Young Double Slit Experiment (Determine wavelength of light source)

[HarleyM]

Homework Statement

Upon using Thomas Young's double slit experiment to obtain measurements, the following data was obtained. Use this data to determine the wavelength of light being used to create the interference pattern. DO this in 3 different ways!

• The Angle to the Eighth maximum is 1.12 deg.
• The distance from the slits to the screen is 3.02 m
• The distance from the first minimum to the fifth minimum is 2.95 cm
• The distance between the slits is 0.00025 m

Homework Equations

sin∅_n=(n-1/2)λ/d
sin∅_m=mλ/d
Δx=Lλ/d
X_n/L=(n-1/2)λ/d
X_m/L=mλ/d

The Attempt at a Solution

sin∅_n=(n-1/2)λ/d
λ=(sin1.12)(0.00025)/(7.5)
λ=651 nm


The distance from the first minimum to the fifth minimum is 2.95 cm .. therefore
4Δx=2.95 cm
Δx=0.0074 m ( i converted it)

Im unsure about 4 x if anyone can shed any light on that that would be cool


Δx=Lλ/d
λ=Δxd/L
λ=(0.0074) (0.00025)/ (3.02)
λ= 612 nm

as for this one I don't know if I should use the X_m or X_n equation..I used

X_m/L=mλ/d

(0.0295)/3.02=4λ/(0.00025)
λ= 610 nm

Number seems plausible but it also does when I use the X_n equation

thanks for any help in advance

 

[technician]

Using Sinθ = 8λ/d for the 8th max I got λ = 6.1 x 10^-7m
Using 4x for the separation of 5 minima then using x = λL/d I got λ = 6.1 x 10^-7m

 

[HarleyM]

Using Sinθ = 8λ/d for the 8th max I got λ = 6.1 x 10^-7m
Using 4x for the separation of 5 minima then using x = λL/d I got λ = 6.1 x 10^-7m

since its a max shouldn't you use sin∅=(n-1/2)λ/d?

 

[technician]

For a max the path difference from the slits must be a whole number of wavelengths.
For the 8th max the path diff = 8 wavelengths.
For minima the path diff must be an odd number of half wavelengths (n+1/2) but the SEPARATION of max and min is given by increases in path diff of whole numbers of wavelengths.
Hope that sounds OK

 

[asteriatic]

I would first commend the student for attempting to use multiple methods to determine the wavelength of the light source. This shows a thorough understanding of the experimental setup and the relevant equations. However, there are a few points that can be clarified and improved upon.

First, in the first attempt, the student has correctly used the equation sin∅n=(n-1/2)λ/d to determine the wavelength. However, the value of n used should be 8 instead of 7.5, as the question states that the angle is to the eighth maximum. This will give a slightly different value for the wavelength.

In the second attempt, the student has correctly used the equation Δx=Lλ/d, but the value of Δx should be divided by 4 instead of using the number 4 in the equation. This is because the distance between the first and fifth minimum represents 4 times the value of Δx. Using the value of Δx=0.0074 m, the calculated wavelength is 612 nm, which is close to the result obtained using the first method.

In the third attempt, the student has used the equation Xm/L=mλ/d, which is also correct. However, the value of Xm used should be 4 times the value of Δx, as mentioned above. Therefore, the equation should be Xm/L=4mλ/d. Using this equation and the value of Δx=0.0074 m, the calculated wavelength is again 612 nm.

Overall, the student has done a good job in using the relevant equations and attempting different methods to determine the wavelength of the light source. The slight discrepancies in the results obtained can be attributed to rounding errors and small differences in the measurements. I would suggest that the student repeats the experiment multiple times and takes more precise measurements to obtain a more accurate value for the wavelength. Additionally, it would be beneficial to compare the results obtained using different methods with the accepted value for the wavelength of the light source, to assess the accuracy of the measurements and calculations.