Question

Laser light of wavelength 632.8 nm falls normally on a slit that is 0.0250 mm wide. The transmitted light is viewed in a distant screen where the intensity at the center of the central bright fringe is 8.50 W/m2 (a) Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all. (b) At what angle does the dark fringe that is most distant from the center occur? (c) What is the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b)? Approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it.

Answer 1

The maximum number of totally dark fringes is 40. The dark fringe that is most distant from the center occurs at an angle of 1.13 degrees. The maximum intensity of the bright fringe before the dark fringe is 6.16 W/m².

Step-by-step explanation:

The number of totally dark fringes can be determined using the formula:

# of fringes = (slit width / wavelength) + 1

In this case, the slit width is 0.0250 mm and the wavelength is 632.8 nm. Plugging in these values, we get:

# of fringes = (0.0250 mm / 632.8 nm) + 1 = 40

Therefore, there are a maximum of 40 totally dark fringes on the screen.

To find the angle at which the dark fringe that is most distant from the center occurs, we can use the formula:

sinθ = (m − 0.5)λ / w

In this case, m is the order of the fringe (which is the number of fringes from the center), λ is the wavelength of the light (632.8 nm), and w is the width of the slit (0.0250 mm). Plugging in these values and solving for θ, we get:

θ = arcsin((40 − 0.5)(632.8 nm) / 0.0250 mm) = 1.13 degrees

To find the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b), we can assume that it occurs at the midpoint between the angles to the dark fringes on either side of it. Therefore, θ = 0.565 degrees. We can use the formula for the intensity of a single-slit diffraction pattern:

I = (I₀(w / λ))²sinc²(πw(sinθ)/λ)

In this case, I₀ is the intensity at the center of the bright fringe (8.50 W/m²), w is the width of the slit (0.0250 mm), λ is the wavelength of the light (632.8 nm), and θ is the angle at which the fringe occurs (0.565 degrees). Plugging in these values, we get:

I = (8.50 W/m²(0.0250 mm / 632.8 nm))²sinc²(π(0.0250 mm)(sin(0.565 degrees))/(632.8 nm)) = 6.16 W/m²

Answer 2

The maximum number of totally dark fringes cannot be determined without knowing the distance from the slit to the screen. The angle at which the dark fringe that is most distant from the center occurs is approximately 25.5°. The maximum intensity of the bright fringe immediately before the dark fringe is approximately 4.36 W/m^2.

Step-by-step explanation:

(a) To find the maximum number of totally dark fringes on the screen, we can use the formula for the number of bright fringes, which is given by:

n = w/Dλ

Where n is the number of fringes, w is the width of the slit, D is the distance from the slit to the screen, and λ is the wavelength of the light.

In this case, the width of the slit is 0.0250 mm, the distance from the slit to the screen is not given, and the wavelength of the light is 632.8 nm. Without the distance from the slit to the screen, we cannot determine the exact number of fringes. We can only say that the maximum number of totally dark fringes is one less than the maximum number of bright fringes, which occurs when the distance from the slit to the screen is equal to an integer multiple of half the wavelength of the light.

(b) To find the angle at which the dark fringe that is most distant from the center occurs, we can use the formula:

θ = sin^-1(mλ/a)

Where θ is the angle, m is the order of the fringe (in this case, m = 1), λ is the wavelength of the light, and a is the width of the slit. Plugging in the values, we get:

θ = sin^-1((1)(632.8 nm)/(0.0250 mm))

θ ≈ 25.5°

(c) To find the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b), we can use the formula:

I = I_0(sin^2(θ))/(sin^2(θ/2))

Where I is the intensity, I_0 is the maximum intensity (which is given), and θ is the angle. We can approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it, so the angle would be (25.5°+0°)/2 = 12.7°. Plugging in the values, we get:

I = (8.50 W/m^2)(sin^2(12.7°))/(sin^2(12.7°/2))

I ≈ 4.36 W/m^2