Question

Light of wavelength 385 nm (in vacuum) is incident on a diffraction grating that has a slit separation of 1.2 × 10-5 m. The distance between the grating and the viewing screen is 0.18 m. A diffraction pattern is produced on the screen that consists of a central bright fringe and higher-order bright fringes (see the drawing). (a) Determine the distance y from the central bright fringe to the second-order bright fringe. (Hint: The diffraction angles are small enough that the approximation tan(θ) ~ sin(θ) can be used.)(b) If the entire apparatus is submerged in water (nwater = 1.33), what is the distance y?

Answer 1

Answer 2

To find the distance y from the central bright fringe to the second-order bright fringe, we can use the equation for the position of bright fringes in a diffraction grating:

mλ = d * sin(θ)

where: m is the order of the bright fringe (m = 0 for the central fringe, m = 1 for the first-order, m = 2 for the second-order, and so on), λ is the wavelength of light, d is the slit separation of the grating, and θ is the angle of diffraction.

(a) For the second-order bright fringe (m = 2), we can approximate tan(θ) ~ sin(θ) for small angles:

sin(θ) = y / L

where: y is the distance from the central bright fringe to the second-order bright fringe, and L is the distance between the grating and the viewing screen.

Now, let's find y using the given values:

Given: λ = 385 nm = 385 x 10^(-9) m (convert nm to meters), d = 1.2 × 10^(-5) m, L = 0.18 m, and m = 2.

Step 1: Convert the wavelength to meters: λ = 385 nm = 385 x 10^(-9) m.

Step 2: Use the equation mλ = d * sin(θ) to find θ for the second-order bright fringe (m = 2): 2 * (385 x 10^(-9) m) = 1.2 × 10^(-5) m * sin(θ)

Step 3: Solve for sin(θ): sin(θ) = (2 * 385 x 10^(-9) m) / (1.2 × 10^(-5) m) sin(θ) ≈ 0.064167

Step 4: Now, find y using the approximation sin(θ) ≈ y / L: 0.064167 ≈ y / 0.18 m

Step 5: Solve for y: y ≈ 0.064167 * 0.18 m y ≈ 0.01155 m

So, the distance y from the central bright fringe to the second-order bright fringe is approximately 0.01155 meters.

(b) If the entire apparatus is submerged in water (nwater = 1.33), we need to account for the change in the wavelength of light in water due to the refractive index.

The new wavelength (λ_water) in water can be calculated using the equation: λ_water = λ_vacuum / nwater

where: λ_vacuum is the wavelength of light in vacuum, and nwater is the refractive index of water.

Step 1: Find λ_water: λ_water = (385 x 10^(-9) m) / 1.33 λ_water ≈ 289.474 x 10^(-9) m

Step 2: Now, use the new wavelength (λ_water) in the previous equation mλ = d * sin(θ) to find θ for the second-order bright fringe (m = 2):

2 * (289.474 x 10^(-9) m) = 1.2 × 10^(-5) m * sin(θ_water)

Step 3: Solve for sin(θ_water): sin(θ_water) = (2 * 289.474 x 10^(-9) m) / (1.2 × 10^(-5) m) sin(θ_water) ≈ 0.0482895

Step 4: Now, find y using the approximation sin(θ_water) ≈ y / L: 0.0482895 ≈ y / 0.18 m

Step 5: Solve for y: y ≈ 0.0482895 * 0.18 m y ≈ 0.008692 m

So, if the entire apparatus is submerged in water, the distance y from the central bright fringe to the second-order bright fringe is approximately 0.008692 meters.