Wavelengths: Length between 2nd-order fringes

AI Thread Summary
The discussion revolves around calculating the distance between second-order fringes for two wavelengths of light passing through slits. The formulas for maxima and minima are provided, but there is confusion regarding which to use without explicit mention of bright or dark fringes. Attempts to solve the problem yield slightly different results depending on the chosen formula, leading to uncertainty about the correctness of the textbook solutions. A suggestion is made to derive a general algebraic formula for the distance between fringes before substituting numerical values. The importance of distinguishing between bright and dark fringes when using different wavelengths is emphasized.
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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