1. Light
Big Idea: Electromagnetic Radiation, which includes light, is a form of radiant energy possessing properties of both waves and zero-mass particles called photons. Photons vary in their energy, which causes them to vary in their frequencies and wavelengths as well. EM radiation can be bent (refracted) or reflected by certain materials, allowing us to manipulate how it travels and what kinds of images it produces.
Topic 7.1: Light basics and the EM spectrum
Topic 7.2: Refraction
Topic 7.3: Reflection and Polarization
Topic 7.4: Diffraction and Wave-Particle Duality

2. I can calculate the magnification and image distance of lenses.
Learning Goal: You will be able to determine how light bends as it travels through a transparent material.
Success Criteria: You will know you have met the learning goal when you can truthfully say:
I can use the index of refraction to determine how light changes directions as it travels through transparent materials.
I can calculate the magnification and image distance of lenses.
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3. Success Criteria 1: I can use the index of refraction to determine how light changes directions as it travels through transparent materials.
In the previous topic, we saw how convex mirrors can be used to focus light. This is what allows reflecting telescopes to work.
In this topic, we’ll see how the path or light can be bent by transparent materials, which is how glasses, contact lenses, your eyes, microscopes, and refracting telescopes are able to focus light or achieve magnification.
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4. Success Criteria 1: I can use the index of refraction to determine how light changes directions as it travels through transparent materials.
First, when light hits an object as we already know, photons are either absorbed as heat or absorbed and re-emitted as light. If the material re-emits most of the photons that strike it, but blocks photons from passing through it, you get reflection. If, however, most of the photons that strike the material are re-emitted and the material doesn’t block photons from passing through it, you get refraction. Refraction is the “bending” of the paths of photons that occurs when light passes through a transparent medium such as glass or water.
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5. Success Criteria 1: I can use the index of refraction to determine how light changes directions as it travels through transparent materials.
The place where an object appears to be located is related to the path that photons have to take to reach your eyes the fastest. Why does it have to be the fastest? That’s because the fastest path allows an image to be produced due to constructive interference (a full description of why this is the case is too complicated for now. See the theory of quantum electrodynamics if you want to know more). Due to the fact that the photons are absorbed and re-emitted many times as they travel through the transparent material, light slows down, the fastest time isn’t a straight line (See the lifeguard analogy below).
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6. Success Criteria 1: I can use the index of refraction to determine how light changes directions as it travels through transparent materials.
The factor by which light slows down is called the index of refraction, and is given the letter n:
n = c/v
n = index of refraction (unitless)
c = speed of light in a vacuum (3.00 x 108 m/s)
v = speed of light through material (m/s)
Water has an index of refraction of about 1.33, while diamond is about Glass is around 1.5 to 1.7, depending on the type of glass. The higher the index of refraction, the slower light travels through the material. Generally, denser materials have greater indices of refraction. As we’ll see in the next section, a higher index of refraction means more bending of light.
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7. Success Criteria 1: I can use the index of refraction to determine how light changes directions as it travels through transparent materials.
This brings us to Snell’s Law, which is the relationship between a material’s index of refraction and the angle that light bends as it travels into and out of the material.
nisinθi = nrsinθr
ni = index of refraction of the incident material (the material the incoming light is traveling through).
θi = the incident angle as measured from the normal (as measured from perpendicular to the surface).
nr = index of refraction of the refracting material (the material that the light goes into).
θr = the refracted angle as measured from the normal.
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8. Task 7.2.1 (5 points) Answer these questions.
Success Criteria 1: I can use the index of refraction to determine how light changes directions as it travels through transparent materials.
Task (5 points) Answer these questions.
If light is measured as travelling at 2.25 x 108 m/s through a piece of clear plastic, what is the index of refraction of the plastic?
If light were to enter a piece of the plastic from part a) at a 30° angle, at what angle would it travel through the plastic? (assume in this section that when you are given an angle, it is relative to the normal unless otherwise stated)
If light were to hit a piece of glass at a 42° angle, what angle would the light travel through the glass if it had an index of refraction of 1.55?
Suppose you measure light travelling through water at a 25° angle. When it leaves the water and goes back into the air, what angle would it travel at (the index of refraction of air is 1.00).
If light goes from a 15.6° angle to a 13.1° angle as it passes from one material to another, what is the index of refraction of the refracting material if the index of refraction of the incident material is 1.44?
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9. sinθc = nr/ni θc= the critical angle.
Success Criteria 1: I can use the index of refraction to determine how light changes directions as it travels through transparent materials.
You may have noticed that when light goes from a material with a higher index of refraction to a lower index of refraction (such as going from water to air), above a certain angle, Snell’s Law gives you a domain error. For example, if light were going from water (n = 1.33) to air (n = 1) at a 60° angle, sinθr would be there is no angle for a sine value greater than 1. This doesn’t mean the equation doesn’t work. It means that the light wouldn’t escape the water and you would get what they call total internal reflection. The maximum angle that would allow light to escape is called the critical angle and is given by:
sinθc = nr/ni θc= the critical angle.
Notice that there is only a critical angle if nr/ni is less than 1 (if ni is greater than nr). If ni is less than nr There is no critical angle and no angle will give total internal reflection.
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10. Task 7.2.2 (3 points) Answer these questions using the chart.
Success Criteria 1: I can use the index of refraction to determine how light changes directions as it travels through transparent materials.
Task (3 points) Answer these questions using the chart.
What is the critical angle for light going from diamond to air?
What is the critical angle for light going from flint glass to water?
Is there are critical angle for light going from glycerin to zircon? How do you know?
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11. Success Criteria 2: I can calculate the magnification and image distance of lenses.
When light is refracted through a curved surface, the light rays that were parallel when they entered the material are no longer parallel. This allows us to create images that are larger or smaller than the actual object.
In converging lenses (those with convex surfaces), light rays are bent toward each other, while diverging lenses (those with concave surfaces) bend light away from each other. We’ll focus on converging lenses due to their greater use in technology.
When light passes through a converging lens, it is focused passed a focal point, then goes on to produce an image. Your eyes have a converging lens that focuses light on your retina, which your brain then interprets. Most cameras, telescopes, microscopes, and contact lenses focus light in this way.
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12. 1/p + 1/q = 1/f or p-1 + q-1 = f-1 M = h’/h = -q/p
Success Criteria 2: I can calculate the magnification and image distance of lenses.
The equations for image formation and magnification are the same as for concave spherical mirrors, except that due to the varying indices of refraction, there is no simple relationship between radius and focal length.
1/p + 1/q = 1/f or p-1 + q-1 = f-1
And
M = h’/h = -q/p
p = object distance from lens (m)
q = image distance from lens (m)
f = focal length (m)
M = magnification (unitless)
h‘ = image height (m)
h = object height (m)
With converging lenses, p, q, and f are all considered positive because the direction of light doesn’t change (never mind if the object is in front of the lens and the image and focal length are behind it. All are considered positive).
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13. Task 7.2.3 (5 points) Answer these questions.
Success Criteria 2: I can calculate the magnification and image distance of lenses.
Task (5 points) Answer these questions.
Read the article at Then, write a one paragraph description of how refracting telescopes work, using the terms “objective lens” and “eyepiece lens” in your description.
What is the focal length of a lens that produces an image that is 0.24 m from the lens when the object is 0.82 m from the lens?
What is the magnification of a lens that produces an image of height 1.2 cm from an object that is 8.4 cm?
A small lens called the eyepiece is placed inside a reflecting telescope to focus the image. What would its magnification be if it took an image that were 1.2 mm and made it appear to be 15 mm? (in this case, the “image” is the object because it is what is produced by the mirror in the telescope)
Why does an image appear upside down after passing through a converging lens?
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14. Task (4 points): Write at least 8 things you learned in this topic (1/2 point each). If you do this in your notebook, please do it in list form rather than paragraph form.
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