A) Solving Interference Minima: Find Wire Diameter

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SUMMARY

The discussion focuses on solving a physics problem involving interference minima caused by a thin wire between two glass slides. The light wavelength is specified as 418 nm, and the problem presents two parts: calculating the wire diameter based on 40 observed minima and determining the index of refraction of a liquid that increases the minima count to 49. The key equation used is d*sin(theta) = n*lambda, which relates the spacing of the minima to the wavelength of light.

PREREQUISITES
  • Understanding of wave interference principles
  • Familiarity with the concept of minima in optics
  • Knowledge of the relationship between wavelength, index of refraction, and phase change
  • Ability to apply trigonometric functions in physical contexts
NEXT STEPS
  • Study the derivation of the interference minima equation d*sin(theta) = n*lambda
  • Explore the impact of varying the index of refraction on interference patterns
  • Learn about phase changes upon reflection in optical systems
  • Investigate practical applications of interference in thin films and coatings
USEFUL FOR

Students and educators in physics, particularly those focusing on optics and wave phenomena, as well as anyone involved in experimental physics or optical engineering.

Vashti
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Homework Statement



A)Two flat slides of glass are separated at one edge by a thin wire. The top surface of the upper slide and the bottom surface of the lower slide have special coatings on them so that they reflect no light. The system is illuminated with light of wavelength 418 nm. Looking down from above you see 40 interference minima. What is the diameter of the wire? Assume that the last minima occurs at the right edge where the wire is placed.

B) Now by submerging the system in a certain liquid we find that the number of interference minima has increased to 49. What is the index of refraction of the liquid?

Homework Equations



For minima: d*sin(theta) = (n)*(lambda)

The Attempt at a Solution



I don't even know where to begin with this one...
I would guess that you find the phase change and then can calculate the space between minima and add them to find the diameter for part A. The only issue is there doesn't seem to be enough information to do this! Help please!
 
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Think about the spacing of the slides at the opposite edge... what is the form of the spacing across the slide? So what is the condition for each minima? (hint: your equation is incomplete). Once you understand that it should be fairly straightforward.
 

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