# Wavelengths: Length between 2nd-order fringes

## 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...

## Answers and Replies

kuruman
Science Advisor
Homework Helper
Gold Member
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:
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

kuruman
Science Advisor
Homework Helper
Gold Member
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.