Solving Dark Fringes Problem with 651nm Light & 5.47x10^-6m Slit

  • Thread starter NotaPhysicsMan
  • Start date
  • Tags
    Fringes
In summary, the question asked is about the number of dark fringes that will be produced on either side of the central maximum when light with a wavelength of 651nm is incident on a single slit that is 5.47 x 10^-6m wide. The formula sin@= m x lambda/width is suggested as a possible solution, with m being the order of the dark fringe. After attempting to solve for m=1 and getting an angle of 6.83 degrees, the conversation turns to discussing the range of values for sin@ and correcting the width to 5.47x10^-6m. It is then concluded that using the formula sin90=m x lambda/W and solving for m would result in
  • #1
NotaPhysicsMan
146
0
Hey, another question I'm working on.

How many dark fringes will be produced on either side of the central maximum if light, (651nm) is incident on a single slit that is 5.47 x 10^-m wide?

Ok, I figure I could use this forumula sin@= m x lambda/Width. Am I trying to locate the maximum m (the order of the dark fringe) that it'll go?

I tried solving for m=1 and got a angle of 6.83 degrees. Now where to?
Any tips/hints welcome.

Thanks.
 
Physics news on Phys.org
  • #2
What do you know about the range of values of sin@ ?
 
  • #3
I missed something, the W is suppose to be 5.47x10^-6 m. Well, I know it can't go past 90 degrees and the min is the central max, which is 0 degrees.
 
  • #4
So if I put sin90=m x lambda/W and solve for m, then I'll get 8.40? so 8 dark fringes?
 
  • #5
can someone verify please and thanks
 

FAQ: Solving Dark Fringes Problem with 651nm Light & 5.47x10^-6m Slit

How does 651nm light help solve the dark fringes problem?

The wavelength of 651nm light is within the visible spectrum, making it ideal for observing and analyzing the dark fringes phenomenon. It allows us to see the interference pattern clearly and accurately measure the distance between fringes.

Why is a 5.47x10^-6m slit used in this experiment?

The size of the slit determines the width of the fringes in the interference pattern. The smaller the slit, the closer the fringes will be and the more distinct the pattern will appear. A slit of 5.47x10^-6m is chosen to create a clear and easily measurable interference pattern.

What is the significance of solving the dark fringes problem?

The dark fringes problem is a fundamental concept in the study of optics and wave interference. By solving this problem, we gain a better understanding of the behavior of light and can apply this knowledge in various fields such as astronomy, microscopy, and telecommunications.

Are there any other factors that can affect the interference pattern besides wavelength and slit size?

Yes, the distance between the light source and the slit, as well as the distance between the slit and the screen where the pattern is observed, can also affect the interference pattern. These distances need to be carefully controlled in order to obtain accurate results.

Can this experiment be replicated with different wavelengths of light and slit sizes?

Yes, this experiment can be replicated with different wavelengths of light and slit sizes. However, the choice of wavelength and slit size may affect the clarity and visibility of the interference pattern, so it is important to carefully select these parameters based on the desired outcome of the experiment.

Similar threads

Back
Top