Go to the phet.colorado.edu site and open the "Bending Light" simulation, all of this lab will use the "Prisms" section. The light source can be moved and rotated as needed and can output a monochromatic beam (laser, the color can be selected), a quintuple-beam, or a white-light beam. To bend the light, we have a selection of prism shapes, We will call these Triangle, Trapezoid, Square, Circle, Convex, and Concave respectively. Note the handles which allow them to be rotated. We can change the composition (index of refraction) of the prisms as well as the surrounding medium ("Environment").

Starting with all the default settings, turn on the laser and drag a triangle prism into the beam's path. Rotate and move the prism to see the various possibilities. We have "Reflections" turned off but you will see reflections when the beam is incident at greater than the critical angle - where we have total internal reflection and no refracted ray. Turn on Reflections to get a better sense of what's "really" going on but then turn off again for simplicity. Q1) What happens if you reduce the index of refraction of the prism down to 1 (Air)? Change the light source to the white beam and make sure the prism is set back to "Glass". For some positions/angles of the prism you should see the outgoing beam spread into a narrow rainbow. This is due to dispersion; different colors of light experience a different index of refraction for the prism and bend different amounts. You could also go back to the laser beam and vary the color; you should see the outgoing ray direction change slightly with changing color. Q2) Which end of the spectrum, red or blue, is experiencing the greater index of refraction? Explain your reasoning. The triangle prism available in the simulation is equilateral, each corner angle is 60. Maybe more common are 45-45-90 prisms, some of their uses are illustrated in the figure below. These all rely on the fact that for a ray within the glass prism, a 45 incident angle is above the critical angle and all the light is reflected. In (a) the beam of light is turned exactly 90, in (b) the light makes a 180 U-turn, and in (c) the beam emerges in the same direction but displaced. Most binoculars have a pair of prisms being used in the (b)-fashion which compactly correct for what would otherwise be an upside-down image from the lenses.

Put away the triangle and pull out a square, rotate the square into a diamond shape and have the light beam hit near the center. You should see a beam refracting twice and emerging parallel to the original direction. By rotating the square, you can vary the amount of deflection. Q3) Does varying the index of refraction of the prism change the amount of deviation (offset) of the beam? Q4) Does varying the index of refraction of the environment change the deviation? Set the prisms and environment back to glass and air respectively. Put away the square and pull out a convex. Change the laser to the quintuple-beam and rotate the convex so the flat side faces the laser. Light coming from a single point on an extremely distant object will be moving along parallel paths when it reaches our eyes or an object like the convex prism. We say such light is coming from an object at infinity (or at least the light is acting that way). On screen, you should see that the light was not bent when passing through the flat side (Snell's Law with incident angle 0). Because the second surface is curved, the incident angles are non-zero and we do have bent rays (temporarily turning on the "Normal" option helps illustrate the varying incident angles of the rays arriving at the second surface). The rays were bent so that they all crossed at a common point, we call this the "image" location. Someone looking at those rays of light from beyond where they crossed perceives them as coming from that image location. That's what we mean by an image, the place the original object appears to be after passing through the optical element (convex prism in this case). This is a "real" image because the light rays really are moving through and coming from the image location. When you look at yourself in a flat mirror you see your image on the other side or behind the mirror; that is a "virtual" image because none of the light rays really came from that location behind the mirror. For the special case of an object at infinity (parallel rays of light), we call the image location the "focal point". The distance from the optical element to the

focal point is called the "focal length". The focal length of an optical instrument indicates how strongly it bends light; lower focal lengths indicate the light was bent more. In general, object distances and image distances can vary although the optical instrument's focal length will not. Only for the special case of an object at infinity does the image distance equal the focal length. Our convex prism can also be called a "lens". Because the rays of light were bent towards each other, the convex prism is a "converging lens". All normal converging lenses are thicker in the middle and thinner at top and bottom. The curvatures of the sides of the lens combined with the index of refraction of the lens material determine its focal length in accordance with the "Lens-maker's equation". Because the first surface of our lens was flat, no bending of light occurred there, all the bending was at the second surface. So, we could also describe this optical instrument as just being a "refracting surface". We'll practice with equations we have for both refracting surfaces and lenses and show that they give consistent results. Equation (23.7) of our textbook is for refracting surfaces: n1/p + n2/q = (n2 - n1)/R Where n1 = index of refraction of incident medium n2 = index of refraction of outgoing medium p = object distance q= image distance R = radius of curvature of the refracting surface Plug in numbers: infinity for p, 1.5 for n1, 1 for n2, and -1.5 cm for R. Explanations: infinity because the rays approaching the refracting surface are parallel, 1.5 and then 1 because at the refracting (curved) surface the rays are going from glass into air, 1.5 cm because on my computer screen I measured a diameter of 3 cm for the curved surface, and negative because that's the proper sign convention for refracting surfaces. Solve for q, because we had an "object at

infinity" this is also the focal length. Show your work (do I have to keep telling you that?). Q5) Visually, given that we are calling the diameter of the lens 3 cm, does your calculated q value match the distance you see of refracting surface to image location? The lens-maker's equation (textbook equation 23.12) is: 1/f = (n - 1) (1/R1 - 1/R2) Put in values of 1.5 for n (the index of refraction of the lens material), infinity for R1 (an infinite radius of curvature is a flat surface), and -1.5 cm for R2 (the radius of curvature of the second surface with the proper sign convention). Solve for the focal length f, you should get the same answer as for the previous part. Q6) If you turn around the convex prism (so that the light goes through the curved side first), does it still behave like a converging lens? Does it still have the same focal length? Explain/discuss. Swap the convex prism for the concave prism. With the quintuple-beam passing through the center, you should see the emerging beams spreading apart. To someone viewing those rays, they seem to be coming from a nearby point behind the prism and not from infinity. This is a virtual image because the light rays did not really pass through the image location. In equations, virtual images are assigned negative image distances. A lens that spreads rays apart, like this one does, is called a "diverging lens". Diverging lenses are thinner in the middle and thicker at the edges. The image location for parallel incident rays is still called the focal point, for diverging devices we assign them negative focal lengths. Q7) A lens can be used to start a fire by concentrating sunlight rays to a single point. Can this be done with a converging lens? A diverging lens? Both? Neither? Put away the concave and take out a circle. Have the quintuple-beam pass through the center of the circle, you'll see that the circle acts as a converging lens. The lens-maker's equation from earlier would not work well for this lens because the equation is designed for "thin" lenses and the separation distance between the refracting surfaces in this case does not qualify as thin.

Q8) Change the lens to air and the environment to water. Is the circle still a converging lens? Return the simulation to glass and air. Move the beams up so that they pass through the top portion of the circle like shown here: You'll see that the rays no longer cross at a single well-defined image location. This is called "spherical aberration". Although the sphere (or circle) shape does fine for rays passing through the center (known as "paraxial" rays), it will give blurry images for "non-paraxial" rays (the spherical aberration). To have all rays focused to the same image location requires a parabolic shape rather than spherical. Change the prism material to water and turn Reflections on. Some of the rays refract into this water drop (like the picture above), then reflect off the right edge of the drop (remaining in the water), and then refract out of the drop near the bottom edge. In fact, there's a bit of a concentration of rays leaving the drop down there. If you change the beam color from red to blue, you'll see a similar concentration of "reflected" light at a very slightly different angle. If you change the light to white, you'll see the outgoing light down there is spread into a narrow rainbow. This is caused by dispersion - the water has a different index of refraction for different colors, so those enhanced reflections end up at slightly different angles. Here's are illustrations (from the Wikipedia page) showing all this in case you got lost.

The increased concentration is right there above the triangle. You'll find additional pictures and explanation in section 22.5 of the textbook. Q9) Rainbows are the result of dispersion. The light going into and back out of the water drops refracts twice and reflects once. How much of the dispersion happens at the refractions relative to the reflection? Hint: The law of reflection does not include any index of refraction terms.