Question

(a) Find the angle of the third diffraction minimum for 633-nm light falling on a slit of width 20.0μm.

(b) What slit width would place this minimum at 85.0º?
Explicitly show how you follow the steps in Problem-Solving Strategies for Wave Optics.
a) (a) 37.2º, (b) 30.0 μm
b) (a) 45.0º, (b) 25.0 μm
c) (a) 52.8º, (b) 20.0 μm
d) (a) 60.0º, (b) 15.0 μm

Answer 1

To find the angle of the third diffraction minimum for 633-nm light falling on a slit of width 20.0 µm, use the equation sin(θ) = (m * λ) / w. The angle is 37.2º. To find the slit width that would place this minimum at 85.0º, rearrange the equation and solve for w. The slit width is approximately 30.0 µm.

Step-by-step explanation:

To find the angle of the third diffraction minimum for 633-nm light falling on a slit of width 20.0 µm, we can use the equation:

sin(θ) = (m * λ) / w

where θ is the angle of diffraction, m is the order of the minimum, λ is the wavelength of light, and w is the width of the slit.

For the third diffraction minimum, m = 3, λ = 633 nm, and w = 20.0 µm.

Plugging in these values, we can solve for θ:

sin(θ) = (3 * 633 nm) / 20.0 µm

Calculating this, we find θ = 37.2º.

To find the slit width that would place this minimum at 85.0º, we can rearrange the equation and solve for w:

w = (m * λ) / sin(θ)

Plugging in the values m = 3, λ = 633 nm, and θ = 85.0º, we can solve for w:

w = (3 * 633 nm) / sin(85.0º)

Calculating this, we find w ≈ 30.0 µm.