Wavelengths: Length between 2nd-order fringes

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SUMMARY

The discussion focuses on calculating the distance between second-order fringes produced by two wavelengths of light (480 nm and 632 nm) passing through slits 0.52 mm apart, with a screen distance of 1.6 m. The formulas used include x/L = nλ/d for maxima and x/L = (n - 1/2)λ/d for minima. The user attempted both approaches but expressed uncertainty regarding the correct application of the formulas. Ultimately, the consensus is that the distinction between bright and dark fringes is crucial when dealing with different wavelengths.

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  • Understanding of wave optics principles, specifically interference patterns.
  • Familiarity with the double-slit experiment setup.
  • Knowledge of the equations for maxima and minima in interference.
  • Ability to manipulate algebraic expressions involving wavelengths and distances.
NEXT STEPS
  • Study the derivation of the double-slit interference equations.
  • Learn about the significance of fringe order in interference patterns.
  • Explore the impact of wavelength differences on fringe spacing.
  • Investigate practical applications of interference in optical devices.
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Students in physics, educators teaching wave optics, and anyone interested in understanding interference patterns in light. This discussion is particularly beneficial for those tackling problems related to the double-slit experiment.

okandrea
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Homework Statement



Light of wavelenghs 4.80x10^2 nm and 632nm passes through two slits 0.52 mm apart. How far apart are the second-order fringes on a screen 1.6m away?

λ₁ = 4.80x10^2 nm = 4.80x10^-7m
λ₂ = 6.32x10^-7m
d = 0.52mm = 5.2x10^-4m
n = 2
L = 1.6

Homework Equations



(Maxima/Bright)
x/L = nλ/d
(Minima/Dark)
x/L = (n - 1/2)λ/d

*subscript of X would be n in both cases

△x = | x₁ - x₂ |

The Attempt at a Solution


I wasn't so sure which of the two formulas I would be using because there doesn't seem to be a clear indication as to whether or not it's bright/dark (this was what I mainly struggled with).

I tried using both but I don't understand if either of them are correct. I rearranged for x in both equations (moving the L variable to the right) and repeated it for each wavelength:

(A) Using dark:
x₁ = ((2 - 1/2)(4.80x10^-7)(1.6))/5.2x10^-4
x₁ = 2.2x10^-3m

x₂ = (2 - 1/2)(6.32x10^-7)(1.6))/5.2x10^-4
x₂ = 2.9x10^-3 m

△x = | 2.2x10^-3 - 2.9x10^-3 |
△x = 7.0x10^-4 m

(B) Using bright:
x₁ = ((2)(4.80x10^-7)(1.6))/5.2x10^-4
x₁ = 3.0x10^-3 m

x₂ = ((2)(1.6)(6.32x10^-7))/5.4x10^4
x₂ = 3.9x10^-3 m

△x = | 3.0x10^-3 - 3.9x10^-3 |
△x = 9.0x10^-4 m

They aren't too far off. I don't quite trust the textbook solutions since plenty of wavelength-related solutions were wrong. They did, however, use the formula for wavelengths with dark fringes...
 
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If neither "bright" nor "dark" is explicitly mentioned, I would assume "bright", but that's my personal interpretation. Also, for better accuracy, I would first find an algebraic formula for Δx using just symbols and then put in the numbers.
 
Last edited:
kuruman said:
I would first find an algebraic formula for Δx using just symbols and then put in the numbers.
Would that mean something like this?
Δx/L = λ/d
Δx = Lλ/d
 
okandrea said:
Would that mean something like this?
Δx/L = λ/d
Δx = Lλ/d
Not really, what happened to the ##n## in the expression? You need two expressions, one for each wavelength. It helps being organized.
1. For wavelength 1 you have ##x_1=nL\lambda_1/d##.
2. Write a similar expression for ##x_2##.
3. Find an algebraic expression for the difference ##x_2-x_1## for the second order fringes.
4. Put in the numbers.

On edit: My earlier statement that it doesn't matter if you use dark or bright fringes is incorrect. It does make a difference if the wavelengths are different. I edited that statement.
 

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