7.3.2: Explorations

Exploration 1: Lens and a Changing Index of Refraction

Light rays from a point source, initially in air, are shown incident on a lens. Restart.

1. How, if at all, would the path of the rays change if the source and lens were placed in another medium with an index of refraction of \(n = 1.2\), which is less than the index of refraction of the lens? Make a prediction and then test your prediction by using the slider to increase the index of refraction of the surrounding medium.
2. Now, what would happen if the index is increased to \(n = 2.0\), so that it exceeds the index of refraction for the lens? Make a prediction and then check your prediction using the slider.

Exploration authored by Melissa Dancy and Wolfgang Christian.

Exploration 2: Snell's law and Total Internal Reflection

Light rays from a beam source, initially in air (\(n = 1\)), are shown incident on material with an index of refraction that you can vary by moving the slider (position is given in meters and angle is given in degrees). You can move the beam source and change the angle of the light from the source by clicking on the beam and click-dragging the hotspot. Restart.

1. Verify that Snell's law holds. Measure the incident angle and refracted angle. You can use the pink protractor to measure angles. You can drag the protractor around and click-drag to adjust the angle. Calculate the value of the index of refraction of the material. Theoretically, what is the maximum angle of incidence (the animation limits the angle of incidence to \(45^{\circ}\), but that is not the maximum)? Given the maximum incidence angle, what is the maximum angle of refraction? This angle is sometimes called the critical angle. Develop a general expression for the critical angle as a function of the indices of refraction of the two materials.
2. Move the light source inside the material and change the beam so it leaves the blue material and goes into the air (black). Measure the angles of incidence and refraction and calculate the index of refraction of the material. What happens if the angle of incidence (from inside the material) is greater than the critical angle of refraction found in (a) above? Why? This is called total internal reflection.
3. Change the index of refraction. Calculate the new critical angle. Measure the critical angle and compare it with your calculated value.
4. Why is it only possible to have total internal reflection when light travels from a medium of higher index of refraction to one of lower index of refraction?

Exploration authored by Anne J. Cox.

Exploration 3: Toward Building a Lens

Light rays from a beam source, initially in air, are incident on a material of different index of refraction (position is given in meters). You can change the curvature of the surface of the material as well as the index of refraction. Restart.

1. Move the slider to decrease the curvature of the blue material. What happens when the edge is curved more (the radius gets smaller)? When the curvature is \(1\), where is the point at which all the rays converge (a focal point)?
2. Increase the index of refraction. When the curvature is \(1\), where is the point at which all the rays converge? If the index of refraction is \(1\), what happens? Why?
3. Mathematically, the relationship between this focal point inside the curved material, the curvature of the surface, and the radius of curvature of the surface is given by \(f = nR/(n - 1)\). Verify this expression with the animation.

How a surface focuses light, then, depends both on the index of refraction as well as on the curvature of the material.

Exploration authored by Anne J. Cox.

Exploration 4: Fermat's Principle and Snell's Law

This animation demonstrates Fermat's principle: Light travels along the path that takes the shortest time. You can click-drag the source (white dot) and the end-points (reflected light, blue, and refracted light, green). The animation will show you the possible paths for the light to take. The white path is the path that takes the shortest time. You can also click on the words at the interface ("air/water") to switch between a light source in air or one in water. Notice that for the reflected light, the angle of incidence equals the angle of reflection. Once the path is completed, you can click in the animation to show the angles for the angles of incidence and refraction for the refracted light.

1. Verify that the angles obey Snell's law.
2. Click on the upper left-hand corner (on the words "Possible paths") to switch to "Real paths." (If you click on the words again, it will switch back to the "Possible paths.") What does the animation show in this mode, and how is it different from the "Possible paths" mode?

(Calculus required): Using the diagram below (and the hints that follow), prove that you can derive Snell's law using Fermat's principle.

Figure \(\PageIndex{1}\)

3. Since the time for the light to travel through the two media (along any arbitrary path) is \(t = s_{1}/v_{1}+s_{2}/v_{2}\), show that you can rewrite the time as
   \[t=\frac{\sqrt{a^{2}+x^{2}}}{v_{1}}+\frac{\sqrt{b^{2}+(d-x)^{2}}}{v_{2}}\nonumber\]
4. To find the path that minimizes the travel time between the two points, solve for \(dt/dx = 0\). Why?
5. When you solve for \(dt/dx = 0\), show that you get
   \[\frac{x}{v_{1}\sqrt{a^{2}+x^{2}}}=\frac{d-x}{v_{2}\sqrt{b^{2}+(d-x)^{2}}}\nonumber\]
6. Show (make the necessary substitutions) that this is the same as Snell's law, \(n_{1}\sin\theta_{1} = n_{2}\sin\theta_{2}\).

Exploration authored by Anne J. Cox.
Applet authored by Fu-Kwan Hwang.

Exploration 5: Index of Refraction and Wavelength

Light rays from a beam source, initially in air, are shown incident on a sphere of water. You can change the wavelength of light by moving the slider. Restart.

1. Move the slider to change the color of the light (by changing the frequency). As you move the slider to the right, does the frequency of light increase or decrease (look up the frequencies of different colors in your book if you need to)?
2. Where does the red light converge? Where does the blue light converge?
3. If the index of refraction of the circular blue region was 1, where would the point of convergence be? Therefore, explain why a higher index of refraction means a convergence point closer to the sphere.
4. For which color light, then, is the index of refraction higher? For which color is it lower?

Exploration authored by Anne J. Cox.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.