How Does Path Difference Affect Interference Patterns in Light Waves?

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SUMMARY

The discussion focuses on calculating the path difference (∆s) for light waves creating interference patterns through two narrow slits. The wavelength of the light is λ = 699 nm, the slit separation is d = 6 μm, and the distance to the screen is L = 80 cm. The calculated path differences at Points P, Q, and R are 0 nm, 699 nm, and 1048.85 nm, respectively. The relationship between path difference and interference maxima and minima is emphasized, confirming that maxima occur at specific path differences.

PREREQUISITES
  • Understanding of wave interference principles
  • Familiarity with the Pythagorean Theorem
  • Knowledge of coherent light sources
  • Basic unit conversions (e.g., nm to μm)
NEXT STEPS
  • Study the derivation of the interference pattern equations for double-slit experiments
  • Learn about the conditions for constructive and destructive interference
  • Explore the effects of varying slit separation on interference patterns
  • Investigate the impact of wavelength changes on path differences in interference
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding wave interference phenomena in light waves.

luke949
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Homework Statement


http://img5.imageshack.us/img5/4605/lchuynh177setphys52awee.png

The figure shows the interference pattern that appears on a distant screen when coherent light is incident on a mask with two identical, very narrow slits. Points P and Q are maxima; Point R is a minimum. The wavelength of the light that created the interference pattern is λ = 699 nm, the two slites are separated by rm d = 6 μm, and the distance from the slits to the center of the screen is L = 80 cm . The difference in path length at a point on the screen is ∆s = |s1 − s2|, where s1 and s2 are the distances from each slit to the point.
1. What is ∆s (in nm) at Point P?
2. What is ∆s (in nm) at Point Q?
3. What is ∆s (in nm) at Point R?




Homework Equations


da=sqr(L^2+(D/2+d/2)^2)
db=sqr(L^2+(D/2-d/2)^2)

The Attempt at a Solution


I know the answer to the first problem is 0.
But i tried plugging in da=sqr(80^2+(699-3)^2) and I am getting it wrong. I tried converting all the units to nm, but then the equation just turns into sqr(80nm^2) because the second term turns into a number close to zero.
 
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luke949 said:

Homework Statement


http://img5.imageshack.us/img5/4605/lchuynh177setphys52awee.png

The figure shows the interference pattern that appears on a distant screen when coherent light is incident on a mask with two identical, very narrow slits. Points P and Q are maxima; Point R is a minimum. The wavelength of the light that created the interference pattern is λ = 699 nm, the two slites are separated by rm d = 6 μm, and the distance from the slits to the center of the screen is L = 80 cm . The difference in path length at a point on the screen is ∆s = |s1 − s2|, where s1 and s2 are the distances from each slit to the point.
1. What is ∆s (in nm) at Point P?
2. What is ∆s (in nm) at Point Q?
3. What is ∆s (in nm) at Point R?




Homework Equations


da=sqr(L^2+(D/2+d/2)^2)
db=sqr(L^2+(D/2-d/2)^2)

The Attempt at a Solution


I know the answer to the first problem is 0.
But i tried plugging in da=sqr(80^2+(699-3)^2) and I am getting it wrong. I tried converting all the units to nm, but then the equation just turns into sqr(80nm^2) because the second term turns into a number close to zero.


No geometry/trigonometry is necessary.

Think about the requirements for those particular maximum and minimum lines!
 
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Hmmmm I am almost positive that the pythagorean Theorem must be used. I can't really see it any way. what do you mean by requirements?
 
Is the answers 1) 0, 2) 699nm, 3) 1048.85? I believe that the the center is a maxima and Q is the next maxima so that is one full wavelength. R is a minima right after the first maxima so it is 699+349.5 = 1048.85nm. Please get back to me thank you.
 
luke949 said:
is the answers 1) 0, 2) 699nm, 3) 1048.85? I believe that the the center is a maxima and q is the next maxima so that is one full wavelength. R is a minima right after the first maxima so it is 699+349.5 = 1048.85nm. Please get back to me thank you.

exactly! They ask for path difference, and the maxima and minima occur for specific path differences

EDIT: I am in East Coast Australia, so time differences will explain any delay in this response.
 

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