Finding Light Intensity with Two Offset Slits

  • Thread starter Thread starter Gee Wiz
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around calculating the light intensity at a specific point (slit O) when light passes through two offset slits (B and C) in a setup involving multiple screens. The problem is situated in the context of wave interference and light intensity measurement, with a given wavelength of 475 nm.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the path difference between the light reaching slit O from slits B and C, questioning the phase relationship of the waves. There are attempts to calculate the intensity using trigonometric relationships and approximations.

Discussion Status

The discussion is active with various participants sharing their calculations and results. Some participants express uncertainty about their answers and the precision required for the calculations. There is acknowledgment of potential issues with the online system affecting the results.

Contextual Notes

Participants note the importance of maintaining precision in calculations and the potential impact of rounding errors on the final results. There is mention of discrepancies in answers and the need for careful tracking of units and decimal places.

Gee Wiz
Messages
137
Reaction score
0

Homework Statement


Consider the above setup, not drawn to scale.

Light of wavelength λ= 475 nm strikes 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.

1) What is the light intensity measured at the point O when both slits B and C are open?
I =?

OffsetSlits.gif

Homework Equations



sin(theta)=(n*lamda)/(d)
Law of cosine

The Attempt at a Solution



I believe that the light enters B and C is out of phase. I tried to find delta using the Pythagorean theorem and got 3.14e-6. Then I used 2*pi*delta/lamba to get what i thought would be phi. Then i did 4*.5*cos^2(phi/2)...but that didn't work. I got something around .256 but that was incorrect
 
Last edited:
Physics news on Phys.org
I think you have the right approach. I'm not getting the same value for the the path difference between path ABO and path ACO. {EDIT: Nevermind! I agree with your value for delta.}
 
Last edited:
I get .253 W/m2 which is very close to your answer.
 
I tried that as well, so only quote unquote hint I get is :
"Note that the waves coming to slits B and C are NOT necessarily in phase with each other. You can either take this into account explicitly, or note that all rays that pass through O also pass through A, and the upper-path and lower-path rays ARE in phase at A. You will need to carry out your calculations carefully, keeping many decimal points. How many? The answer needs to be good to within 1%. How accurately does phi need to be specified? And how accurately then does the path length difference need to be specified, and how many decimal places out of the total path length is this? Make sure you keep track of units!"

I have not been rounding anything in my calculations. My calculator gives me about 10 spots/places after the decimal place...
 
Oh wait, I just got an email from my professor saying the system is having issues with presicion...so i guess i'll wait a little while until it is back up to continue (maybe i actually have the right answer)
 
OK. Now I'm getting .256 in agreement with you. Good.
 
So apparently the .256 is not correct, but one of the random answers i put earlier .23305 is correct. Although I'm not sure how i got that.
 
Interesting. I don't see why your answer isn't correct. I'll try to get some of the other helpers around here to look at it.
 
With a taylor approximation to second and third order, I can confirm .25627 W.
Using only first order, I get 0.25632 W. To get a wrong answer of .23305, you need some significant rounding errors.
 
  • #10
Thank you guys so much. It turns out that .2563 is correct (there was just apparently a lot of issues with the online system)
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
9K
  • · Replies 10 ·
Replies
10
Views
9K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
6
Views
3K
Replies
3
Views
3K