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A laser light with wavelength of, 630nm, allowed to pass through double slit interference wilh a separalion distance 81, a U.0Amm. The distance between the slits and the board is x-1.5 meter. What is the distance in cm for the fourth bright maxima, y4 from center of the screen?

A laser light with wavelength of, 630nm, allowed to pass through double slit interference wilh a separalion distance 81, a U.0Amm. The distance between the slits and the board is x-1.5 meter. What is the distance in cm for the fourth bright maxima, y4 from center of the screen?

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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A laser light with wavelength of, 2-630nm, allowed to pass through double slit interference with a separation distance of, d= 0.03mm. The distance between the slits and the board is x-1.5 meter. What is the distance in cm for the fourth bright maxima, y4 from center of the screen?
Transcribed Image Text:A laser light with wavelength of, 2-630nm, allowed to pass through double slit interference with a separation distance of, d= 0.03mm. The distance between the slits and the board is x-1.5 meter. What is the distance in cm for the fourth bright maxima, y4 from center of the screen?
Expert Solution
Step 1

Given

  • The wavelength of laser light is λ=630 nm=630×10-9 m.
  • The seperation distance of interference is d = 0.03 mm = 0.03 x 10-3 m.
  • The distance between slit and board is x =1.5 m.

For the bright fringe, the sin angle of light for fourth interference m=4 is calculated as,

sinθ4=mλdsinθ4=4×630×10-9 m0.03×10-3 msinθ4=0.084θ4=sin-10.084=4.8185°

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