Distance between two bright fringes

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The discussion revolves around calculating the distance between the second order bright fringe of light at 683 nm and the third order bright fringe at 400 nm, given a screen distance of 2.2 m and slit spacing of 0.12 mm. The calculations for the angles using the formula sin∅=mλ/d yield fringe distances of approximately 0.0249 m and 0.0219 m, respectively. The participant believes the difference of 0.003 m should represent the distance between the two fringes, but this answer is repeatedly rejected. Additional advice suggests considering the width of each fringe to determine the smallest possible value, indicating that the current calculations may not fully account for the problem's requirements. The participant is encouraged to re-evaluate their approach to ensure accuracy.
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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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