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Andrea Houck

Geometric optics: understanding refraction of light and Snell’s law

Andrea Houck, Lawrence Technological University

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Understanding Refraction of Light
Determining the path of a light ray through different media to find the apparent position of an object
Author: Andrea Houck and Bhubanjyoti Bhattacharya

This essay discusses light traveling from one point in one medium to another point in another medium. We show first, how to find the potential light ray paths, then how to determine the allowed path by calculating the Euclidean optical path and finding the extremum with a single surface of interface. We verify the path follows Snell’s Law. We then create a final figure that shows the light path as the index of refraction of both media 1 and 2, the x and y coordinates of both points A and B, and the location of the surface of interface are manipulated. We include an application of this process as an observer viewing a fish in a pool of water from above. Following the same process, we add a second ray of light. We allow for manipulable input for the observer’s eye aperture in addition to the manipulable position. From this and the fixed actual position of the fish, we determine the fish’s apparent position. This situation is analyzed using single-surface refraction techniques since light from the fish passes through two media - water and air. We then expand our methods to investigate refraction with two surfaces. We follow a similar approach as with single-surface refraction where we first show how to find the potential light ray paths. We modify our solution method for the allowed Euclidean optical path and verify the solution by checking that Snell’s Law holds for all medium changes. Again, we display a manipulable final figure where we can adjust the indices of refraction for each media, the thickness of the second medium, and the x and y coordinates of point A. We include a second application of refraction with a fish. In this application, the fish is in a glass tank of water. We use the same refraction techniques described when looking at refraction with two surfaces and in the fish with a single surface of interface to find the path of the light ray and show the apparent position of the fish in the tank.

Running the code to see the output

The code in this notebook can be ran by clicking the "Make Your Own Copy" button on the top right of the screen in the Wolfram menu. Click the outermost bracket on the right hand side that encompasses all the blocks of code and use the keyboard to hold down the "Shift" key and at the same time press the "Enter" key. This will evaluate all the code in the notebook.

Single-surface refraction

In this computational essay, we will discuss a principle that will let us determine how a ray of light can travel from a point in one medium to another point in another medium. We include manipulable graphics to aid in understanding. The different manipulable figures can then be interacted with by adjusting the values in the sliders alongside the output figure.

The computational essay format is beneficial in helping students better understand their subject matter. The text explains the concepts that are then expressed as code, which then generates output that creates a visual representation of the ideas described in the text [1]. Manipulating the figures to see how the variables in the problem relate to one another makes the concept much more clear. This process of using computational thinking to solve a problem can also be extended beyond the current problem by modifying the code and using it to investigate other aspects that may be either time consuming or impossible to do by hand.

To begin, we determine point A in medium 1 to be the object and point B in medium 2 to be the observer. We will fix the x and y values of both points as well as the y-value of the flat surface of interface between the two media.

Defining points A, B, and surface of interface

Consider point A in medium 1 (refractive index

) to be at

(

)

. We consider point A to be the object.

In[]:=

n1=1.5;x1=-10;y1=-10;

Consider point B in medium 2 (refractive index

) to be at

(

)

. We consider point B to be the observer.

In[]:=

n2=1.3;x2=10;y2=10;

Finally, consider the surface of interface between medium 1 and medium 2 to be at

In[]:=

y=0;

Creating a canvas with possible rays

Below, we will create a canvas that shows the possible paths of the rays of light using the function Graphics. For the options on Graphics we will use Frame (True), GridLines (Automatic), and provide directive for GridLinesStyle to produce dashed gray lines. We will also join the points A and B using a red line, which we should be able to manipulate depending on the location of the point of refraction given by its

-coordinate. We label these points and their respective mediums using the primitive Text and specifying the point at which the text should appear. A dashed normal line will be shown at the point of refraction. We include rotated text to label the normal line by using the primitive Text and the option Rotate, specifying a 90° (or

radian) rotation. We will manipulate the value of

to trace out different rays. Finally, we fix PlotRange to include the two points A and B. We use the function Labeled around the entire Manipulate function to add a caption beneath the figure.

In[]:=

DynamicModule{},LabeledManipulateGraphicsDisk[#,0.2]&/@{{x1,y1},{x2,y2}},{Thick,Line[{{x1,y},{x2,y}}]},{Red,Dashed,Line[{{{x1,y1},{x,y},{x2,y2}},{{x,y1},{x,y2}}}]},{Text["Medium 1",{8,y1+1}]},{Text["Medium 2",{-8,y2-1}]},Red,RotateText["Normal",{x-0.5,4}],

,{Text["A",{x1+0.5,y1}]},{Text["B",{x2-0.5,y2}]},FrameTrue,GridLinesAutomatic,GridLinesStyleDirective[Gray,Dashed],PlotRange{{x2+1,x1-1},{y2+1,y1-1}},{x,x1,x2},"Figure 1: Potential light ray paths. The x-intercept can be adjusted to trace out the possible paths for a light ray to take between points A and B. The red line shows these potential paths.",SaveDefinitionsTrue

Out[]=

Figure 1: Potential light ray paths. The x-intercept can be adjusted to trace out the possible paths for a light ray to take between points A and B. The red line shows these potential paths.

We see by manipulating the figure above, the point at which the light ray intersects the surface of interface affects how much the ray bends. When the ray intersects the surface of interface close to B or close to A, the ray bends much more than when the ray intersects the surface of interface closer to the midpoint between A and B. When the ray goes through the midpoint between these two points, the ray does not bend at all. However, light cannot follow all of these paths, as certain principles must be followed.

Solving for the allowed ray of light

The above figure does not fully represent how a ray of light travels between points A and B because there is only one path a ray of light can take between two points. In order to complete this picture, we need to obtain a specific value for

. This can be determined in a couple of ways. First, we know that the path of a ray of light is extremal. This means that the optical distance from point A to point B will become an extremal path for the correct value of

Let us begin by calculating the Euclidean optical path between points A and B.

To find the distance light travels in vacuum, we multiply the speed of light c by the amount of time that passes Δt.

distance=cΔt

(

)

We can write the speed of light in terms of it's refractive index with the following equation

(

)

where n is the refractive index, c is the speed of light in vacuum (in units of

), and v is the speed of light in the medium (in units of

). We can re-write equation (2) as

(

)

We then can plug v into equation (1) in place of c to find the speed of light in a medium instead of in vacuum.

distance=

Δt

(

)

Because we know that c Δt is equal to the distance from equation (1), we can re-write this equation again as

distance=

Δx

(

)

Multiplying by n, we see that Δx, or the location of x, is equal to n multiplied by the distance.

Δx=ndistance

(

)

where distance is the Euclidean distance. We know from geometry the formula for Euclidean distance to be

EuclideanDist=

(

-x)

(

-y)

(

)

Therefore, to find the Euclidean optical path between points A and B, we will multiply the Euclidean distance in each medium with the refractive index of the medium.

In[]:=

dist[x_]:=n1

(x1-x)

(y1-y)

+n2

(x2-x)

(y2-y)

;

The first derivative test from calculus tells us where we find the extremum. We can use this to determine a condition for the extremum, and set it to zero to find

. Instead of performing this calculation by hand because computers do these methodical calculations much quicker, we use code to calculate the derivative with respect to x. To do this in code, we must input the command in a way the Wolfram Language can understand. The function D takes the derivative of the function, in this case dist[x], with respect to the specified variable, x. We use the double equal sign to signify that the derivative equals zero. We then set this quantity equal to condition to make it easier to use in future steps.

In[]:=

condition=D[dist[x],x]0

Out[]=

1.5(-10-x)

100+

(-10-x)

1.3(10-x)

100+

(10-x)

0

We can solve the above equation represented by condition for

by using the Solve function.

In[]:=

xval=x/.Solve[condition,x][[1]]

Out[]=

-1.40752

Let us now check if this value of

is correct, by checking if Snell’s law is satisfied! This will entail constructing sines of the angles in the two media. θ1 is the angle of incidence in medium 1 and θ2 is the angle of refraction in medium 2.

In[]:=

sθ1=

x-x1

(x1-x)

(y1-y)

;sθ2=

x2-x

(x2-x)

(y2-y)

;

From Snell’s Law, we know that

sin(θ1)=

sin(θ2)

(

)

Therefore, we know that Snell’s law is satisfied if equation (9) is true.

sin(θ1)-

sin(θ2)=0

(

)

We verify this with the following code.

In[]:=

(n1sθ1-n2sθ2)/.{xxval}

Out[]=

Since we have determined the value of

, we may now modify our plot to add the correct light ray path in solid blue (normal in dashed blue).

In[]:=

DynamicModule{},LabeledManipulateGraphicsDisk[#,0.2]&/@{{x1,y1},{x2,y2}},{Thick,Line[{{x1,y},{x2,y}}]},{Red,Dashed,Line[{{{x1,y1},{x,y},{x2,y2}},{{x,y1},{x,y2}}}]},Thick,Blue,Line[{{x1,y1},{xval,y},{x2,y2}}],Dashed,Line[{{xval,y1},{xval,y2}}],RotateText["Normal",{xval-0.5,4}],

,{Text["Medium 1",{8,y1+1}]},{Text["Medium 2",{-8,y2-1}]},Red,RotateText["Normal",{x-0.5,4}],

,{Text["A",{x1+0.5,y1}]},{Text["B",{x2-0.5,y2}]},FrameTrue,GridLinesAutomatic,GridLinesStyleDirective[Gray,Dashed],PlotRange{{x2+1,x1-1},{y2+1,y1-1}},{x,x1,x2},"Figure 2: Possible and allowed light ray paths. The red line shows the possibilities for the paths that the ray of light might take. Not all of these paths are possible because they do not all satisfy Snell's Law. The blue line shows the actual path the light ray takes based on the principle that light travels along the path that takes the least amount of time. The blue path satisfies Snell's Law.",SaveDefinitionsTrue

Out[]=

Figure 2: Possible and allowed light ray paths. The red line shows the possibilities for the paths that the ray of light might take. Not all of these paths are possible because they do not all satisfy Snell's Law. The blue line shows the actual path the light ray takes based on the principle that light travels along the path that takes the least amount of time. The blue path satisfies Snell's Law.

Because the difference in indices of refraction is modest, the blue line showing the allowed path of the light ray intersects the surface of interface close to the midpoint. The index of refraction of Medium 2 is less than that of Medium 1, which causes the angle of refraction to be smaller than the angle of incidence and the light to bend towards the normal line. Therefore, we see that the ray of light intersects the surface of interface further from B than A. The larger the difference in the indices of refraction between the two media, the more the light ray will bend and the closer the intersection point on the surface of interface will be to either point A (if the index of refraction of Medium 1 is greater than that of Medium 2) or to point B (if the index of refraction of Medium 2 is greater than that of Medium 1).

Creating the final figure

Now that we know how to get the correct ray of light using computation, let us try to put everything together into a single line of code. We do not need to Manipulate the value of

anymore, but we need to solve for its value before we plot it. This is best done by using a Module. We will produce a function of all the input variables which we should now be able to play with/Manipulate.

In[]:=

fig[med1_List,med2_List]:=Module{n1,n2,x1,x2,y1,y2,dist,x,xval},{n1,x1,y1}=med1;{n2,x2,y2}=med2;dist[x_]:=n1

(x1-x)

(y1-y)

+n2

(x2-x)

(y2-y)

;xval=x/.Solve[D[dist[x],x]0,x][[1]];GraphicsThick,Line[{{{-10,y},{10,y}}}],{Blue,Line[{{x1,y1},{xval,y},{x2,y2}}]},Dashed,Blue,Line[{{xval,-10},{xval,10}}],RotateText["Normal",{xval-0.5,4}],

,{Text["Medium 1",{8,y1+1}]},{Text["Medium 2",{-8,y2-1}]},{Text["A",{x1+0.6,y1+0.2}]},{Text["B",{x2-0.3,y2-0.6}]},PlotRange{{-12,12},{-12,12}};Labeled[fig[{n1,x1,y1},{n2,x2,y2}],"Figure 3: Light ray between points A and B. The solid blue line shows the path the light ray takes when going between points A and B. The dashed blue line is the normal to the surface of interface at the point the light ray intersects the surface."]

Out[]=

Figure 3: Light ray between points A and B. The solid blue line shows the path the light ray takes when going between points A and B. The dashed blue line is the normal to the surface of interface at the point the light ray intersects the surface.

Manipulating the final figure

Let us now manipulate the final figure by changing the input values. nA is the refractive index of medium 1 and nB is the refractive index of medium 2. (xA, yA) and (xB, yB) are the Cartesian coordinates of the points in medium 1 and medium 2, respectively. The y-value of the surface of interface between the two mediums is yint. We can adjust the variable’s values by adjusting the slider next to the graphic that is output. This will show us how each variable affects the ray of light passing from one medium to another.

In[]:=

DynamicModule[{},Labeled[Manipulate[fig[{nA,xA,yA},{nB,xB,yB}],{{nA,1,Style["Medium 1 Index of Refraction",12]},1,2.5},{{nB,1,Style["Medium 2 Index of Refraction",12]},1,2.5},{{xA,0,Style["X-Position of Point A",12]},0,10},{{yA,7,Style["Y-Position of Point A",12]},y+0.1,10},{{xB,0,Style["X-Position of Point B",12]},-10,xA},{{yB,-5,Style["Y-Position of Point B",12]},-10,y-0.1}],"Figure 4: Light ray path given manipulable input. The path the light ray takes between points A and B changes depending on the input value for the indices of refraction of the two media, the x and y coordinates of the points A and B, and the location of the surface of interface."],SaveDefinitionsTrue]

Out[]=

Medium 1 Index of Refraction

Medium 2 Index of Refraction

X-Position of Point A

Y-Position of Point A

X-Position of Point B

Y-Position of Point B

Figure 4: Light ray path given manipulable input. The path the light ray takes between points A and B changes depending on the input value for the indices of refraction of the two media, the x and y coordinates of the points A and B, and the location of the surface of interface.

From this diagram, we can see that when points A and B have the same x-position, the light ray follows the path of the normal line and does not bend. When the index of refraction of Medium 1 is equal to the index of refraction of Medium 2, the angle of incidence (or the angle between point B and the normal line) is equal to the angle of refraction (or the angle between point A and the normal line). In this scenario, the light ray does not bend when passing through the surface of interface, but does not necessarily follow the path of the normal line.

We can also adjust the diagram to show how light bends when moving between mediums with different indices of refraction. When light moves from high index of refraction to low index of refraction, the ray bends away from the normal. From this, we also can see that going from low index of refraction to high index of refraction bends the light towards the normal.

Apparent fish in a pond

Now that we understand how ray diagrams work for refraction from a single surface, let us use this to figure out a way to find the apparent image of a fish in a pond for an observer. We will allow great freedom to the observer: variable height, position, and eye aperture. However, we will fix the fish position and the pool depth. Also, since this is a standard pool of water, the refractive indices of the media are also going to be fixed.

We acknowledge that the observer’s ability to view the fish from any position without adding an additional medium is not necessarily realistic. However, we assume that the fish can be viewed from any position, including directly above, for the purposes of showing the light ray’s path and finding the apparent position of the fish.

Position of the fish, the pool surface, and refractive indices

We will fix the values for our problem to those given below. Whenever a quantity can have units (such as

fish

, etc.) we will assume that the units are meters.

In[]:=

nair=1.;nwater=1.33;xfish=10.;yfish=-10.;ypool=0;

The first ray of light

Our variables for the first ray of light in this problem are

(x,y)

obs

. We will use the method laid out in the previous section to find

pool

, which is the x-value at which the light ray intersects the surface of interface between the air and water. Then, we plot the ray of light. We also extend the ray of light from air into water using a dashed line to show the apparent ray under water. Finally, we will manipulate

(x,y)

obs

using the sliders for Observer X-Position and Observer Height.

In[]:=

ray[xobs_,yobs_]:=Module{dist,x,xpool,slope},dist[x_]:=nair(

(xobs-x)

(yobs-y)

)+nwater(

(xfish-x)

(yfish-y)

);xpool=x/.Solve[D[dist[x],x]0,x][[1]];slope=

yobs-ypool

xobs-xpool

;Graphics[{Line[{{{-10,ypool},{10,ypool}}}],{{Blue,Line[{{{xfish,yfish},{xpool,ypool},{xobs,yobs}}}],Dashed,Line[{{xpool,ypool},{xfish,ypool+slope(xfish-xpool)}}]},{Text["Water",{-9,-9}]},{Text["Air",{-9,9}]}}},PlotRange{{-10,10},{-10,10}}];

In[]:=

DynamicModule[{},Labeled[Manipulate[ray[xx,yy],{{xx,0,Style["Observer X-Position",12]},-10,10},{{yy,2,Style["Observer Height",12]},1,10}],"Figure 5: Observer viewing fish in a pool of water from above. The path the light ray takes from the fish to the observer is shown with the solid blue line. The dashed blue line shows the apparent path the light ray takes since it appears light travels in a straight line. The x-position of the observer and the observer's height can be adjusted. This will change the path the light ray takes by changing the point at which it intersects the surface of interface and will also change the apparent position line."],SaveDefinitionsTrue]

Out[]=

Observer X-Position

Observer Height

Figure 5: Observer viewing fish in a pool of water from above. The path the light ray takes from the fish to the observer is shown with the solid blue line. The dashed blue line shows the apparent path the light ray takes since it appears light travels in a straight line. The x-position of the observer and the observer's height can be adjusted. This will change the path the light ray takes by changing the point at which it intersects the surface of interface and will also change the apparent position line.

The dashed apparent ray under water is an extension of the straight line from the observer to the point the surface of interface is intersected, as seen in the figure above. This line is straight because the human eye perceives light to travel in a straight line. This causes the actual position of an object, such as a fish, to be different from the apparent position of the same object. The values of the indices of refraction of the media determine if the apparent position of the object is closer to, as in this case, or farther from the observer. Because light bends toward the normal line when going from low index of refraction to high index of refraction, the apparent position of the fish is closer to the surface than the actual position of the fish.

The second ray of light

We will need a second ray of light to pin point the location of the apparent fish. To do this, we will simply allow the

-coordinate of the observer’s eyes to have a margin given by

eye

. This value can be manipulated by adjusting the Eye Aperture slider. The

-coordinate of the second ray will still be the same as the first ray.

In[]:=

DynamicModule[{},Labeled[Manipulate[Show[ray[xx,yy],ray[xx,yy+yeye]],{{xx,-5,Style["Observer X-Position",12]},-10,10},{{yy,5,Style["Observer Height",12]},1,10},{{yeye,1,Style["Eye Aperture",12]},0,2}],"Figure 6: Using 2 rays of light to determine the apparent positon of the fish for the observer above a pool of water. The actual path of the light rays are shown with solid blue lines. The dashed lines represent the apparent position since light appears to travel in a straight path. The point of intersection of the two dashed blue lines is the apparent position of the fish. The eye aperture and observer positon can be adjusted to show how the apparent position of the fish changes with these variables, even though the actual position of the fish remains fixed."],SaveDefinitionsTrue]

Out[]=

Observer X-Position

Observer Height

Eye Aperture

Figure 6: Using 2 rays of light to determine the apparent positon of the fish for the observer above a pool of water. The actual path of the light rays are shown with solid blue lines. The dashed lines represent the apparent position since light appears to travel in a straight path. The point of intersection of the two dashed blue lines is the apparent position of the fish. The eye aperture a