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Total Internal Reflection

When light is passing from one material to another it can be reflected back, trapping the light in the first material.

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Total Internal Reflection

Students will learn about total internal reflection. More specifically, they will learn what the critical angle is, how it is derived and how to solve for it in real life applications.

Key Equations

\begin{align*} \theta_C = sin^{-1}\frac{n_2}{n_1} \; \end{align*} ; where \begin{align*} \theta_C \end{align*} is the critical angle, n1 is the index of refraction of the material where the light emanates from and n2 is the index of the material outside.

Total internal reflection occurs when light goes from a slow (high index of refraction) medium to a fast (low index of refraction) medium. With total internal reflection, light refracts so much it actually refracts back into the first medium. This is how fiber optic cables work: no light leaves the wire.

Example 1

You have some unknown material and you would like to determine it's index of refraction. You find that you are able to create total internal reflection when the material submerged in water, but not when submerged in cooking oil. (a) Can you give a range for the index of refraction? (b) you are able to determine the critical angle in water to be 71.8 degrees; what is the index of refraction of this material?


(a): Since it is not possible to create total internal refraction when going from a material with a higher index of refraction to a lower index of refraction, we know that the index of refraction of this material must be between 1.33 (water) and 1.53 (cooking oil).

(b): We can use the equation given above to determine the index of refraction of the unknown material.

\begin{align*} \theta_c&=sin^{-1}(\frac{n_2}{n_1})\\ n_1&=\frac{n_2}{\sin(\theta_c)}\\ n_1&=\frac{1.33}{\sin(71.8)}\\ n_1&=1.40\\ \end{align*}

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  1. Imagine a thread of diamond wire immersed in water. Can such an object demonstrate total internal reflection? If so, what is the critical angle? Draw a picture along with your calculations.
  2. A light source sits in a tank of water, as shown.
    1. If one of the light rays coming from inside the tank of water hits the surface at \begin{align*}35.0^\circ\end{align*}, as measured from the normal to the surface, at what angle will it enter the air?
    2. Now suppose the incident angle in the water is \begin{align*}80^\circ\end{align*} as measured from the normal. What is the refracted angle? What problem arises?
    3. Find the critical angle for the water-air interface. This is the incident angle that corresponds to the largest possible refracted angle, \begin{align*}90^\circ\end{align*}.


  1. \begin{align*}33.3^\circ\end{align*}
  2. .
    1. \begin{align*}49.7^\circ\end{align*}
    2. No such angle
    3. \begin{align*}48.8^\circ\end{align*}

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