Calculating Micrometer Reading Change for 50 Fringes at λ = 550 nm

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SUMMARY

The discussion focuses on calculating the change in micrometer reading when 50 fringes pass, given a spectrometer calibration constant of 0.5 and a wavelength of λ = 550 nm. The formula used is delta(l) = delta(n) * λ / 2, leading to delta(m) = delta(l) / K. The final calculation results in a micrometer reading change of 27,500 nm. This problem is referenced from "Modern Physics for Engineers," indicating its relevance in engineering physics contexts.

PREREQUISITES
  • Understanding of fringe patterns in optics
  • Familiarity with micrometer calibration techniques
  • Knowledge of wavelength calculations in nanometers
  • Basic principles of interference in light waves
NEXT STEPS
  • Study the principles of optical interference and fringe formation
  • Learn about spectrometer calibration methods
  • Explore advanced calculations involving wavelength and fringe shifts
  • Review applications of micrometer readings in experimental physics
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Students and professionals in physics and engineering, particularly those involved in optical experiments and spectrometry, will benefit from this discussion.

Benzoate
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Homework Statement




Suppose the calibration constant of your spectrometer is 0.5, and suppose the wavelength of the light is λ = 550 nm. How much does the micrometer reading change during the movement of M2 that generates the passage of 50 fringes?

Homework Equations



possibly delta(l)= delta(n)*lambda /2
delta(m)= delta(l)/K , K being some constant to find the distance

The Attempt at a Solution



possibly delta(l)=50* 550 nm/ 2 delta(m)= delta(l)/K = 13750 nm/(.5) =27500 nm
 
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Benzoate said:

Homework Statement




Suppose the calibration constant of your spectrometer is 0.5, and suppose the wavelength of the light is λ = 550 nm. How much does the micrometer reading change during the movement of M2 that generates the passage of 50 fringes?

Homework Equations



possibly delta(l)= delta(n)*lambda /2
delta(m)= delta(l)/K , K being some constant to find the distance

The Attempt at a Solution



possibly delta(l)=50* 550 nm/ 2 delta(m)= delta(l)/K = 13750 nm/(.5) =27500 nm

Is this from Modern Physics for Engineers? I have seen this problem before. When I get home in about an hour or two I'll try to digg up the answer.
 

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