Check out the trippy effect when you put a diffraction grating over white lights! Learn how this effect happens!

https://www.youtube.com/watch?v=mr8A3qNHYDc

Pre-lecture Study Resources

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

BoxSand Videos

Learning Objectives

Summary

Summary

Atomistic Goals

Students will be able to...

  1.  

BoxSand Introduction

Wave Optics  |  Multi and Single-Slit Interfernce

Multi-Slit Interference

A preferred experiment to measure the wavelength of light is the multi-slit experiment. Instead of only two openings, you have multiple slits, which has the effect of sharpening the constructive interference peaks. The multi-slit apparatus is called a diffraction grating when light passes through it and a reflection grating when light reflects off of it. Below is a diagram for a diffraction grating.

This is an image of a wall with multiple slits of equal vertical distance between each other denoted as d. There are rays of light that enter the slit parallel to one another and are shifted downward at some angle theta from the horizontal. There are dashed perpendicular lines that go up from the point of entry of the light rays and the distance between the dashed lines are labeled as delta l where delta l is equal to d sine theta.

Here the first slit interferes with the second and the second with the third, and so on. Each subsequent slit has the same PLD condition as for a double slit. It is a bunch of double slits slightly shifted from each other. The result is a sharper set of interfrence fringes and the same conditions and equations as for the double slit.

Single Slit Interference

Once you have a wave picture of light and understand superposition, the inteference patterns of the double slit and muti-slit setups are not wierd. But what's truely a mind bender is when passing light through a single slit you also observe an interference pattern, although one different from the double slit. To understand this effect you must consider Huygens Principle which states that each point on a wavefront can act as a new spherical source (left figure).

This is an image of a plane wave where there multiple sources that are all lined up vertically which is labeled as the initial wave front. As the source emits light or sound, it moves outwards in a spherical shape where the radius of the circle are all equal. At the edge of the circle, the wave front at a later time is tangent to all the wavelets.            This is an image of a plane wave where there are multiple sources all lined up vertically and are connected to a wall on either side. The distance between the wall and the center source is labeled as a divided by two and each source emits some light at some angle to the horizontal in parallel lines. There are six lines total that are labeled from top to bottom as one, three, five, two, four, six. At the same angle theta from the vertical on source one, the point at which it meets line two and the distance between the source of two to that intersection is labeled as theta r of one and two.

Each new spherical source interfere to create the next wavefront, which can then be a set of new spherical sources. This may seem counter intuitive and we don't know if light really behaves this way, but if modeling it this way is consistent with observation, we can't throw it out as a possibility. In fact this model then helps us understand the single slit interference pattern. Now waves generated at the top of the source can interfere with waves on the bottom of the source (figure above on the right).     

The interference pattern is similar in that there are light and dark regions but it differs in the locations of the maxima and minima.

This is an image of a wall with a single slit that has a width of a. The distance between the slit and the second wall is labeled as L from the slit at some angle theta to the first band of no light is labeled as angle theta. On the other wall, there are bands of light with different intensities and bands of no light. The central band of light directly from the slit is the most intense and is called the central maximum. Then there are bands lower in intensity. The distance between the areas of no light is labeled as width w and the adjacent bands of light on either side of the central maximum is labeled as p equals one and the next lower intensity band of light labeled as p equals two. The distance between the central maximum and the area of dark band is labeled as y. On the right shows several equations. The slit width a multiplied by the sine of theta p is equal to p lambda. Theta p is approximately equal to p lambda divided by a with the assumption that the distance between the slit and the screen is much larger than the slit width. Y of p is equal to p lambda multiplied by L all divided by a. W is equal to two lambda multiplied by L all divided by a.

The interference is also often discribed by the locations of the destructive interference points. Here the integer $p$ is describing the order of the dark fringes. The width of the cental maximum ($w$) can be found by doubling $y_1$.

To see the difference between the single and the double slit interference patterns, along with the effects of sharpening the fringes by adding more slits, refer to the figure below.

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.


Single-Slit Interference Mathematical Model

To explain why the mathematical model for single slit interference uses destructive interference fringes, unlike either of the double or multi-slit models, we need to consider several special cases illustrated below.

We will split the single slit into successively more equal sections. The first of which we will consider is the slit as a whole. We see that light incident on the slit can simply shine directly through the slit onto a screen opposite the slit. Since these rays are all parallel with no path length difference between them, they will interfere constructively and we will find a central bright spot on the screen directly opposite the slit.

This is the first image in a series of single slit interference images. This image shows rays of light shining directly through the open slit. The angle of diffraction is zero.

To find the angle at which the first destructive interference, dark, fringe occurs, we will divide the single slit of width a into equal sections of width a/2. Additionally, we will consider waves which have been diffracted up at a slight angle. We will examine three waves, one from the top edge of the slit (labeled with a number 1), another from the middle of the slit (labeled with a number 2), and a final wave from the bottom edge of the slit (labeled with a number 3). We will choose an angle such that there is a path length difference of one half of a wavelength between the top and middle waves. We can then examine the triangle made near the start of waves 1 and 2. The sine of the shown angle is the opposite side (1/2 lambda) divided by the hypotenuse (a/2). Simplifying this, we find the equation given between the images. The "1" before the lambda is emphasized because this will eventually be our variable labeled "p". This is the p = 1 dark fringe, which is the first fringe on either side of the central bright spot.

This is the second set of images in a series of single slit interference images. This image shows rays of light shining through the open slit. This image shows parallel rays which are diffracted upwards at a slight angle. The slit is divided into two equal lengths. There are three rays, one emanating from the top of the slit, one from the middle, and one from the bottom. The middle ray travels an extra distance when compared with the top ray. This distance is equal to half a wavelength. A triangle is drawn from the start of the top ray, to the start of the middle ray, to the point on the middle ray that is even with the start of the top ray. The sine of the angle near the start of the top ray is equal to one half of a wavelength (the     This is an animated GIF which shows that a ray just underneath the top ray will have a path length difference of one half lambda with a ray just underneath the middle ray. This will result in destructive interference. For every ray between the top and middle rays, there is a corresponding ray between the middle and bottom rays with which it will interfere destructively.

To see that this gives a destructive interference fringe, we can look at the animated gif. For every wave between waves 1 and 2, there is a corresponding wave between 2 and 3 which starts a distance a/2 away, and has a path length difference of one half of a wavelength, thus causing destructive interference.

Next we look at what happens when we divide the single slit into three equal sections of width a/3 and perform the same procedure as above. The diffraction angle needs to increase in order to give the same path length difference of lambda / 2. We find that the resulting equation might represent a "p" value of 3/2. This doesn't fit the mathematical model presented above, so this is a hint that this situation doesn't quite work out!

This image shows the case where the single slit is divided into three equal sections. There are four waves starting at equal intervals between and including the top and bottom edges of the single slit. This image looks very similar to the case where we divided the single slit into two sections. The wave are diffracted up at a higher angle to allow the path length difference to remain one half of a wavelength. The rays are labeled 1, 2, 3, and 4 from top to bottom. Evaluating the sine of the angle gives the equation 3/2 lambda = a times the sine of theta.     This image shows an illustration of waves which interfere destructively when the single slit is divided into three equal parts. We see that waves from the top section each have a corresponding wave in the middle section with which they will interfere destructively. We also see that the bottom set of waves then do not have counterparts with which to interfere destructively. Some light thus gets through the slit, and we do not have a dark fringe at this angle.

To see that this case is a little funny, look at the image on the right, which shows that as in the first case, for every wave between waves 1 and 2 there is a wave between 2 and 3 which results in destructive interference. However, there is another whole section between waves 3 and 4 which do not have corresponding destructive interference. This means that some light does get through the slit at this angle, and we do not see a destructive interference at this angle. Note that in actuality it is not just the bottom set of waves that get through, there is actually partial destructive interference all around. But that image is a useful conceptual picture to show that there is not complete destructive interference happening as there would be for a dark fringe.

Finally, we look at what happens when we divide the single slit into four equal sections of width a/4. We see that for this case, we arrive at a nicer mathematical model with a "p" value of 2. We also see, in the image on the right, that there are now two sets of two regions which each destructively interfere with each other. This results in the p = 2 dark fringe.

This is the image which shows the case of the single slit divided into four sections. The angle of diffraction has increased yet again to maintain the path length difference of one half of a wavelength between rays 1 and 2. We see that the mathematical model derived from looking at the sine of the angle results in an equation of 2 times lambda = a times the sine of theta.    This image shows the case of the single slit divided into four equal sections. We can see that each wave between waves 1 and 2 has a corresponding wave between waves 2 and 3 which has a path length difference of one half of a wavelength, thus causing destructive interference. The same can be said for waves between waves 3 and 4 and their destructive counterparts between waves 4 and 5.

We can repeat this process to find that the p = 3 case will come from dividing the single slit into 6 even sections. From here we can generalize this process to describe the angle of dark fringes from a single slit. We find that the mathematical model is p lambda = a sin(theta), where p = 1, 2, 3, etc. Note that p starts at 1, where m started at 0 for the double and multi-slit mathematical model. The single slit mathematical model also finds the locations of the dark fringes.




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 27.4  |  Multiple Slit Diffraction

Openstax College Textbook Icon

OpenStax Section 27.5  |  Single Slit Diffraction

Openstax College Textbook Icon

 

Equations, definitions, and notation icon Concept Map Icon
Key Terms Icon Student Contributed Content Icon

Additional Study Resources

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

YouTube Videos

Doc Schuster covers wave interference,

https://www.youtube.com/watch?v=_N5Y9PVxHLY

Doc Shuster covers single slit diffraction,

https://www.youtube.com/watch?v=ps8IBv5_VeM

Some mathematics behind double slit diffraction,

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

Simulations


Phet - wave interference, 

Phet Interactive Simulations Icon

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

Simulation Icon

Demos


 

For additional demos involving this subject, visit the demo repository

Demos Icon

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=DfPeprQ7oGc

Other Resources


Resource Repository

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

Content Icon

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.

Clicker Questions Icon

 

Practice Problems

BoxSand Practice Problems

Conceptual Problems

BoxSand's multiple select problems

BoxSand's quantitative problems

Recommended example practice problems 


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.

Example Problems Icon