A wavelength of 625 nm is used in a Young's double-slit experiment. The separation between the slits is d = 1.3 10-5 m. The total width of the screen is 0.20 m. In one version of the setup, the separation between the double slit and the screen is LA = 0.35 m, whereas in another version it is LB = 0.60 m. On one side of the central bright fringe, how many bright fringes lie on the screen in the two versions of the setup? Do not include the central bright fringe in your counting. Verify that your answers are consistent with your answers to the Concept Questions.

bright fringes on one side (version A)

bright fringes on one side (version B)

I think it is asking you for n when x is .20 on each screen.

To find the number of bright fringes on one side of the central bright fringe in each version of the setup, we can use the formula for the fringe spacing:

Δy = λ * L / d

where Δy is the fringe spacing, λ is the wavelength of light, L is the separation between the double slit and the screen, and d is the separation between the slits.

First, let's calculate the fringe spacing for version A:

Δy_A = (625 nm) * (0.35 m) / (1.3 * 10^-5 m)
= 0.0168 m

Now, let's calculate the fringe spacing for version B:

Δy_B = (625 nm) * (0.60 m) / (1.3 * 10^-5 m)
= 0.0307 m

Next, we can calculate the number of bright fringes on one side of the central bright fringe by dividing the total screen width by the fringe spacing:

Number of bright fringes on one side (version A) = (0.20 m) / (2 * Δy_A)
= (0.20 m) / (2 * 0.0168 m)
≈ 5.95

Number of bright fringes on one side (version B) = (0.20 m) / (2 * Δy_B)
= (0.20 m) / (2 * 0.0307 m)
≈ 3.26

Therefore, in version A, there are approximately 5.95 bright fringes on one side of the central bright fringe, and in version B, there are approximately 3.26 bright fringes on one side of the central bright fringe.

These answers are consistent with the answers to the concept questions, which stated that increasing the distance between the double slit and the screen would result in fewer bright fringes.

To determine the number of bright fringes on one side of the central bright fringe in each version of the setup, we can use the formula:

n = (m * λ * L) / d

where n is the order of the bright fringe, λ is the wavelength of light, L is the distance from the double slit to the screen, d is the separation between the slits, and m is an integer representing the order of the fringe.

Given:
- λ = 625 nm = 625 * 10^-9 m
- d = 1.3 * 10^-5 m
- total width of the screen = 0.20 m
- LA = 0.35 m (version A)
- LB = 0.60 m (version B)

Let's first calculate the number of bright fringes on one side (version A):
For version A, we will be placing LA = 0.35 m in the formula. We need to calculate the maximum value of m such that the fringe falls within the width of the screen (0.20 m).

m * λ * L / d = 0.20 / 2 = 0.10 m

Solving for m:

m = (0.10 * d) / (λ * L)
= (0.10 * 1.3 * 10^-5 m) / (625 * 10^-9 m * 0.35 m)
= 0.037

So, in version A, there are approximately 0.037 bright fringes on one side (excluding the central bright fringe).

Now, let's calculate the number of bright fringes on one side (version B):
Using the same formula, we are now placing LB = 0.60 m in the formula. We need to calculate the maximum value of m such that the fringe falls within the width of the screen (0.20 m).

m * λ * L / d = 0.20 / 2 = 0.10 m

Solving for m:

m = (0.10 * d) / (λ * L)
= (0.10 * 1.3 * 10^-5 m) / (625 * 10^-9 m * 0.60 m)
= 0.034

So, in version B, there are approximately 0.034 bright fringes on one side (excluding the central bright fringe).

These results indicate that there are slightly fewer bright fringes on one side in version B compared to version A, which is consistent with the concept that increasing the distance between the double slit and the screen reduces the interference pattern's fringe spacing.

6 bright fringes