Interference of Light: Calculating Intensity at Point O with Two Offset Slits

  • Thread starter Thread starter dopher
  • Start date Start date
dopher
Messages
2
Reaction score
0
I'm having trouble getting started with this problem. I know the interference formulas and how it works, but I'm having trouble determining the phase difference (if B and C were in phase i could do it easily, but they aren't, how do I find out the phase they are initially apart?)
|----------|------------|
|----------B------------|
|----------|------------|
|----------C------------|
|----------|------------|
A----------|------------O
|----------|------------|
|----------|------------|
|<---x1-->|<-----x2--->|


Consider the above setup, not drawn to scale.

Light of wavelength λ= 475 nm is shined at normal incidence to the first screen with slit A.
The second screen, x1 = 0.7 meters behind the first screen, has two slits, B and C .
The third screen is x2 = 1.5 meters behind the second screen. It has slit O, which is level with slit A. A lightmeter measures the light intensity at the slit O.

When light is sent through slit A and measured at the slit O with either slit B or slit C open one slit at a time, the intensity at the point O is the same: I0 = 0.5 W/m2. (The slit widths can always be adjusted so that this is true, but for this problem you can/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 both slits B and C are open?
 
Physics news on Phys.org
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.

...this is not from me...
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?

Similar threads

Back
Top