Single Slit Question: Linear Distance & Wavelength Calculation

  • Thread starter Thread starter MetsFan09
  • Start date Start date
  • Tags Tags
    Single slit Slit
Click For Summary

Homework Help Overview

The problem involves a single slit diffraction scenario where a slit is illuminated with light of wavelength 660 nm, and the resulting diffraction pattern is observed on a screen 2.3 m away. The task is to determine the width of the slit based on the linear distance between the first and second dark fringes, which is given as 12 cm.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the diffraction formula to calculate the slit width but questions the validity of their approach after receiving feedback suggesting a need to halve the linear distance. Other participants inquire about the meaning of variables in the equation and explore the relationship between slit width and fringe spacing.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been provided regarding the application of the diffraction equation, and questions about the implications of changing slit width on the diffraction pattern are being raised.

Contextual Notes

Participants are discussing the implications of assumptions regarding the slit width and the behavior of the diffraction pattern as the slit width changes. There is a lack of consensus on the correct application of the formula and the interpretation of the linear distance between dark fringes.

MetsFan09
Messages
5
Reaction score
0

Homework Statement


A single slit is illuminated with 660 nm and the resulting diffraction pattern is viewed on a screen 2.3 m away.
a. If the linear distance between the 1st and 2nd dark fringes of the pattern is 12 cm. What is the width of the slit?


Homework Equations


W(X/L) = m(lambda)


The Attempt at a Solution


W(.012m/.0023m = 1(660x 10^-9)
W = 1.27 x 10^-7

Someone else tells me that's wrong and that I have to divide the linear distance by two to get 6 cm instead of 12. Is this true? And if so why? I really don't understand why we would have to do that.
 
Physics news on Phys.org
Explain me the meaning of the letters in the equation you used, please.

ehild
 
What happens to the distance between fringes as the width of the single slit become larger? Does the distance between fringes increases? What happens to the centre bright fringe? Does it become larger with an increasing single slit width? My hunch is: Increasing the width increases the size of the bright central fringe, but applets on the net show otherwise. Can someone help?
 
It is a single slit, so a pair of rays which cancel each other come out from the same slit. For a dark fringe, each rays emerging from the slit has to get an other one to cancel with. So the angles at which a dark fringe occurs are those for which

W/2 sin(α)=(2m+1)λ /2 ---->W sin(α)=(2m+1)λ

If the screen is at distance L from the slit and the fringe is at distance X from the centre, tan(a)=X/L, but tan(α)=sin(α) for small angles, so

W *X/L=(2m+1)λ , (m=0,1,2...)

for the dark fringes.

Find X for the first and second dark fringes and see what happens if the width of the slit W increases.

ehild
 

Attachments

  • slit.JPG
    slit.JPG
    10.3 KB · Views: 419

Similar threads

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