Finding the theoretical value of the wavelength for a double slit experiment?

In summary, the conversation is about finding the experimental and theoretical wavelengths for a double slit interference pattern. The given variables are m (maximum), h (distance from center to mth maximum), D (distance from central bright peak to slits), d (distance between two slits), and wavelength. The equation used to find the experimental wavelength is wavelength = dh/mD. The speaker is unsure of how to find the theoretical wavelength without an angle measurement. They are also confused about the number of light sources used and the definition of experimental and theoretical wavelengths. A clearer understanding of the setup or a drawing is needed for further assistance.
  • #1
ronsa
1
0

Homework Statement


m=maximum h=distance on the screen from the center of the pattern to the mth maximum D=distance from the central bright peak to the slits d=distance between two slits
I have m=1 d(mm)=0.25mm D=2000mm h=0.5mm wavelength= 6.25E-5mm

Homework Equations


So I used this equation to find the experimental wavelength for the first 5 wavelengths
Wavelength= dh/mD for experimental wavelength

The Attempt at a Solution


To find the experimental wavelength I took the average of the five wavelengths (other four wavelengths are not shown), but now I am lost in trying to find the theoretical value for the wavelengths. Since we don't have an angle because we traced the interference patterns for a double slit, I couldn't use this equation (wavelength=dsin(theta)/m) for the theoretical wavelength. I am not sure which equation to use to find the theoretical wavelength. Please help I am so confused...
 
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  • #2
Hello Ronsa, and welcome to PF :)

Glad to help if I can, but you have to help me in getting a clear picture what this is about. The problems statement is sort of mssing: you start with a list of variables.

What have you measured ? h = 0.5 mm from the center of the pattern to the first maximum ? That's quite small, but who knows.

How do you calculate the wavelength ? Visible red is more like ten times longer wavelengths...

Then: "To find the experimental wavelength I took the average of the five wavelengths" Do you have five light sources ?
What is your definition of experimental wavelength ? and of theoretical wavelength ?

Why do you think you don't have an angle ?
Can you post a drawing of the setup ? Or find a link that has a clear drawing ?
 

1. What is the equation for finding the theoretical value of the wavelength for a double slit experiment?

The equation for finding the theoretical value of the wavelength for a double slit experiment is: λ = d sin(θ), where λ represents the wavelength, d is the distance between the two slits, and θ is the angle of the interference pattern.

2. How do I determine the distance between the two slits in a double slit experiment?

The distance between the two slits can be measured using a ruler or a caliper. It is important to ensure that the measurement is accurate, as it directly affects the calculated wavelength.

3. Can the theoretical value of the wavelength be different from the actual value observed in a double slit experiment?

Yes, the theoretical value of the wavelength is an ideal value, and it may differ from the actual value observed in an experiment due to experimental error, limitations of equipment, and other factors.

4. How does changing the angle of the interference pattern affect the theoretical value of the wavelength?

Changing the angle of the interference pattern, also known as the angle of diffraction, affects the value of the wavelength in the equation λ = d sin(θ). As the angle increases, the wavelength decreases, and vice versa.

5. Can the theoretical value of the wavelength be calculated for any type of interference pattern in a double slit experiment?

Yes, the equation λ = d sin(θ) can be used to calculate the theoretical value of the wavelength for any type of interference pattern, as long as the distance between the slits and the angle of diffraction can be measured accurately.

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