Optics (double slit experiment)

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SUMMARY

The discussion focuses on the calculation of the wavelength of light in Young's double slit experiment, specifically with slits 0.20 mm apart and a screen distance of 1.5 m. The distance between the 5th minima on either side of the zeroth-order maximum is measured at 34.73 mm. The wavelength is calculated using the formula λ = (y * d) / (L * (m + 1/2)), resulting in a wavelength of approximately 420.96 nm, which falls within the blue/violet spectrum. Participants clarified the interpretation of the minima and the correct value of m to use in the calculations.

PREREQUISITES
  • Understanding of Young's double slit experiment
  • Familiarity with the concepts of minima and maxima in wave interference
  • Knowledge of trigonometric approximations (sinθ ≈ tanθ)
  • Ability to manipulate equations involving wavelengths and fringe positions
NEXT STEPS
  • Study the derivation of the double slit interference formula
  • Learn about the significance of the zeroth-order maximum in interference patterns
  • Explore the effects of slit width and distance on interference patterns
  • Investigate the application of the double slit experiment in modern physics
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding wave interference phenomena in light. This discussion is particularly beneficial for those tackling problems related to the double slit experiment and wavelength calculations.

roam
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Homework Statement



I have some difficulty understanding a part of the following problem:

In Young’s experiment, narrow double slits 0.20 mm apart diffract monochromatic light onto a screen 1.5 m away. The distance between the 5th minima on either side of the zeroth-order maximum is measured to be 34.73 mm. Determine the wavelength of the light.

Homework Equations



Here is a diagram of the double slit experiment:

http://imageshack.us/scaled/landing/850/doubleslit.jpg

Condition for minima: ##d \ \sin \theta = (m+\frac{1}{2}) \lambda##

Linear positions measured along the screen: ##\tan \theta = \frac{y}{L}##

##\therefore \ y_{dark} = \frac{L (m+\frac{1}{2})\lambda}{d}##

The Attempt at a Solution



So, in the question what is meant by "distance between the 5th minima on either side of the zeroth-order maximum"? I'm not sure if I understand this. :confused:

Does this mean the distance from the 5th minima on one side of the maxima to the 5th minima on the other side like this:

http://imageshack.us/scaled/landing/706/52496318.jpg

Did I understand the question correctly? If this is correct, then

##34.73=2y \implies y=17.37 \ mm##

And I can find the λ by rearranging the above equation:

##\lambda=\frac{y \ d}{L(m+\frac{1}{2})} = \frac{(17.365\times 10^{-3})\times (0.2 \times 10^{-3})}{1.5(5+\frac{1}{2})} = 4.2096 \times 10^{-7}##

Is this correct? The answer looks reasonable (about 420.96 nm in the blue/violet region of spectrum), but I doubt it is correct. Any help would be greatly appreciated.
 
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roam said:
So, in the question what is meant by "distance between the 5th minima on either side of the zeroth-order maximum"? I'm not sure if I understand this. :confused:

Does this mean the distance from the 5th minima on one side of the maxima to the 5th minima on the other side like this:

Yes, it is meant the distance between the fifth minimum at one side of the central maximum (at O) to the fifth minimum at the other side.

You approximated sinθ by tanθ, it is all right in this case.

But reconsider what number you should use for m. The first minimum is when the dsinθ=λ/2. m starts from zero.

ehild
 
ehild said:
Yes, it is meant the distance between the fifth minimum at one side of the central maximum (at O) to the fifth minimum at the other side.

You approximated sinθ by tanθ, it is all right in this case.

But reconsider what number you should use for m. The first minimum is when the dsinθ=λ/2. m starts from zero.

ehild

Thank you for the reply. I used m=5 because we want the fifth-order dark fringe. Personally I think that in a double-slit experiment there should be no zeroth-order minimum (i.e., m=0), but my book also says ##m=0, \pm1, \pm2, ...##. Does this mean I have to use m=4 instead (pretending m=0 is m=1)? :confused:

Also I am wondering when do we use the negative values of m?
 
You get destructive interference, that is a dark fringe when the path-length difference δ=r2-r1 is odd multiple of the half wavelength: dsinθ=(2m+1)λ/2. If m is 0 or positive integer you get the dark fringes above O, with δ= λ/2, 3λ/2, 5λ/2,7λ/2, 9λ/2. For the dark fringes on the other side, m is negative (-1,-2,..) and δ= -λ/2, -3λ/2, -5λ/2, -7λ/2, -9λ/2. The fifth dark fringe corresponds to δ=9 λ/2.


ehild
 
Thank you very much, that makes perfect sense now.
 
Last edited:

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