Interference

[kd9485]

Can someone tell me the equations we need to know related to single/double-slit interference and thin-film interference etc.?

Also on the TMSCA 2012 Science Test, #58, how would you do it?

Two slits each have a width of [unparseable or potentially dangerous latex formula] are separated by a distance [unparseable or potentially dangerous latex formula] and are illuminated by light of wavelength [unparseable or potentially dangerous latex formula]. How many bright fringes are seen in the central diffraction maximum?

Answer: 7

[nsguy1350]

Umm.. I believe this works:
[unparseable or potentially dangerous latex formula]
For one slit. I think you solve for x, then multiply by the number of fringes.
Wait, I'm rusty.
d = .015E-3
x = .060E-3
wavelength = 6.25E-7
Solve for l....
hmm... disregard that, I'm a failure.

[Salvadoramerican]

other than that i have a bit difficulty but i try [unparseable or potentially dangerous latex formula] and combining it with another formula

[Salvadoramerican]

or we could do [unparseable or potentially dangerous latex formula] which gives 7 if you want explanation just ask and i can prove it

[Salvadoramerican]

took me a moment just devising the method to solve this it only works if the wavelength is fixed not changing

[kd9485]

I think the wavelength is fixed.

Would you mind showing me how you got there? Just in case another similar question shows up?

[Salvadoramerican]

sure first of lets begin with the fundamental two-slit formula [unparseable or potentially dangerous latex formula] anf the single slit [unparseable or potentially dangerous latex formula] its gonna get confusing so bear with me first lets start of with the dark fringes
The first dark fringes for diffraction are located at [unparseable or potentially dangerous latex formula]then [unparseable or potentially dangerous latex formula] where Y is their distance from the center of the central diffraction maximum, x is the distance of the two slits from the screen which reflects the interference pattern, l is the wavelength, and a is the width of the slits. Since these diffraction minima occur on either side of the diffraction central maximum, we have the total distance between them equal to [unparseable or potentially dangerous latex formula]

The interference maxima are given by [unparseable or potentially dangerous latex formula]

The number of maxima inside a given width [unparseable or potentially dangerous latex formula] centered on the central maximum for interference (which coincides with the central maximum for diffraction) is [unparseable or potentially dangerous latex formula] with [unparseable or potentially dangerous latex formula]. being the largest value of m for which the interference maximum still fits inside [unparseable or potentially dangerous latex formula]. This occurs since we have [unparseable or potentially dangerous latex formula]. interference maxima on either side of the central maximum then [unparseable or potentially dangerous latex formula]
then we substitute y with an equation above thus [unparseable or potentially dangerous latex formula]

where we note the strictly less than sign results since a maximum that occurs right at the diffraction minimum would not appear. Since [unparseable or potentially dangerous latex formula]. must be an integer, we have [unparseable or potentially dangerous latex formula]

then finally with nearest integer meaning the nearest integer smaller than or equal to the quantity in the square bracket. Hence the number of maxima is
[unparseable or potentially dangerous latex formula]
wow thats was longer than i thought

[sxk1693]