How Do You Calculate the Width of a Slit in a Diffraction Experiment?

  • Thread starter Thread starter BuBbLeS01
  • Start date Start date
  • Tags Tags
    Slit Width
Click For Summary
SUMMARY

The width of a slit in a diffraction experiment can be calculated using the formula W = 2tL/a, where t is the wavelength of the laser (590 nm), L is the distance from the slit to the screen (1.7 m), and a is the distance between the first and second minima (6.78 mm). The correct interpretation of the distance is crucial; it represents the difference between the m=1 and m=2 minima, not the distance from the center to the second minimum. By correctly applying the formula and understanding the setup, the slit width can be accurately determined.

PREREQUISITES
  • Understanding of single-slit diffraction principles
  • Familiarity with the wavelength of light (e.g., helium-neon laser at 590 nm)
  • Ability to manipulate equations involving distances and minima in diffraction patterns
  • Basic knowledge of trigonometry as it applies to wave phenomena
NEXT STEPS
  • Study the derivation of the single-slit diffraction formula
  • Learn how to draw and interpret diffraction patterns
  • Explore the effects of varying slit widths on diffraction patterns
  • Investigate the relationship between wavelength and diffraction in different materials
USEFUL FOR

Students in physics, particularly those studying wave optics, educators teaching diffraction concepts, and anyone involved in experimental physics setups.

BuBbLeS01
Messages
602
Reaction score
0
Width of Slit...HeLP on setup Please!

Homework Statement


A helium-neon laser (λ=590 nm) illuminates a single slit and is observed on a screen 1.7 m behind the slit. The distance between the first and second minima in the diffraction pattern is 6.78 mm. What is the width (in mm) of the slit?


Homework Equations





The Attempt at a Solution


I think I am supposed to use the equation...

W = 2tL/a
t = wavelength
t = 0.00000059 m
L = 1.7 m
a = .00678 m

but that gets me 0.2956 mm
If I divide that by 2 I get the right answer so I am a little confused on my setup
 
Physics news on Phys.org
The distance 6.78 mm is the difference between the m=2 and m=1 minima. It is not the distance from the centre of the screen to the second minima (if that's what you are doing). Draw a diagram. You need to find one expression for the distance for m=1 minimum and one for the m=2 minimum. You know the difference between these two is 6.78 mm. This will allow you to solve for the slit width.
 
The width of the slit is going from m=1 on 1 side of the central maxima to m=1 on the other side right? And fringe spacing between all fringes is equal correct?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
1
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K