# Slit Width & Single Slit Diffraction

• CalinDeZwart
In summary, we are asked to consider ultrasound with a frequency of 40 kHz and answer the following questions: (a) What is the wavelength and period of a 40 kHz sound wave? (b) What is 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 slit widths of 3.5λ, 5λ, and 2λ. The formula a sin θ = λ is used to calculate the first minimum for each slit width, and the equations for the second and third minimums can be determined by substituting the appropriate values for a.

## 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:

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.

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:

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.

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:

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?

## 1. What is the significance of slit width in single slit diffraction?

The slit width is a crucial factor in single slit diffraction as it determines the amount of diffraction that occurs. A narrower slit will result in a wider diffraction pattern, while a wider slit will produce a narrower diffraction pattern. Additionally, the slit width also affects the intensity of the diffraction pattern.

## 2. How does changing the slit width affect the diffraction pattern?

Changing the slit width alters the diffraction pattern by changing the width and intensity of the central maximum and the width of the secondary maxima and minima. A narrower slit will result in a wider central maximum and narrower secondary maxima and minima, while a wider slit will produce a narrower central maximum and wider secondary maxima and minima.

## 3. Is there an optimal slit width for producing the most distinct diffraction pattern?

There is no one optimal slit width for producing the most distinct diffraction pattern as it depends on the specific experiment and the desired outcome. Generally, a narrower slit will produce a more distinct diffraction pattern, but it may also result in a decrease in intensity. It is important to carefully choose the slit width based on the specific experiment and its objectives.

## 4. How does the wavelength of light affect the diffraction pattern in single slit diffraction?

The wavelength of light plays a significant role in single slit diffraction. The diffraction pattern is directly proportional to the wavelength of light, meaning that as the wavelength increases, the diffraction pattern becomes wider. This is because longer wavelengths diffract more than shorter wavelengths.

## 5. Can single slit diffraction occur with light of any wavelength?

Yes, single slit diffraction can occur with light of any wavelength. However, the amount of diffraction and the resulting pattern will vary depending on the wavelength. Light with longer wavelengths will result in a wider diffraction pattern compared to light with shorter wavelengths.