Optimizing 4-slit Interference Pattern for θ1st Min with Homework Solution

  • Thread starter Thread starter Thefox14
  • Start date Start date
  • Tags Tags
    Interference Slit
Click For Summary

Homework Help Overview

The discussion revolves around a problem involving a 4-slit interference pattern, specifically determining the angle at which the intensity first goes to zero when light of a certain wavelength passes through slits spaced a specific distance apart.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the two-slit interference equation to a four-slit scenario, leading to confusion about the resulting angle. Some participants clarify that the approach must consider the contributions from all four slits to achieve destructive interference.

Discussion Status

Participants are actively discussing the differences between two-slit and four-slit interference patterns. Some guidance has been offered regarding the correct approach to determine the angle for the first minimum, with acknowledgment of the original poster's misunderstanding.

Contextual Notes

There is a mention of the original poster's frustration with the topic, indicating a potential challenge in grasping the underlying concepts of interference patterns.

Thefox14
Messages
40
Reaction score
0

Homework Statement



A 4-slit interference pattern is created by 4 slits spaced 2μm apart. At what approximate angle θ does the intensity go to zero the first time, if the incoming light has a wavelength of 450 nm?
θ1st min =

Homework Equations


d*sinθ = m\lambda

The Attempt at a Solution



I thought this problem was as simple as using the above equation, setting dsinθ = \lambda/2 and solving for θ. When I did, I got 6.45 degrees; but the correct answer is supposed to be 3.22 degrees.

Why would the angle I got be 2x the angle I'm looking for?
 
Physics news on Phys.org
Thefox14 said:
... dsinθ = \lambda/2 ...
That is for two-slit interference.

For four slits, you need to combine waves from all four slits and have sum to zero.
 
Redbelly98 said:
That is for two-slit interference.

For four slits, you need to combine waves from all four slits and have sum to zero.

Oh duh! haha thanks I got it now.

For anyone else who comes across this, though, here is what I did:

I_{f} = I(\frac{sin(N\theta/2)}{sin(\theta/2)})^{2}

So intensity will go to zero if we get the top part in the fraction to be 0. Then we know sin is zero in multiples of pi. So after that you get this equation:

N\theta/2 = m*pi

Simplify that and you end up with sin\theta = \frac{m\lambda}{Nd}
 
Kind of off-topic, this had to be the worst part of class for me to be honest. Completely uninteresting and overall hard concept to grasp (for me).

@above that's a good idea because I hated it when I needed help asap and finally found a thread with my question but no help :p
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 34 ·
2
Replies
34
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
6K