Slit Width & Single Slit Diffraction

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SUMMARY

The discussion focuses on single slit diffraction involving ultrasound at a frequency of 40 kHz. The wavelength is calculated as 8.575 x 10-3 nm, and the period is determined to be 2.5 x 10-5 seconds. The required slit width for the first order diffraction minimum at 30 degrees is calculated to be 0.01715 µm. Participants also explore calculating angles for the first three minima using different slit widths, specifically three and a half, five, and twice the wavelength.

PREREQUISITES
  • Understanding of wave properties, specifically wavelength and frequency.
  • Familiarity with the diffraction equation a sin θ = λ.
  • Knowledge of inverse sine functions for angle calculations.
  • Basic principles of sound wave propagation and speed of sound (343 m/s).
NEXT STEPS
  • Learn how to derive the diffraction angles for multiple orders using the equation a sin θ = nλ.
  • Study the implications of varying slit widths on diffraction patterns.
  • Explore the relationship between frequency, wavelength, and period in sound waves.
  • Investigate the effects of different mediums on sound wave propagation.
USEFUL FOR

Students studying physics, particularly those focusing on wave mechanics and diffraction, as well as educators seeking to clarify concepts related to sound waves and their behavior in various scenarios.

CalinDeZwart
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Homework Statement


Consider ultrasound with a frequency of 40 kHz.

(a) What is the wavelength and period of a 40 kHz sound wave?
(b) The slit width required to give the first order diffraction minimum at thirty degrees.
(c) Calculate the angles at which the first three minimums should be observed for single slit diffraction with the following slit widths:
1. three and a half times the wavelength:
2. five times the wavelength:
3. twice the wavelength:


Homework Equations



a sin θ = λ
sin θ = λ/a

The Attempt at a Solution


(a)

40kHz = 40,000Hz
v = speed of sound = 343m/s
f = 40,000Hz
f = 1/T
T = 1/f = 1/40,000 = [2.5*10^(-5)]
λ = v/f = 343/40000 = [8.575*10^(-3)nm]

I'm not sure if this is correct.

(b)

slit width = a
slit separation = d
angle = θ

a sin θ = λ
sin θ = λ/a
sin 30 = (8.575*10^(-3)) / a
0.5 = (8.575*10^(-3)) / a
a = λ/sin 30 = [0.01715um]

I am sure that this is wrong.

(c)

I am not sure how to calculate this with the limited variables that I have to work with.
 
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CalinDeZwart said:

Homework Statement


Consider ultrasound with a frequency of 40 kHz.

(a) What is the wavelength and period of a 40 kHz sound wave?
(b) The slit width required to give the first order diffraction minimum at thirty degrees.
(c) Calculate the angles at which the first three minimums should be observed for single slit diffraction with the following slit widths:
1. three and a half times the wavelength:
2. five times the wavelength:
3. twice the wavelength:


Homework Equations



a sin θ = λ
sin θ = λ/a

The Attempt at a Solution


(a)

40kHz = 40,000Hz
v = speed of sound = 343m/s
f = 40,000Hz
f = 1/T
T = 1/f = 1/40,000 = [2.5*10^(-5)]
λ = v/f = 343/40000 = [8.575*10^(-3)nm]

I'm not sure if this is correct.

(b)

slit width = a
slit separation = d
angle = θ

a sin θ = λ
sin θ = λ/a
sin 30 = (8.575*10^(-3)) / a
0.5 = (8.575*10^(-3)) / a
a = λ/sin 30 = [0.01715um]

I am sure that this is wrong.

(c)

I am not sure how to calculate this with the limited variables that I have to work with.
The period and wavelength are correct, but you need to include the units.
You gave the formula for the first diffraction minimum, what is it for other orders?
There is a mistake in your value of a.
 
ehild said:
The period and wavelength are correct, but you need to include the units.
You gave the formula for the first diffraction minimum, what is it for other orders?
There is a mistake in your value of a.

Thanks for your reply.

I am good with a) and b), but still struggling with c).

My lecturer says I need the diffraction equation from b) and to work out the inverse sine function. I understand what he is saying, but the question seems strangely worded.

(c) Calculate the angles at which the first three minimums should be observed for single slit diffraction with the following slit widths:
1. three and a half times the wavelength:
2. five times the wavelength:
3. twice the wavelength:

I am being asked to calculate NEW angles, that explains why I use the inverse sine function.

but then the question states '...with the following slit widths:', then proceeds to give multipliers of the wavelength. Which one is it? What do I keep constant?
 
CalinDeZwart said:
Thanks for your reply.

I am good with a) and b), but still struggling with c).

My lecturer says I need the diffraction equation from b) and to work out the inverse sine function. I understand what he is saying, but the question seems strangely worded.

(c) Calculate the angles at which the first three minimums should be observed for single slit diffraction with the following slit widths:
1. three and a half times the wavelength:
2. five times the wavelength:
3. twice the wavelength:

I am being asked to calculate NEW angles, that explains why I use the inverse sine function.

but then the question states '...with the following slit widths:', then proceeds to give multipliers of the wavelength. Which one is it? What do I keep constant?

You have to determine those angles for each slit widths.
The equation you wrote for the first minimum is
a sin(θ)=λ
If a= 3.5λ,
3.5λsin(θ)=λ.
What is sin(θ) ? What is θ?
What is θ if a= 5λ? if a=2λ?
What are the equation for the second and third minimums?
 

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