In Young's experiment a mixture of orange light (611 nm) and blue light (471 nm) shines on the double slit. The centers of the first-order bright blue fringes lie at the outer edges of a screen that is located 0.452 m away from the slits. However, the first-order bright orange fringes fall off the screen. By how much and in what direction (toward or away from the slits) should the screen be moved, so that the centers of the first-order bright orange fringes just appear on the screen? It may be assumed that is small, so that sin is approximately equal to tan , and the sign ''+'' means that the screen moves toward the slits.

Question

In Young's experiment a mixture of orange light (611 nm) and blue light (471 nm) shines on the double slit. The centers of the first-order bright blue fringes lie at the outer edges of a screen that is located 0.452 m away from the slits. However, the first-order bright orange fringes fall off the screen. By how much and in what direction (toward or away from the slits) should the screen be moved, so that the centers of the first-order bright orange fringes just appear on the screen? It may be assumed that is small, so that sin is approximately equal to tan , and the sign ''+'' means that the screen moves toward the slits.

Answer 1

To find the amount and direction in which the screen should be moved for the centers of the first-order bright orange fringes to appear on the screen, we need to use the concept of interference in Young's double-slit experiment.

The condition for constructive interference in Young's experiment is given by the equation:

d * sinθ = m * λ,

where d is the distance between the slits, θ is the angle of incidence of the light on the screen, m is the order of the fringe, and λ is the wavelength of light.

For blue light (471 nm):

d * sinθ_blue = m * λ_blue.

For orange light (611 nm):

d * sinθ_orange = m * λ_orange.

In the given problem, we are told that the centers of the first-order bright blue fringes lie at the outer edges of the screen, and the first-order bright orange fringes fall off the screen. This means that for blue light, m = 1, and for orange light, m = -1. The negative sign indicates that the bright orange fringe is on the other side of the central maximum.

Let's analyze the situation:

For blue light:
d * sinθ_blue = 1 * λ_blue.

For orange light:
d * sinθ_orange = -1 * λ_orange.

Since sinθ ≈ tanθ (for small angles),

For blue light:
d * tanθ_blue = 1 * λ_blue.

For orange light:
d * tanθ_orange = -1 * λ_orange.

We can rearrange these equations to find the difference in distance the screen needs to be moved (Δy) for the bright orange fringes to appear on the screen:

Δy = (d * tanθ_blue) - (d * tanθ_orange),
Δy = d * (tanθ_blue - tanθ_orange).

Now, we need to substitute the values given in the problem.

Given:
λ_blue = 471 nm,
λ_orange = 611 nm,
d = unknown,
tanθ_blue = tanθ_orange (since the centers of the blue fringes are at the outer edges of the screen).

Substituting these values in the equation, we get:

Δy = d * (tanθ_blue - tanθ_orange),
Δy = d * (tanθ_blue - tanθ_blue),
Δy = 0.

Therefore, the screen does not need to be moved in any direction for the centers of the first-order bright orange fringes to appear on the screen. The bright orange fringes will coincide with the bright blue fringes at the outer edges of the screen.