I\'m having trouble getting started with this problem. I know theinterference fo
ID: 2115557 • Letter: I
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 %u03BB= 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?
b)What is the smallest wavelength greater than 475 nm for which the intensity at point O will be zero?
Explanation / Answer
a) .2563
b) 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).
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