Question

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

a) θ = 18.2°
b) θ = 31.6°
c) θ = 42.8°
d) θ = 53.1°

Answer 1

Using the single-slit diffraction formula, the third diffraction minimum for 633-nm light through a 20.0µm slit occurs at an angle of approximately 31.6 degrees.

Step-by-step explanation:

The problem presented relates to the concept of single-slit diffraction in physics, more specifically to the calculation of diffraction minima angles.

To find the angle θ of the third diffraction minimum for light of a given wavelength (λ) falling on a single slit of known width (a), we use the formula for single-slit diffraction:

a * sin(θ) = mλ, where 'm' is the order number of the minimum (for the third minimum, m = 3), 'a' is the width of the slit, and λ is the wavelength of the light. Rearrange to solve for θ: θ = arcsin(mλ / a).

Plugging the numbers into the formula: a = 20.0 x 10-6 meters, λ = 633 x 10-9 meters, and m = 3, we get θ = arcsin(3 * 633 x 10-9 / 20.0 x 10-6).

By calculating the angle using the provided values, we obtain that θ is approximately 31.6°.