Check out the Original Double Slit Experiment

https://www.youtube.com/watch?v=Iuv6hY6zsd0

Pre-lecture Study Resources

Watch the pre-lecture videos and read through the OpenStax text before doing the pre-lecture homework or attending class.

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Learning Objectives

Summary

Summary

Atomistic Goals

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BoxSand Introduction

Wave Optics  |  Young's Double Slit Experiment


Superposition of Waves

When two waves enter the same region of space at the same time they interfere in a way that obeys the Superposition of Waves. This addition of waves creates places where the peaks line up and the resultant wave is larger, which would manifest as a bright spot for light waves. There are also regions where the peak from one wave is lined up with the trough from the other and the resultant is wave cancelation, where their amplitudes add to zero and there would be a dark spot. These constructive and destructive areas are seperated by an entire gradient of partially constructive and partially destructive. The resulting interference pattern is one of the defining features of a wave.

This is a gif of two sources at some distance from each other emitting same frequency. The frequency is shown as circles emitted from the source and there are areas of overlap from the other source.

Wave Interpretation of Light - Young's Double Slit Interference

The theory of Electromagnetic Radiation Wave Theory is one of the most tested and confirmed theories in physics. It relies heavily on the experimental fact that light has wave-like properties. These wave-like properties are displayed by the interference effects that has only been observed in wave systems. The hallmark experiment that enabled us to observe the effect was Young's Double Slit Experiement.

To observe the interference of two sources you need two, coherent, single frequency sources. Essentially you would like two identical sources. The clever way Young achieved this was by isolating a single color of light then sending that light through two small slits. Each slit acts like a new wave source due to the diffraction of light, which is the effect that light "bends" around corners and spreads out when passing through an opening. Since each slit acts like a new source, and each originated from the same source, they are two coherent, single frequency (same color) sources. The diagram below shows a snapshot of the waves as they interfere.

This is an image of a light behind a wall with a single slit. The light emitted then goes through another wall with two slits some vertical distance between each other. The waves emitted from the two different slits have some interference with one another and show an interference pattern with bands of white light and bands of no light.

The modern technology of a LASER (Light Amplification through the Stimulation of Electromagnetic Radiation) has made this experiment much easier due to the very high intensity soures they provide. The experiment also enables one to measure the wavelength of visable light, something on the order of hundreds of nanometers, a size only an order of magnitude greater than the size of atoms. For nearly 100 years this was the only way to probe scales this size scale and it still remains as one of the best.

The geometry of Young's Double Slit is below.

This is an image of a thin wall with two slits with some vertical distance between each other denoted by d and each slit has some width denoted by a. From the top slit there is a line that goes at some angle from the lower vertical labeled as theta prime. From the center of d is another dashed line that is at some angle theta from the horizontal that ends at the center of no light and there is a horizontal distance D to the back wall from the slits. If there was a light shined on the other side of the slit wall, there would be band of light and no light and the distance between the areas of no light are denoted by y. There are notes that says the an assumption of infinite source distance gives plane wave at the slit so that all amplitude elements are in phase. And, for when the distance between the two slits and the back wall is much greater than the width of the slit, this approaches the right angle and theta prime is approximately equal to theta. And, tangent of theta is equal to the distance between the the areas of no light divided by the horizontal distance, for distance screen assumption, this is approximately equal to sine theta which is approximately equal to theta. And, for a maximum condition, the vertical distance between the slits multiplied by sine theta is equal to m multiplied by lambda.

Here the dstance between the slits is $d$ and the screen observing the interference effect is a distance $D$ away from the slits. The central maximum is located along the perpendicular bisector between the two sources where the Path Length Difference (PLD) between the two sources is zero. The condition for constructive interference is for the PLD to be an interger ($m$) multiple of the wavelength. So as you move away from the central maximum and go from constructive to destructive and back to constructive, you've increased the PLD by one whole wavelength. These bright spots are called interference fringes. Using the condition for constructive interference $PLD = m \lambda$ and the geometery that $PLD=d sin(\theta)$, you arrive at the overall condition for constructive interference.

$d sin{\theta}=m \lambda$

The distance $y$ is measured from the central maximum and can be related to the angle ($\theta$) and the distance $D$ by the equation $tan (\theta) = \frac{y}{D}$. In most cases, where $\lambda << d$ the angle is so small that $sin (\theta) \approx \theta$. Since $tan (\theta) = \frac{sin(\theta)}{\cos(\theta)}$ and $\cos(\theta) \approx 1$ for very small angles, $tan(\theta) \approx sin(\theta) \approx \theta$. This allows a more simple connection between the variables shown in the equation below, but note, this only if the angles are very small.

$y \approx \frac{m \lambda D}{d}$,   if $\lambda << d$


Key Equations and Infographics

 

A representation with the words Young’s double slit interference on the top. There is an equation that shows that the multiplication of the distance between slits and the sine of the diffraction angle of the mth order fringe which is equal to the product of the mth order fringe value and the wavelength. This is also written in words below.


A representation with the words Young’s double slit interference on the top. There is an equation that shows that the distance from the central maxima to the mth order bright fringe is approximately equal to the mth order value multiplied by the wavelength and the distance between the slits and the screen, divided by the distance between the slits. This is also written in words below.


A representation with the words single slit interference on the top. There is an equation that shows that the width of the slit multiplied by the sine of the diffraction angle of the pth order dark fringe is equal to the product of the order of the dark fringe and the wavelength. This is also written in words below.


A representation with the words single slit interference on the top. There is an equation that shows that the diffraction angle of the pth order dark fringe is approximately equal to the order of the dark fringe multiplied by the wavelength divided by the slit width. This is also written in words below.


A representation with the words single slit interference on the top. There is an equation that shows that the distance to the dark fringe from the central maxima for the pth order dark fringe is equal to the order of the dark fringe multiplied by the wavelength and distance between the slit and screen, divided by the slit width. This is also written in words below


A representation with the words single slit central maxima width on the top. There is an equation that shows that the width of the central maxima is equal to twice the wavelength multiplied by the distance between the slit and screen, divided by the slit width. This is also written in words below.

Now, take a look at the pre-lecture reading and videos below.

OpenStax Reading


OpenStax Section 27.2  |  Huygens's Principle: Diffraction

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OpenStax Section 27.3  |  Young's Double Slit Experiment

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Additional Study Resources

Use the supplemental resources below to support your post-lecture study.

YouTube Videos

Here is another good introduction to the Double Slit Experiment

https://www.youtube.com/watch?v=uva6gBEpfDY

Some mathematics behind double slit diffraction,

https://www.youtube.com/watch?v=KeHry37evb4

Simulations


Phet - wave interference, 

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For additional simulations on this subject, visit the simulations repository.

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Demos


Lake Experiment #2 - Wave Interference(1min)

Lake Experiment #2 - Wave Interference(1min)

For additional demos involving this subject, visit the demo repository

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History


Oh no, we haven't been able to write up a history overview for this topic. If you'd like to contribute, contact the director of BoxSand, KC Walsh (walshke@oregonstate.edu).

Physics Fun


Check out Dr. Quantum explore the double slit experiment from a quantum physics perspective.

https://www.youtube.com/watch?v=btImof4nyzo

Other Resources


Resource Repository

This link will take you to the repository of other content related resources.

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Problem Solving Guide

Use the Tips and Tricks below to support your post-lecture study.

Assumptions

 

Checklist

1. Read the problem carefully and visualize the physical setup.

2. Determine if it is indeed a problem best analyzed with wave optics and not ray optics.

3. Determine what physical model best fits the problem. Is it a double, multi, or single slit interference or is a thin film effect?

4. For the type of system, draw the appropriate physical representation.

5. Organize the relevant equations for the specific system.

6. Determine what variables are known or unknown. Perhaps what variables are changing - is it a proportional reasoning problem?

7. If it is a slit problem, is it in the small angle approximation? If so, how does this simplify the equations?

8. Solve the system of equations, perhaps with the use of geometry (this is a big step and mastery comes with practice).

9. Check your answer for reasonableness. Magnitude, dimensions, limiting cases?

Misconceptions & Mistakes

  • ... students mistakenly assume all the interference patterns are the same
  • ... students mistakenly make the small angle approximation for all cases
  • ... students mistakenly ignore the $\pi$ phase shifts in certain reflections in a thin film interference

Pro Tips

  • Draw the physical representation - being able to see the fringes and the geometry is very useful

Multiple Representations

Multiple Representations is the concept that a physical phenomena can be expressed in different ways.

Physical

Physical Representations describes the physical phenomena of the situation in a visual way.

The diagram for Young's double slit experiment depicts the important features used to describe the phenomena. 

This is an image of a thin wall with two slits with some vertical distance between each other denoted by d and each slit has some width denoted by a. From the top slit there is a line that goes at some angle from the lower vertical labeled as theta prime. From the center of d is another dashed line that is at some angle theta from the horizontal that ends at the center of no light and there is a horizontal distance D to the back wall from the slits. If there was a light shined on the other side of the slit wall, there would be band of light and no light and the distance between the areas of no light are denoted by y. There are notes that says the an assumption of infinite source distance gives plane wave at the slit so that all amplitude elements are in phase. And, for when the distance between the two slits and the back wall is much greater than the width of the slit, this approaches the right angle and theta prime is approximately equal to theta. And, tangent of theta is equal to the distance between the the areas of no light divided by the horizontal distance, for distance screen assumption, this is approximately equal to sine theta which is approximately equal to theta. And, for a maximum condition, the vertical distance between the slits multiplied by sine theta is equal to m multiplied by lambda.

From a single slit diffraction to mutli slit diffraction the single slit forms an envelope where the sharpness of bright peaks become more sharp as the number of slits is increased.

This is an image of four different kinds of slit apparatuses. The first shows a single slit diffraction where there is a very intense central maximum and two faint with lower intensity fringes. The double slit shows the central maximum labeled as the single slit envelope and two slightly lower intense peaks on either side and lower intense fringes. The three slit shows the central maximum labeled as the single slit envelope and two slightly lower intense peaks on either side and even slightly lower intense peaks on either side of those with less intense fringes. The five slit shows the central maximum labeled as the single slit envelope and two slightly lower intense peaks on either side with more fringes on either side. This is to show that with increasing slits, the greater the sharpening of the fringes.

Mathematical

Mathematical Representation uses equation(s) to describe and analyze the situation.

A representation with the words Young’s double slit interference on the top. There is an equation that shows that the multiplication of the distance between slits and the sine of the diffraction angle of the mth order fringe which is equal to the product of the mth order fringe value and the wavelength. This is also written in words below.


A representation with the words Young’s double slit interference on the top. There is an equation that shows that the distance from the central maxima to the mth order bright fringe is approximately equal to the mth order value multiplied by the wavelength and the distance between the slits and the screen, divided by the distance between the slits. This is also written in words below.


A representation with the words single slit interference on the top. There is an equation that shows that the width of the slit multiplied by the sine of the diffraction angle of the pth order dark fringe is equal to the product of the order of the dark fringe and the wavelength. This is also written in words below.


A representation with the words single slit interference on the top. There is an equation that shows that the diffraction angle of the pth order dark fringe is approximately equal to the order of the dark fringe multiplied by the wavelength divided by the slit width. This is also written in words below.


A representation with the words single slit interference on the top. There is an equation that shows that the distance to the dark fringe from the central maxima for the pth order dark fringe is equal to the order of the dark fringe multiplied by the wavelength and distance between the slit and screen, divided by the slit width. This is also written in words below


A representation with the words single slit central maxima width on the top. There is an equation that shows that the width of the central maxima is equal to twice the wavelength multiplied by the distance between the slit and screen, divided by the slit width. This is also written in words below.

Graphical

Graphical Representation describes the situation through use of plots and graphs.

 

Descriptive

Descriptive Representation describes the physical phenomena with words and annotations.

 

Experimental

Experimental Representation examines a physical phenomena through observations and data measurement.

Check out this easy to do at home experiment to observe Young's double slit experiment.

https://www.youtube.com/watch?v=kKdaRJ3vAmA

Practice

Use the practice problem sets below to strengthen your knowledge of this topic.

Fundamental examples

 

(1) A red laser with wavelength $\lambda = 700$ nm shined on a double-slit with width $d=10$ mm. The screen is located $D= 2 $ m away from the slit. How far up from the central maximum is the $m=2$ bright fringe?

(2) How does the position of the 2nd bright fringe in example 1 change if the wavelength used is $\lambda_{new} = 350$ nm instead of $\lambda_{old} = 700 $ nm?

(3) In a single-slit experiment, the width of the central maximum is $1.5 $ mm. The screen is 1 meter from the slit and the wavelength of light used in this experiment is $360 $ nm. (a) What is the size of the slit? (b) How much wider would the central maximum be if the wavelength was tripled?

(4) In a single-slit experiment, the second dark fringe is located 7.5 cm from the center of the central maximum on a screen located $L = 3 m$ away from the slit. What is the distance between the second dark fringe and the third dark fringe that is located on the other side of the central maximum? The wavelength of the illuminating light is $\lambda = 600 $ nm.

Solutions found HERE

Short foundation building questions, often used as clicker questions, can be found in the clicker questions repository for this subject.

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Practice Problems

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Conceptual Problems

BoxSand's multiple select problems

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For additional practice problems and worked examples, visit the link below. If you've found example problems that you've used please help us out and submit them to the student contributed content section.

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