I\'m having trouble getting started with this problem. I know theinterference fo

Question

I'm having trouble getting started with this problem. I know theinterference formulas and how it works, but I'm having troubledetermining the phase difference (if B and C were in phase i coulddo it easily, but they aren't, how do I find out the phase they areinitially apart?)
|----------|------------|
|----------B------------|
|----------|------------|
|----------C------------|
|----------|------------|
A----------|------------O
|----------|------------|
|----------|------------|
|<---x1-->|<-----x2--->|


Consider the above setup, not drawn to scale.

Light of wavelength = 475 nm is shined at normal incidence tothe first screen with slit A.
The second screen, x1 = 0.7 meters behind the first screen, has twoslits, B and C .
The third screen is x2 = 1.5 meters behind the second screen. Ithas slit O, which is level with slit A. A lightmeter measures thelight intensity at the slit O.

When light is sent through slit A and measured at the slit O witheither slit B or slit C open one slit at a time, the intensity atthe point O is the same: I0 = 0.5 W/m2. (The slit widths can alwaysbe adjusted so that this is true, but for this problem youcan/should ignore the width of all slits.)

Slit B is at height y1 = 2 mm above slit A.

Slit C is at height y2 = 1 mm above slit A.

Note that the drawing is not drawn to scale.

a) What is the light intensity measured at the point O when bothslits B and C are open?

Explanation / Answer

a) .2563 b) .2563 c) 479 For a, you have to find the path difference between ABO and ACO. This is 3.142851 E-6. Plug this into phi = 2pi (delta) / lambda. This gets you 41.572874. Now you should have everything you need to solve I = 4(I1) cos^2 (phi/2). B should be the same, because you're increasing both ABO and ACO by the same amount, therefore not changing the path difference.

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