Maximums on a Diffraction Grating

[MicahP]

Homework Statement

Consider an array of parallel wires with uniform spacing of 1.40 cm between centers. In air at 20.0°C, ultrasound with a frequency of 35.6 kHz from a distant source is incident perpendicular to the array. (Take the speed of sound to be 343 m/s.)

(a) Find the number of directions on the other side of the array in which there is a maximum of intensity. (State the number of angles for which there is a maximum of intensity. An angle above the horizontal and an angle below the horizontal count as two separate angles. If there is a maximum at the horizontal, it also counts as an angle.)


(b) Find the angle for each of these directions relative to the direction of the incident beam. (State the angles corresponding to maxima of intensity for each value of m. If there does not exist a maximum of intensity for a given value of m, enter 'NONE' in the answer blank.)

Homework Equations

d sin θ = mλ

The Attempt at a Solution

I really have no idea. I know that this is a diffraction problem, but that's about it. I feel like giving the speed of sound is the hint in the question, but I don't see the relevance...

 

[ehild]

How do you get the wavelength λ from the speed of the wave and from its frequency?


ehild

 

[technician]

You are on the right track. It is useful to know the wavelength of the waves... Can you get that from what you have been given?

 

[MicahP]

I got it, that helped me along. What was throwing me off was that for some reason I thought I needed to know how many lines there were. Thanks!

 

[Nickie]

As a scientist, it is important to have a strong understanding of the concepts and principles involved in a problem before attempting to solve it. In this case, we are dealing with diffraction, which is the bending of waves around obstacles or through openings. The maximums on a diffraction grating refer to the points where constructive interference occurs, resulting in a higher intensity of the wave.

To solve this problem, we can use the equation d sin θ = mλ, where d is the spacing between the wires, θ is the angle of diffraction, m is the order of the maximum, and λ is the wavelength of the wave. In this case, we are dealing with ultrasound, so we can use the speed of sound (343 m/s) to calculate the wavelength using the formula λ = v/f, where v is the speed of sound and f is the frequency.

(a) To find the number of directions on the other side of the array with a maximum of intensity, we can use the equation d sin θ = mλ. We know the spacing between the wires (1.40 cm), the wavelength of the ultrasound (343/35.6 = 9.63 mm), and we can assume that the incident beam is perpendicular to the array, so sin θ = 0. Using these values, we can solve for m and find that there are 5 possible directions for the maximum intensity (m = 0, ±1, ±2).

(b) To find the angle for each of these directions, we can rearrange the equation d sin θ = mλ to solve for θ. Plugging in the values for d, m, and λ, we can find the corresponding angles for each maximum intensity. For m = 0, we get θ = 0, which corresponds to the horizontal direction. For m = ±1, we get θ = ±sin^-1 (λ/d) ≈ ±38.9°, and for m = ±2, we get θ = ±sin^-1 (2λ/d) ≈ ±71.6°. These angles correspond to the directions of the maximum intensity relative to the direction of the incident beam.

In conclusion, by using the principles of diffraction and the given values, we can determine the number of directions on the other side of the array with maximum intensity and the corresponding angles for each direction. It is important to have a strong understanding of