Calculating Average Distance Between Bright Fringes in a Double-Slit Experiment

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SUMMARY

The discussion focuses on calculating the average distance between bright fringes in a double-slit experiment using specific parameters. The first scenario involves a monochromatic light source with a wavelength of 5.00 × 10-7 m, slits spaced 0.10 mm apart, and a screen 150 cm away, resulting in bright fringes 0.30 cm apart. When the screen distance is increased to 3.0 m, the average distance between bright fringes changes proportionally. The second scenario examines the effect of changing the wavelength to 7.50 × 10-7 m, requiring the application of the formula nλ = d sin(θ) to determine the new fringe spacing.

PREREQUISITES
  • Understanding of double-slit experiment principles
  • Knowledge of wave interference and fringe patterns
  • Familiarity with the formula nλ = d sin(θ)
  • Ability to perform calculations involving wavelength and distance
NEXT STEPS
  • Research the derivation and application of the double-slit interference formula nλ = d sin(θ)
  • Learn how to calculate fringe spacing for varying wavelengths in double-slit experiments
  • Explore the effects of changing screen distance on interference patterns
  • Investigate destructive interference in wave phenomena and its mathematical representation
USEFUL FOR

Students and educators in physics, particularly those studying wave mechanics and interference patterns in optics, as well as researchers conducting experiments involving light and sound waves.

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1. A double-slit experiment is performed using a monochromatic light source, two slits spaced 0.10 mm apart, and a screen located 150 cm away. The bright fringes are located 0.30 cm apart. If the screen distance was changed to 3.0 m from the sources, what would the average distance between bright fringes become?

2. Again a double-slit experiment is performed using a monochromatic light source with a wavelength of 5.00 ´ 10–7 m. The pattern appears on a screen 150 cm away and the bright fringes are 0.40 cm apart. If the wavelength of the light used is changed to 7.50 ´ 10–7 m, what would the average distance between bright
 
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I don't think that i have the right formula to use for this question, but i know that for question 1
ë_1=5.00*10^-7m
L=1.5m
X= 0.004m
ë_2=7.5*10^-7m

Whats the formula? I think that if i have the formula for question 1 I can solve for 2.
 
Another question
Two speakers producing exactly the same frequency and always in phase are located on an east–west line. The sound is going north. The speakers are 50 cm apart and a microphone is located 80 cm due north of one speaker. Which of the following wavelengths would produce sound with the least intensity at the location of the microphone?

I drew a diagram and i understand how the speakers and mic are set up but i think that all i need is the formula for destructive interference. i have the formula sin(theta)=(n - 1/2)ë/d, but i don't think that its the right one b/c I don't have an angle of a nodal line #. What is the formula that I'm looking for?
 
Just so you know, there's an "Edit" button on your post. Instead of double posting and wasting space, simply press the edit button, and put your questions in your original post. Thanks.
 
For the first one you have to use an equation that looks like

n\lambda = dSin(\theta)

NOTE: those terms may not be correct, check your formula sheet

the n is the fringe number, lambda is wavelength, d is the distance between the lines in the diffraction grating (the inverse of the lines/m rating the grating has) and theta is the angle made between light going straight and going to one of the fringes.

To find theta, you have to use the inverse tangent of the distance between fringes divided by the distance between the diffraction grating and the screen.
 

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