Optics (double slit experiment)

In summary: So I should use m=2 for the fifth-order dark fringe, which gives##\lambda=\frac{y \ d}{L(m+\frac{1}{2})} = \frac{(17.365\times 10^{-3})\times (0.2 \times 10^{-3})}{1.5(2+\frac{1}{2})} = 6.629\times 10^{-7}##Which is 662.9 nm in the red region of spectrum. That is more reasonable. Thank you again for the help.In summary, the question was asking for the distance between the fifth dark fringe on one side of the central maximum to the fifth dark fringe on the other side. The
  • #1
roam
1,271
12

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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  • #2
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
 
  • #3
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?
 
  • #4
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
 
  • #5
Thank you very much, that makes perfect sense now.
 
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Related to Optics (double slit experiment)

1. What is the double slit experiment and why is it important in optics?

The double slit experiment is a classic experiment in optics that demonstrates the wave-like properties of light. It involves passing a beam of light through two parallel slits and observing the resulting interference pattern on a screen. This experiment is important because it provides evidence for the wave nature of light, which was a major discovery in the field of optics.

2. How does the double slit experiment work?

In the double slit experiment, a beam of light is shone through two parallel slits and the resulting interference pattern is observed on a screen placed behind the slits. This pattern is created by the constructive and destructive interference of the light waves passing through the two slits. The spacing between the slits and the distance to the screen are important factors that determine the pattern observed.

3. What is diffraction and how does it relate to the double slit experiment?

Diffraction is the bending of waves around obstacles or through narrow openings. In the double slit experiment, the light waves passing through the two slits diffract and interfere with each other, creating the characteristic interference pattern on the screen. Diffraction is an essential concept in understanding the behavior of light in the double slit experiment.

4. Can the double slit experiment be performed with other types of waves?

Yes, the double slit experiment can be performed with other types of waves such as sound waves and water waves. In fact, the experiment was originally performed with water waves by Thomas Young in 1801. This further supports the idea that light behaves like a wave, as it exhibits similar properties to other types of waves in this experiment.

5. What are some real-world applications of the double slit experiment?

The double slit experiment has many real-world applications, including the study of interference and diffraction in optics, the development of technologies such as holography and laser technology, and the testing of theories related to the wave-particle duality of light. It has also been used in fields such as acoustics, fluid dynamics, and quantum mechanics.

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