6.3.2: Refraction Simulation
- Page ID
- 26189
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)The ratio of the speed of light in a material to the speed in a vacuum (\(\text{c} = 3.0\times 10^{8}\text{ m/s}\)) is called the index of refraction; \(n=c/v\) where \(v\) is the speed of light in the medium. In this simulation we will investigate the effects of a change in the speed of a wave as it moves from one material to another. Although our example is for light, the same behavior can be demonstrated with other waves. For example, much of what we know about the interior of the earth, the sun and other planets comes from tracking earthquake waves when they refract as they pass through layers of material that have different densities.
The relationship between the index of refraction and the change in the direction angle of the a ray as it goes across a medium is given by Snell's law: \(n_{1}\sin\theta _{1}=n_{2}\sin\theta _{2}\) where \(n_{1}\) and \(\theta _{1}\) are values measured on side of the boundary and the subscript 2 is for the material on the other side of the boundary.
Simulation Questions:
- The simulation shows a single ray of light passing through a medium surrounded by air on both sides. You can move the protractor around and use it to measure angles by dragging the tip of the arrow. The source can also be moved. What can you say about light striking the medium perpendicular to its surface (\(\theta =0\))? What if the source is inside the medium; what happens to a perpendicular beam leaving the medium? Are these observations still true if the index of refraction is something other than \(1.00\)?
- Set the index of refraction to \(1.40\). Change the angle of the rays from the source. The angle of incidence is the angle the ray makes with a perpendicular to the surface (not the angle between the ray and the surface). Use the protractor to measure the angle of incidence (put the bottom left edge of the protractor where the ray enters the glass and line up the arrow with the ray). Move the protractor to the boundary and line up the arrow with the rays inside the material to get the refracted angle (the angle inside the medium).
- The index of refraction of air is approximately one (light travels about as fast in air as it does in a vacuum). Use Snell's law to calculate the index of refraction in the medium. (\(n_{1}=1;\:\theta _{1}=\) the incident angle; \(\theta _{2}\) is the refracted angle; you are solving for \(n_{2}\) ). NOTE: It is very difficult to get the protractor arrow to line up exactly with the rays so your answer may be off a little.
- For the same angle you used in the previous two questions, move the protractor once again to the right edge of the material and measure the angle the ray leaves the material. How does this compare with the incident angle on the left?
- Choose a different angle and use Snell's law to find the index of refraction outside the medium on the right. This time the incident angle will be inside the material at the right boundary, and \(n_{1}\) will be the index inside the material (your answer to question three).
- Drag the source inside the medium and try different angles. What do you notice?
- When light goes from a material where the index is higher to a lower index there are angles for which the rays cannot leave the material. The angle at which a ray starts to reflect off the boundary instead of passing through is called the angle of total internal reflection. You can see this if you look at the surface of the water in a pool from under water; at a certain angle you see a reflection of the bottom of the pool instead of objects above the water. Total internal reflection is also why light remains in a fiber optic cable instead of being absorbed by the coating. Find the angle for total internal reflection two ways. First experiment with the angle of incidence (source inside the medium) for a fixed index of refraction of \(1.40\). At what incident angle does the ray reflect? You can also find the angle using \(n_{1}\sin\theta _{1}=n_{2}\sin\theta _{2}\) where \(n_{1}\) is the index inside the medium and \(\theta _{2}=90^{\circ}\). Do your two values match?
- Verify the reflected beam obeys the law of reflection using the protractor (Note: You may have to slightly move the source to get the program to update the reflected ray position).