How Do Different Wavelengths Overlap in Double-Slit Interference?

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SUMMARY

The discussion focuses on the double-slit interference experiment involving two light sources with wavelengths of 450 nm and 600 nm. The slit separation is 8.40 x 10-6 m, and the distance to the screen is 2.80 m. Participants analyze how to determine overlapping bright fringes by equating the interference equations for both wavelengths, specifically addressing the conditions under which fringes coincide. The key insight is that the first overlap occurs when the difference in fringe orders is one, and further exploration of higher orders is encouraged.

PREREQUISITES
  • Understanding of double-slit interference principles
  • Familiarity with the equations dsin(θ) = m and dsin(θ) = (m + 0.5)
  • Knowledge of wavelength measurement in nanometers
  • Ability to manipulate trigonometric functions and inequalities
NEXT STEPS
  • Explore the concept of fringe order in double-slit experiments
  • Learn how to derive conditions for overlapping fringes using different wavelengths
  • Investigate the implications of varying slit widths on interference patterns
  • Study the effects of changing slit separation on fringe visibility and overlap
USEFUL FOR

Students and educators in physics, particularly those studying wave optics and interference phenomena, as well as anyone interested in practical applications of light behavior in experimental setups.

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Homework Statement


In a double-slit interference experiment the slit separation is 8.40 x 10-6 m and the slits are 2.80 m from the screen. Each slit has a width of 1.20 x 10-6 m.

2 lights of wavelength 450nm and 600nm are shone on the same slits.

If both of the above light sources are shined on the slits at the same time, which order bright fringes from the two light sources would occur at the same locations on the screen? Include the orders of all possible fringes that would overlap.

Homework Equations


dsin
img32.gif
= m
img7.gif

dsin
img32.gif
= (m +0.5)
img7.gif

The Attempt at a Solution


Ok so this is part of a multi-part assignment. Using the equations above I've determined "how far (in meters) from the center of the interference pattern on the screen will the third order (m = 3)" be for the bright (450) and dark (600) fringes separately. The next part of the question asks about shining them at the same time and about bright fringe orders.

My guess was to set both equations equal to each other without m and having both dsin
img32.gif
cancel out leaving m
img7.gif
= (m + 0.5)
img7.gif
and calculating for the order, but I realize this will only give me potentially one answer so I feel like my methods are incorrect.

Could someone give me some guidance about how to approach this question?
 
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Why are you using the equation with ##m+0.5## in it? Does it even apply to this problem? Don't you also need to keep track of the wavelengths? You seem to be using ##\lambda## to represent both at the same time.
 
I agree with Vela, where does this 0.5 come from?
The question is: what is the value of n for which a fringe due to (λ1) will coincide with that a fringe of λ2). The first occurrence will occur when one of the wavelength has created an extra fringe compared to the other... (So a difference in fringe order of 1...). Rewrite your equation taking this into account, and find the value of n that will satisfy such condition.
"I realize this will only give me potentially one answer...". You can also explore the case where the difference of order is more than 1, and see if such solution is possible by checking that sin([PLAIN]http://theory.uwinnipeg.ca/physics/light/img32.gif) remains <1
 
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