- #1

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...