Calculating Single Slit Diffraction Width with 630 nm Light

Click For Summary

Homework Help Overview

The problem involves calculating the width of a single slit using the diffraction pattern produced by 630 nm light. The setup includes a screen positioned 2.8 m away from the slit, with a specified linear distance between the first and second dark fringes of 12 cm.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the angle of diffraction and the positions of the dark fringes. There is an exploration of the equations related to the diffraction pattern, with some questioning the presence of a factor in the calculations.

Discussion Status

Some participants have provided feedback on attempts, with one noting a calculation error related to a factor in the formula. Others have expressed confusion about the values used for the order of the dark fringes, indicating an ongoing exploration of the problem.

Contextual Notes

Participants are navigating through the implications of their calculations and the definitions of variables involved in the diffraction equations. There is a mention of common mistakes regarding the order of dark fringes and the use of specific factors in the equations.

muffintop
Messages
13
Reaction score
0

Homework Statement


A single slit is illuminated with 630 nm light, and the resulting diffraction pattern is viewed on a screen 2.8 m away.
If the linear distance between the first and second dark fringes of the pattern is 12 cm, what is the width of the slit?

Homework Equations


y=Ltan[tex]\theta[/tex]
W = [tex]\lambda[/tex] m / sin [tex]\theta[/tex]

The Attempt at a Solution


theta = arctan (.12/2.8) = 2.45
W = 630 x 10^-9 x 2 / sin 2.45
 
Physics news on Phys.org
Looks good.
 
I got 29.4 micrometers. It's wrong though.
 
muffintop said:
W = 630 x 10^-9 x 2 / sin 2.45
Oops... I missed that factor of 2 before. Why is it there?

Hint: Write expressions for the first dark fringe position and the second dark fringe position, then subtract.
 
I got it thank you so much!
I always get my "m" values wrong.
 

Similar threads

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