have been combined into one slit of width 'a'.