A double-slit experiment is performed with light from a Helium Neon Laser on two slits
80 micrometers apart. The first bright interference fringe left of the center is spaced 2m from the bright center.
How far away from the double slit is the screen? wavelength =632nm
What will the fringe spacing be on the same screen if the light is changed to a wavelength of 532nm? Show your work!
To determine the distance between the double slit and the screen, we can use the formula for fringe spacing in a double-slit experiment:
d*sinθ = mλ
where:
- d is the distance between the slits (80 micrometers)
- θ is the angle between the central bright fringe and the desired fringe
- m is the order of the fringe (1 in this case since we're looking at the first bright fringe)
- λ is the wavelength of light
First, let's find the angle θ for the given scenario.
Using the formula for calculating the fringe spacing, we have:
2m = (mλ) / sinθ
Since we're interested in the angle θ for the first bright fringe to the left of the center, we set m = 1:
2 = λ / sinθ
Rearranging the equation:
sinθ = λ / 2
Now, we can find the value of θ using the arcsin function:
θ = arcsin(λ / 2)
Given that the wavelength of the light is 632nm, we substitute this value into the equation to find θ:
θ = arcsin(632nm / 2) = arcsin(316nm) ≈ 18.13°
Next, we need to find the distance between the double slit and the screen.
Using trigonometry, we can determine the distance x between the double slit and the screen:
x = d*tanθ
Substituting the values we have:
x = 80μm * tan(18.13°)
Now, let's calculate the value of x.
tan(18.13°) ≈ 0.3249
x = 80μm * 0.3249 = 25.99μm
Therefore, the distance between the double slit and the screen is approximately 25.99 micrometers.
Now, let's calculate the fringe spacing for a different wavelength of light.
Using the same formula for fringe spacing:
d*sinθ = mλ
Substituting the new wavelength (532nm) and solving for the fringe spacing (d):
d = (mλ) / sinθ
For the first bright fringe (m = 1) and the new wavelength (532nm), we substitute the values into the equation:
d = (1 * 532nm) / sinθ
Using the previously calculated value for θ (18.13°), we can find the new fringe spacing:
d = (1 * 532nm) / sin(18.13°)
Using a calculator, we find:
sin(18.13°) ≈ 0.309
d = (1 * 532nm) / 0.309 ≈ 1722nm
Therefore, the fringe spacing on the same screen for a wavelength of 532nm is approximately 1722 nanometers.