AP Physics 2 Exam Question:

A laser beam with a wavelength of 532 nm is incident on a diffraction grating with a slit separation of 1.5 μm and a slit width of 25 μm. The resulting diffraction pattern is observed on a screen placed 1.5 m away from the diffraction grating.

a) Calculate the number of slits that contribute to the first-order maximum of the diffraction pattern. b) Determine the angle at which the first-order maximum occurs. c) Calculate the width of the central maximum of the diffraction pattern on the screen. d) If the screen is replaced with a screen that is 5 m away from the diffraction grating, how will the width of the central maximum change?

Answer:

a) To calculate the number of slits that contribute to the first-order maximum, we can use the formula:

n * λ = d * sin(θ)

Where n is the order of diffraction, λ is the wavelength of light, d is the slit separation, and θ is the angle of diffraction.

For the first-order maximum, n = 1:

1 * (532 nm) = (1.5 μm) * sin(θ)

Converting the units to meters:

1 * (5.32 × 10^(-7) m) = (1.5 × 10^(-6) m) * sin(θ)

Simplifying the equation:

sin(θ) = (5.32 × 10^(-7) m) / (1.5 × 10^(-6) m)

sin(θ) = 0.3547

Now, we can calculate the angle of diffraction:

θ = sin^(-1)(0.3547)

θ ≈ 20.05°

b) The angle at which the first-order maximum occurs is approximately 20.05°.

c) To calculate the width of the central maximum, we use the formula:

w = λ * D / d

Where w is the width of the central maximum, λ is the wavelength, D is the distance between the diffraction grating and the screen, and d is the slit width.

Given values: λ = 532 nm, D = 1.5 m, and d = 25 μm.

Converting the units to meters:

w = (532 × 10^(-9) m) * (1.5 m) / (25 × 10^(-6) m)

Simplifying the equation:

w ≈ 0.03192 m = 31.92 mm

Therefore, the width of the central maximum is approximately 31.92 mm.

d) If the screen is replaced with a screen that is 5 m away from the diffraction grating, the distance D will increase. Let's calculate the new width of the central maximum using the same formula.

Given values: λ = 532 nm, D = 5 m, and d = 25 μm.

Converting the units to meters:

w = (532 × 10^(-9) m) * (5 m) / (25 × 10^(-6) m)

Simplifying the equation:

w ≈ 0.1064 m = 106.4 mm

Therefore, the width of the central maximum increases to approximately 106.4 mm.