Double slit problem, finding λ of second laser given data about first laser

Click For Summary
SUMMARY

The discussion centers on calculating the wavelength λ2 of a second laser in a double slit experiment, where the first laser has a wavelength of d/8 and a slit separation of 0.500 mm. The goal is to find λ2 such that its second maximum aligns with the fourth minimum of the first laser. The equations used include dsin(θ) = 2λ1 for the first laser and dsin(θ) = (9/2)λ2 for the second laser. The correct approach involves equating these two expressions to derive λ2 accurately, as previous attempts yielded incorrect results.

PREREQUISITES
  • Understanding of double slit interference patterns
  • Familiarity with the relationship between wavelength and interference maxima/minima
  • Knowledge of trigonometric functions in the context of wave physics
  • Ability to manipulate and solve algebraic equations
NEXT STEPS
  • Study the principles of wave interference in double slit experiments
  • Learn how to derive relationships between wavelengths and interference patterns
  • Explore the mathematical derivation of maxima and minima in wave physics
  • Investigate the impact of varying slit separation on interference patterns
USEFUL FOR

Students in physics, particularly those studying wave mechanics and optics, as well as educators looking to enhance their understanding of interference phenomena in laser applications.

aliaze1
Messages
173
Reaction score
1

Homework Statement



A laser with wavelength d/8 is shining light on a double slit with slit separation 0.500 rm mm. This results in an interference pattern on a screen a distance L away from the slits. We wish to shine a second laser, with a different wavelength, through the same slits.

What is the wavelength λ2 of the second laser that would place its second maximum at the same location as the fourth minimum of the first laser, if d = 0.500 mm?


Homework Equations



http://photo.ringo.com/233/233184793O806463150.jpg

http://photo.ringo.com/233/233184793O806463150.jpg

The Attempt at a Solution



I have only one attempt left, so I used those equations and made them both sin(θ)= ..., and set them equal to each other. I got an answer of 1/56 but I do not want to use it and loose my last attempt and get the problem wrong. can anybody confirm this?

Thanks
 
Last edited by a moderator:
Physics news on Phys.org
i should also add that i got 1/36, but it said that the first minimum corresponds to m=0 not m=1, would this also apply to the maximum? if so, that is how i got 1/56...
 
Since the maxima and the minima are at the same point, dsin\theta for both cases is the same.

For the first case, dsin\theta=2\lambda _1.

For the second case, dsin\theta=\frac{9}{2}\lambda _2 (m=4)

Equate the two and solve, your answer seems to be incorrect.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 9 ·
Replies
9
Views
8K
  • · Replies 6 ·
Replies
6
Views
5K
Replies
1
Views
4K
  • · Replies 11 ·
Replies
11
Views
4K
Replies
6
Views
2K
Replies
3
Views
9K
  • · Replies 3 ·
Replies
3
Views
11K
  • · Replies 14 ·
Replies
14
Views
6K