When conducting an interference experiment with monochromatic light ofwavelength 6000 angstroms, it is known that the two slits are 0.5 millimetersapart, and the distance between the slit and the screen is 2 meters. Then, onthe screen where the interference fringes are formed, (1) what is the width ofthe first bright fringe? (2) How far is the third dark line from the central axis? (3)If monochromatic light with a wavelength of 4000 angstroms is used instead,how far is the centerline of the third bright fringe from the central axis?

To answer these questions, we need to use the formula for the position of bright and dark fringes in a double-slit interference experiment.

The formula for the position of bright fringes (constructive interference) is:

y_bright = (m * λ * L) / d

where:

• m is the order of the fringe (m = 0 for the central bright fringe, m = 1 for the first bright fringe, etc.)
• λ is the wavelength of the light
• L is the distance from the slits to the screen
• d is the distance between the slits

The formula for the position of dark fringes (destructive interference) is:

y_dark = ((m + 1/2) * λ * L) / d

where m is the order of the dark fringe (m = 0 for the first dark fringe, m = 1 for the second dark fringe, etc.)

(1) To find the width of the first bright fringe, we need to find the position of the first bright fringe (m = 1) and the position of the second bright fringe (m = 2), and then subtract the two.

First, convert the given values to the correct units: λ = 6000 angstroms = 6000 * 10^-10 meters, d = 0.5 mm = 0.5 * 10^-3 meters, and L = 2 meters.

The position of the first bright fringe (m = 1) is:

y_bright1 = (1 * 6000 * 10^-10 * 2) / (0.5 * 10^-3) = 0.0024 meters = 2.4 mm

The position of the second bright fringe (m = 2) is:

y_bright2 = (2 * 6000 * 10^-10 * 2) / (0.5 * 10^-3) = 0.0048 meters = 4.8 mm

So, the width of the first bright fringe is y_bright2 - y_bright1 = 4.8 mm - 2.4 mm = 2.4 mm.

(2) To find the position of the third dark line, we use m = 2 in the formula for y_dark:

y_dark3 = ((2 + 1/2) * 6000 * 10^-10 * 2) / (0.5 * 10^-3) = 0.006 meters = 6 mm

So, the third dark line is 6 mm from the central axis.

(3) If the wavelength of the light is changed to 4000 angstroms = 4000 * 10^-10 meters, the position of the third bright fringe (m = 3) is:

y_bright3 = (3 * 4000 * 10^-10 * 2) / (0.5 * 10^-3) = 0.0048 meters = 4.8 mm

So, with the shorter wavelength light, the centerline of the third bright fringe is 4.8 mm from the central axis.

To answer these questions, we need to use the formula for the position of bright and dark fringes in a double-slit interference experiment.

The formula for the position of bright fringes (constructive interference) is:

y_bright = (m * λ * L) / d

And the formula for the position of dark fringes (destructive interference) is:

y_dark = ((m + 1/2) * λ * L) / d

where:

• m is the order of the fringe (0 for the central bright fringe, 1 for the first bright fringe, etc.)
• λ is the wavelength of the light
• L is the distance from the slits to the screen
• d is the distance between the slits

(1) To find the width of the first bright fringe, we need to find the position of the first bright fringe (m=1) and subtract the position of the central bright fringe (m=0).

Using the formula for bright fringes, with λ = 6000 angstroms = 6000 * 10^-10 meters, L = 2 meters, and d = 0.5 mm = 0.5 * 10^-3 meters:

y_bright_1 = (1 * 6000 * 10^-10 * 2) / 0.5 * 10^-3 = 0.0024 meters = 2.4 mm

The position of the central bright fringe (m=0) is always y_bright_0 = 0.

So, the width of the first bright fringe is 2.4 mm.

(2) To find the position of the third dark line, we use the formula for dark fringes with m=2 (since the first dark line is m=0, the second is m=1, etc.):

y_dark_3 = ((2 + 1/2) * 6000 * 10^-10 * 2) / 0.5 * 10^-3 = 0.0036 meters = 3.6 mm

So, the third dark line is 3.6 mm from the central axis.

(3) To find the position of the third bright fringe using light with a wavelength of 4000 angstroms, we use the formula for bright fringes with m=3:

y_bright_3 = (3 * 4000 * 10^-10 * 2) / 0.5 * 10^-3 = 0.0048 meters = 4.8 mm

So, with light of wavelength 4000 angstroms, the centerline of the third bright fringe is 4.8 mm from the central axis.

To answer these questions, we need to use the formula for the interference pattern created by two slits, which is given by:

dsin(θ) = mλ

where:

• d is the distance between the slits,
• θ is the angle between the central axis and the mth bright fringe,
• m is the order of the fringe, and
• λ is the wavelength of the light.

1. To find the width of the first bright fringe (m=1), we need to find the angle θ first. Rearranging the formula gives:

θ = arcsin((m*λ)/d)

Substituting the given values (remember to convert the units appropriately so they all match):

θ = arcsin((1600010^-10)/(0.5*10^-3)) = 0.0012 radians

The width of the fringe on the screen can be found using simple trigonometry. The width is the opposite side of a right triangle, where the adjacent side is the distance between the slit and the screen (L), and the angle is θ. So:

Width = Ltan(θ) = 2tan(0.0012) = 0.0024 meters = 2.4 mm

2. The third dark line corresponds to m=2.5 in the formula (since dark lines fall between bright fringes). So:

θ = arcsin((2.5600010^-10)/(0.5*10^-3)) = 0.003 radians

The distance from the central axis is then:

Distance = Ltan(θ) = 2tan(0.003) = 0.006 meters = 6 mm

3. If the wavelength changes to 4000 angstroms, the distance to the third bright fringe (m=3) is:

θ = arcsin((3400010^-10)/(0.5*10^-3)) = 0.0024 radians

The distance from the central axis is then:

Distance = Ltan(θ) = 2tan(0.0024) = 0.0048 meters = 4.8 mm