In an experiment conducted using the Young's double-slit apparatus, the separation between the slits is 20 µm. A first-order constructive interference fringe appears at an angle of 2.5o from the zeroth order (central) fringe.
A. What wavelength of light is used in the experiment?
B. At what angle would the fourth-order (m = 4) bright fringe appear?
C. At what angle would the fourth-order (m=4) dark fringe appear?
Use the formula on these, I can double check them for you.
To answer these questions, we need to use the formula for the location of the bright or dark fringes in the Young's double-slit experiment:
λ = (d * sinθ) / m
Where:
λ is the wavelength of light used in the experiment,
d is the separation between the double-slit (20 µm in this case),
θ is the angle of the fringe,
and m is the order of the fringe.
A. To find the wavelength of light (λ), we can rearrange the formula:
λ = (d * sinθ) / m
We are given that the separation between the slits (d) is 20 µm, and the angle of the first-order bright fringe (θ) is 2.5 degrees. Since it is a first-order fringe (m = 1), we can substitute these values into the formula:
λ = (20 µm * sin(2.5 degrees)) / 1
Calculate sin(2.5 degrees) = 0.0436 approximately.
Substituting this value into the formula:
λ = (20 µm * 0.0436) / 1
λ = 0.872 µm
So, the wavelength of light used in the experiment is approximately 0.872 µm.
B. Now, let's find the angle at which the fourth-order bright fringe would appear. We'll use the same formula, but with m = 4:
λ = (d * sinθ) / m
We know the separation between the slits (d) is 20 µm, and we need to find θ when m = 4. Rearranging the formula:
θ = arcsin((λ * m) / d)
Substituting the values we have:
θ = arcsin((0.872 µm * 4) / 20 µm)
Calculate (0.872 µm * 4) / 20 µm = 0.1744.
Taking the arcsin of 0.1744:
θ ≈ arcsin(0.1744)
θ ≈ 10 degrees
So, the fourth-order bright fringe will appear at an angle of approximately 10 degrees.
C. Finally, let's find the angle at which the fourth-order dark fringe would appear. We can use the same formula, but with m = 4 again:
λ = (d * sinθ) / m
Substituting the given values:
θ = arcsin((λ * m) / d)
θ = arcsin((0.872 µm * 4) / 20 µm)
Calculating (0.872 µm * 4) / 20 µm = 0.1744.
Taking the arcsin of 0.1744:
θ ≈ arcsin(0.1744)
θ ≈ 10 degrees
So, the fourth-order dark fringe will also appear at an angle of approximately 10 degrees.