light diffraction help? a single monochromatic indigo light source is shined through an etched, flat prism with a slit separation of .0250mm. the resulting interference pattern is viewed on a screen 1.25 m away. the third Maximum is found to be 6.6 cm from the central maximum.

What is the wavelength of the indigo light?
What are the first and second maxima?

To find the wavelength of the indigo light and the positions of the first and second maxima in the interference pattern, we can use the formula for the position of the maxima in a double slit experiment:

y = (m * λ * L) / (d)

Where:
- y is the distance from the central maximum to the mth maximum.
- λ is the wavelength of the light.
- L is the distance between the screen and the double slit.
- d is the separation between the slits.

In this case, we know:
- y = 6.6 cm = 0.066 m (for the third maximum)
- L = 1.25 m (distance to the screen)
- d = 0.0250 mm = 0.0250 * 10^-3 m (separation of the slits)

To solve for the wavelength (λ), we can rearrange the formula to isolate λ:

λ = (y * d) / (m * L)

Substituting the known values:

λ = (0.066 * 0.0250 * 10^-3) / (3 * 1.25)

Simplifying and calculating:

λ = 0.00165 / 3.75

λ ≈ 0.00044 m

Therefore, the wavelength of the indigo light is approximately 0.00044 meters (or 440 nm).

To find the positions of the first and second maxima, we can use the same formula by substituting the values for m:

For the first maximum (m = 1):
y1 = (1 * 0.00044 * 0.0250 * 10^-3) / (1 * 1.25)
y1 ≈ 8.80 * 10^-6 m (approximately)

For the second maximum (m = 2):
y2 = (2 * 0.00044 * 0.0250 * 10^-3) / (1 * 1.25)
y2 ≈ 1.76 * 10^-5 m (approximately)

Therefore, the first maximum is approximately 8.80 * 10^-6 meters from the central maximum, and the second maximum is approximately 1.76 * 10^-5 meters from the central maximum.

To find the wavelength of the indigo light, we can use the formula for the slit separation in the diffraction pattern:

d * sin(θ) = m * λ

Where:
- d is the slit separation
- θ is the angle of diffraction
- m is the order of the maximum
- λ is the wavelength of light

In this case, we are given the slit separation (d = 0.0250 mm) and the position of the third maximum (m = 3) as 6.6 cm from the central maximum. We can convert the distance to meters:

distance = 0.066 m

Next, we need to find the angle of diffraction, which is given by:

tan(θ) = distance / distance to screen

distance to screen = 1.25 m

By substituting the values and solving for θ:

tan(θ) = 0.066 m / 1.25 m

θ ≈ 0.0528 radians

Now, we can calculate the wavelength of the indigo light by rearranging the formula and solving for λ:

λ = d * sin(θ) / m

λ = (0.0250 mm * sin(0.0528 radians)) / 3

Converting mm to meters:

λ ≈ (0.0250 * 10^-3 m * sin(0.0528 radians)) / 3

Calculating λ:

λ ≈ 4.13 x 10^-7 m

Therefore, the wavelength of the indigo light is approximately 4.13 x 10^-7 meters (413 nm).

Now, let's find the positions of the first and second maxima:

For the first maximum, since m = 1, we can use the same formula:

d * sin(θ) = m * λ

θ1 = sin^-1(m * λ / d)

θ1 = sin^-1(1 * λ / d)

Calculating θ1 using the previously calculated values:

θ1 = sin^-1(4.13 x 10^-7 m / 0.0250 * 10^-3 m) = sin^-1(0.0165)

θ1 ≈ 0.0165 radians

Next, we need to find the position of the first maximum on the screen. Using the small-angle approximation, we can approximate:

Position of the first maximum = θ1 * distance to screen

Position of the first maximum ≈ 0.0165 * 1.25 m = 0.0206 m (or 2.06 cm)

For the second maximum (m = 2), we use the same formula:

θ2 = sin^-1(2 * λ / d)

Calculating θ2 using the previously calculated values:

θ2 = sin^-1(2 * 4.13 x 10^-7 m / 0.0250 * 10^-3 m) = sin^-1(0.033)

θ2 ≈ 0.033 radians

Using the small-angle approximation:

Position of the second maximum ≈ 0.033 * 1.25 m = 0.04125 m (or 4.12 cm)

Therefore, the first maximum is approximately 2.06 cm from the central maximum, and the second maximum is approximately 4.12 cm from the central maximum.