Optics (double slit experiment)

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Homework Help Overview

The discussion revolves around a problem related to the double slit experiment in optics, specifically focusing on the calculation of the wavelength of light based on the positions of minima observed on a screen. The setup involves narrow double slits and monochromatic light, with specific measurements provided for the distance between minima.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to clarify the meaning of the "distance between the 5th minima on either side of the zeroth-order maximum" and questions their understanding of the setup. They explore the implications of using different values for the order of minima (m) in their calculations.

Discussion Status

Participants are actively engaging in clarifying the definitions and assumptions related to the minima in the double slit experiment. Some guidance has been offered regarding the correct interpretation of the order of minima and the use of m values in calculations. There is an ongoing exploration of how to apply these concepts correctly.

Contextual Notes

Participants are discussing the definitions of minima in the context of the double slit experiment and the implications of using different values for m, including the consideration of negative values for m in the context of dark fringes.

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.
 
Last edited by a moderator:
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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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