Double Slit Help: 2nd Order Dark Fringe Location

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SUMMARY

The discussion focuses on calculating the location of the second order dark fringe in a double-slit experiment using light of wavelength 490nm and slit separation of 0.44m. The correct formula for finding the location is Xn = (n-1/2)Lλ/d, which yields a result of 0.23cm. The alternative method using ΔX = Lλ/d does not provide the correct answer when multiplied by 2, as it misinterprets the relationship between dark fringes and the central maximum. The confusion arises from the distinction between the two formulas and their respective applications in calculating fringe locations.

PREREQUISITES
  • Understanding of double-slit interference patterns
  • Familiarity with the formulas Xn = (n-1/2)Lλ/d and ΔX = Lλ/d
  • Knowledge of wavelength measurement in nanometers
  • Basic trigonometry for angular calculations
NEXT STEPS
  • Study the derivation of the dark fringe location formula Xn = (n-1/2)Lλ/d
  • Learn about the significance of the central maximum in interference patterns
  • Explore graphical representations of interference patterns to visualize fringe locations
  • Investigate the impact of varying slit separation on fringe spacing
USEFUL FOR

Students studying wave optics, physics educators explaining interference patterns, and anyone seeking to deepen their understanding of double-slit experiments.

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Homework Statement


A light source shines light of wavelength 490nm onto a pair of slits separated by 0.44m. Calculate the angular location and the location in cm of the second order dark fringes on a screen 1.4m from the slit

Homework Equations


Xn = (n-1/2)L lambda/d
delta x = L lambda/d

The Attempt at a Solution


So I understand how to find the angular location, but I don't understand why only one method for finding the location works. The method which works is using the formula containing Xn which gives the answer of 0.23cm. I don't understand why it is not possible given that this is double slit to use delta X = L lambda/d then multiply that answer by 2 to get your answer as it gives a different value.

X2=(2-1/2)(1.4)(4.9x10-7)/0.44x10-3 = 0.23cm (the answer in the back of the textbook)

Using ΔX = Lλ/d
ΔX = (1.4)(490x10-9)/0.44x10-3
= 0.0015
as it is to the second dark fringe, shouldn't 0.0015x2 get me the same answer as my previous Xn calculation?
 
Last edited:
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Please show your work step by step.
 
chrisahn97 said:

Homework Statement


A light source shines light of wavelength 490nm onto a pair of slits separated by 0.44m. Calculate the angular location and the location in cm of the second order dark fringes on a screen 1.4m from the slit
Is the sit separation supposed to be 0.44 mm?

Homework Equations


Xn = (n-1/2)lambda/d
delta x = L lambda/d

Should the first equation have a factor of L?

The Attempt at a Solution


So I understand how to find the angular location, but I don't understand why only one method for finding the location works. The method which works is using the formula containing Xn which gives the answer of 0.23cm. I don't understand why it is not possible given that this is double slit to use delta X = L lambda/d then multiply that answer by 2 to get your answer as it gives a different value.

Think about the meaning of "delta x" and think about the meaning of Xn. In particular, "delta x" represents the distance between what two points? Xn represents the distance between what two points?
 
I thought it wasn't necessary as it was more an issue with the concept as I was able to reach the correct answer on my own using the Xn formula, so sorry!
Updated
 
Hi chrisahn97. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

Sketch the expected pattern of fringes, and mark in those 2 distances precisely using dimensioning lines such as:

|<--------------- Xn --------------->|
 
Last edited by a moderator:
Is the first dark fringe a distance Δx away from the center? If not, should the second dark fringe be a distance 2Δx away?
 
Orodruin said:
Is the first dark fringe a distance Δx away from the center? If not, should the second dark fringe be a distance 2Δx away?
oh I see now thank you!
I think my method would have only worked if the question was asking for the second bright fringe, as the distance from the central maximum to the first dark band would not be delta X but would be 1/2ΔX
 
Yes. In fact, this is how you would derive your other formula, which gives the location of an arbitrary dark fringe.
 
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Careless error on my part thanks for the help. This question can be marked as solved if that's a thing on these forums
 

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