Double Slits, Calculating Wavelength

  • Thread starter Thread starter STEMucator
  • Start date Start date
  • Tags Tags
    Wavelength
AI Thread Summary
The discussion focuses on determining the wavelength of light from a double-slit experiment using provided data. The user has calculated values using different methods but is uncertain about one approach due to a perceived lack of information. They suggest that knowing the distance from a nodal line to the right bisector would help solve the problem, but they are struggling to derive that variable. The user proposes using half the distance between nodal points to estimate the distance from the central maximum to the nearest node. Overall, they seek clarification on their calculations and the missing variable needed for a complete solution.
STEMucator
Homework Helper
Messages
2,076
Reaction score
140

Homework Statement



The following data was obtained upon using a double-slit experiment. Use this data to determine the wavelength of light being used to create the interference pattern. Do this in three different ways.

- The angle to the eight maximum is 1.12°
- The distance from the slits to the screen is 302 cm
- The distance from the first minimum to the fifth minimum is 2.95cm
- The distance between the slits is 0.025cm

Homework Equations


The Attempt at a Solution



My work is shown in this image: http://gyazo.com/b9625064b97100febb726a6f6a12236a

I believe I have interpreted the information correctly. I also believe methods 1 and 3 are correct as they were the most obvious. I have placed a series of question marks on method 2, which I am having trouble understanding. I don't believe they have given me sufficient information to solve this problem as is.

After I have 3 values for wavelength, I plan to average them out to get a final answer.

Does anyone have any insight to this? I seem to be missing a very important variable and I'm not sure how to go about it.
 
Physics news on Phys.org
Friendly bump, Is it safe to say this problem is not solvable?

If I knew the distance from a nodal line to the right bisector I could solve this, but I can't see how to get it for some reason.

All I seem to know is the distance between two consecutive nodal points, namely ##\Delta x = 0.7375cm##.

Would that mean half of that distance, ##\frac{\Delta x}{2}##, is the distance between a node and an antinode?

If so, could I not consider the distance between the central maximum (which is exactly where the right bisector would happen to be) and the closest node to it? That would give me a workable value of ##x_1 = \frac{\Delta x}{2}## and ##n = 1##.

This doesn't feel right to me though, does anyone have any thoughts?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top