How Far is the Third Interference Minimum in a Double Slit Experiment?

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The discussion revolves around calculating the position of the third interference minimum in a double slit experiment with light of wavelength 655 nm and slits 245 µm apart. The initial calculations mistakenly identified the third maximum instead of the minimum. It was clarified that for the third maximum (m=3), the corresponding minimum is at m=2.5. The correct distance from the center to the third minimum was ultimately determined to be 72.2 mm after adjusting the m value. This highlights the importance of accurately identifying maxima and minima in interference patterns.
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Homework Statement



Light of wavelength = 655 nm passes through a pair of slits that are 245 µm apart and travels to a screen 9.0 m away. How far from the center of the pattern is the third interference minimum? (in mm)

d=245e-6
λ=655e-9
m=3
D=9
y=distance from center of 3rd interference minimum

Homework Equations



sinθ=λm/d
tanθ=y/D

The Attempt at a Solution



sinθ=λm/d=(655e-9)(3)/(245e-6)
θ=0.4595 degrees

tanθ=y/D
y=Dtanθ=9tan(0.4595)
y=0.722m=72.2mm (incorrect)
 
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Hi Wahsup, welcome to PF!
It seems to me you have found the 3rd maximum, but the question asks for the MINIMUM. You'll just have to adjust your m value.
 
Delphi51 said:
Hi Wahsup, welcome to PF!
It seems to me you have found the 3rd maximum, but the question asks for the MINIMUM. You'll just have to adjust your m value.

thanks!

if m=3 is the max, would that mean m=2 is the min?
 
nevermind. i got it.

Since m=3 is the max, the min is m=2.5

thank you for the help
 
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