Distance between two bright fringes

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SUMMARY

The discussion focuses on calculating the distance between the second order bright fringe of light with a wavelength of 683 nm and the third order bright fringe of light with a wavelength of 400 nm, using a screen positioned 2.2 m from the slits with a spacing of 0.12 mm. The calculations involve the equation sin∅=mλ/d, leading to fringe distances of D_683=0.0249 m and D_400=0.0219 m. The participant's final answer of 0.003 m was rejected, indicating a potential misunderstanding of the problem's requirements regarding the smallest possible distance between fringes.

PREREQUISITES
  • Understanding of wave interference and diffraction patterns
  • Familiarity with the equation sin∅=mλ/d
  • Basic trigonometry, particularly tangent functions
  • Knowledge of fringe spacing in double-slit experiments
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  • Review the derivation and application of the equation sin∅=mλ/d in double-slit experiments
  • Learn about calculating fringe widths and their significance in interference patterns
  • Explore the concept of absolute values in the context of distance measurements
  • Investigate common mistakes in fringe distance calculations and how to avoid them
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Students studying optics, physics educators, and anyone involved in experimental physics or wave mechanics who seeks to understand fringe calculations in interference patterns.

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Homework Statement


A new experiment is created with the screen at a distance of 2.2 m from the slits (with spacing 0.12 mm). What is the distance between the second order bright fringe of light with l = 683 nm and the third order bright fringe of light with l = 400 nm? (Give the absolute value of the smallest possible distance between these two fringes: the distance between bright fringes on the same side of the central bright fringe.)



Homework Equations


sin∅=mλ/d


The Attempt at a Solution


For first wavelength,
∅=arcsin(2(683*10^-9)/1.2*10^-4)
∅=.65
Setting up triangle, I get D_683=2.2tan(.65)=.0249

For second wavelength,
∅=arcsin(3(400*10^-9)/1.2*10^-4)
∅=.57
Setting up triangle, I get D_400=2.2tan(.57)=.0219
I thought that D_683-D_400 should give me the distance between the two fringes but my answer of .003m keeps getting rejected. I have done this problem over and over and I cannot for the life of me figure out what I am doing wrong.
 
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The math looks pretty solid to me. Not quite sure why that answer would be rejected. Sorry if that's not helpful.
 
csinger1 said:

Homework Statement


A new experiment is created with the screen at a distance of 2.2 m from the slits (with spacing 0.12 mm). What is the distance between the second order bright fringe of light with l = 683 nm and the third order bright fringe of light with l = 400 nm? (Give the absolute value of the smallest possible distance between these two fringes: the distance between bright fringes on the same side of the central bright fringe.)



Homework Equations


sin∅=mλ/d


The Attempt at a Solution


For first wavelength,
∅=arcsin(2(683*10^-9)/1.2*10^-4)
∅=.65
Setting up triangle, I get D_683=2.2tan(.65)=.0249

For second wavelength,
∅=arcsin(3(400*10^-9)/1.2*10^-4)
∅=.57
Setting up triangle, I get D_400=2.2tan(.57)=.0219
I thought that D_683-D_400 should give me the distance between the two fringes but my answer of .003m keeps getting rejected. I have done this problem over and over and I cannot for the life of me figure out what I am doing wrong.

Might want to find the width of each fringe too, it does state smallest possible value, but with your calculation it is not smallest possible value.
 

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