A 3rd method of finding wavelength in a double slit

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Homework Help Overview

The discussion revolves around determining the wavelength of light used in a double slit experiment, utilizing given measurements such as angles, distances, and slit separation. The original poster attempts to find the wavelength using three different methods based on the provided data.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster describes two methods for calculating the wavelength and proposes a third method, questioning the validity of their friend's suggestion regarding the calculation of a distance related to antinodal lines.
  • Some participants seek clarification on variable definitions and express concerns about the independence of the proposed methods, suggesting that the third method may not be sufficiently distinct from the others.
  • There is discussion about the similarity between the second and third methods, with some participants noting that they differ only by a factor of 8.

Discussion Status

The conversation is ongoing, with participants providing feedback on the original poster's methods and questioning the assumptions made. Some guidance has been offered regarding the similarity of methods, and there is acknowledgment of the challenges in finding three distinct approaches given the available data.

Contextual Notes

Participants note the lack of certain values needed for independent methods and the implications of the provided measurements on the calculations. There is an emphasis on the need for clarity in variable definitions and the constraints of the problem setup.

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Homework Statement



Upon using Thomas young’s double slit experiment to obtain measurements, the following data were obtain. Use these data to determine the wavelength of light being used to create the interference pattern. Do this in three different ways (6)
• the angle to the eighth maximum is 1.12◦
• the distance from the slit to the screen is 302cm
• the distance from the first minimum to the fifth minimum is 2.95cm
• the distance between the slit is 0.00025cm

Homework Equations

The Attempt at a Solution


I found 2 ways of finding the wavelength using the formulas

Change in x = L (Lambda) / d

and

m(Lambda)= (d)(sin Theta subscript m)

For my third method i thought about using the formula

m(Lambda)= (d)xm/ L

the problem is I was not given xm sincexm is the distance from a point on a antinodal line to the right bisector.

However, I was told by my friend that since the 8th antinodal is provided and that "change in x" is easily obtainable, xmcan be found by just multiplying the 2 together.

which would look like

change in x = 0.007375 (Converted to Meters from cm)
8th antinodal = 8

0.007375(8) = 0.059

therefore
xm= 0.059 meters

Is my friend right?
 

Attachments

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Please define the variables you use. L, ##\lambda## and d are clear, what is everything else?

What is 0.059 meters? Certainly not the wavelength.

The third method has to be very similar to one of the other two methods, you don't have enough given values for three completely independent methods. Distance between slits and distance to the screen alone doesn't tell you anything about the wavelength, the given angle leads to one method, the given separation between minima leads to the second method.
 
mfb said:
Please define the variables you use. L, ##\lambda## and d are clear, what is everything else?

What is 0.059 meters? Certainly not the wavelength.

The third method has to be very similar to one of the other two methods, you don't have enough given values for three completely independent methods. Distance between slits and distance to the screen alone doesn't tell you anything about the wavelength, the given angle leads to one method, the given separation between minima leads to the second method.
Sorry, I was in a rush and I thought I attached my work when I posted. My work should be in the attachments now.
 
Method 2 and 3 are nearly identical, they just differ by a factor 8 in both numerator and denominator.
 
mfb said:
Method 2 and 3 are nearly identical, they just differ by a factor 8 in both numerator and denominator.

should I use a different one ?
 
See post #2: No matter what you do, two methods will look very similar. I guess the shown "third method" is as good as it gets.
 
mfb said:
See post #2: No matter what you do, two methods will look very similar. I guess the shown "third method" is as good as it gets.

Ahh ok.

Thank you so much for reviewing my work! I really appreciate it.
 

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