Homework Helpers: Physics

8 Waves and Light

Refraction makes objects appear to be bent when placed in water. If you ever used a net on a pole to clean leaves out of the bottom of a pool, or a net to remove a fish from a fish tank, you probably noticed that these items appear to bend right at the surface of the water. If you haven”t had such an experience, you have probably noticed that people”s legs appear very squat when they are standing in a body of water, when viewed from outside the water. Refraction involves the bending of waves as they enter a new medium.

The key to understanding how and why refraction occurs begins with a discussion of material media. Light (and all forms of electromagnetic waves) has a maximum speed of 3.00 × 10⁸ m/s in a vacuum. When light enters a new medium, its speed changes based on the index of refraction(n) of that particular medium. The index of refraction (n) is simply the ratio of the speed of light in a vacuum (C) to the speed of light in that particular medium (v).

or

Light from the sun slows down when it hits our atmosphere, but it slows down by such a small amount that the speed of light in our atmosphere still rounds to 3.00 × 10⁸ m/s. So, the index of refraction for air (n_air) is equal to

Notice that all of the units are cancelled in our calculation, so the index of refraction has no units.

Example 1

Light has a speed of 1.80 × 10⁸ m/s in a particular piece of glass. Calculate the index of refraction for this glass.

Example 2

The index of refraction for a sample of water is 1.33. How fast does light travel through this water?

Recall our example of a golf cart driving at a 45° angle into slippery grass, and turning slightly as the wheels on one side of the cart slip. When light passes at an angle into a new medium, it will bend because one side of the wave changes speed before the other side. The direction the light will bend towards when it enters at an angle into a new medium is given by the law of refraction.

When light passes at an angle from a medium with a lower index of refraction to one with a higher index of refraction, it will bend toward the normal to the boundary.

When light passes at an angle from a medium with a higher index of refraction to one with a lower index of refraction, it will bend away from the normal to the boundary.

The extent to which a wave of light bends when it passes obliquely into a new medium is given by Snell”s law.

n₁ sin θ₁ = n₂ sin θ₂

where θ₁ is the angle of incidence, and θ₂ is the angle of refraction.

Example 3

A ray of light passes through air and strikes a piece of glass at an angle of 45.0° from the normal. If the index of refraction for the glass is 1.60, what will be the angle of refraction?

Figure 8.6

Given: n₁ = 1.00 θ₁ = 45.0° n₂ = 1.60

Find: θ₂

Example 4

Light passes through air (n₁ =1.00) and strikes the surface of a piece of glass at an angle of 58.0° from the normal. The angle of refraction is 33.0° measured from the normal. Calculate the index of refraction for the glass.

Given: n₁ = 1.00 θ₁ = 58.0° θ₂ = 33.0°

Find: n₂

Critical Angle of Refraction

Did you ever look through a fish tank containing water and notice that one or two of the other sides of the tank look like mirrors? You can look straight through a fish tank and see that the back wall is actually clear glass, and yet when you look through from the side, you can no longer see through the back wall. In such a situation, you are observing a phenomenon known as total internal reflection.

One of the implications of the law of refraction is that when light passes from a medium with a greater index of refraction to a medium with a lower index of refraction, the angle of refraction will be bigger than the angle of incidence. That means that if the angle of incidence is large enough, the angle of refraction would be made to exceed a 90° angle! Such a ray would not refract into the next medium at all, but would reflect back into the original medium. The effect, total internal reflection, occurs when the critical angle of incidence has been exceeded.

The formula for calculating the critical angle of incidence is:

Because the index of refraction of air is 1.00, when the second medium involved in a problem is air, this equation may simply be written as:

Example 5

The index of refraction for water is 1.33. Above the water is air (n = 1.00). What is the critical angle of incidence for light starting inside the water and passing into the air above it?

Given: n₁ = 1.33 n₂ = 1.00

Find: θ_c

Example 6

A diamond with an index of refraction of 2.42 is surrounded by air (n =1.00). If a ray of light originating inside the diamond strikes the surface of a facet at an angle of 27.0° from the normal, will the ray be refracted, or will it experience total internal reflection?

The key to this problem is to figure out if the angle of incidence (27.0°) exceeds the critical angle for total internal reflection. If it is greater than the critical angle, the ray is reflected. If it is less than the critical angle, it will be refracted according to Snell”s law.

Because the angle of incidence exceeds the critical angle, the light will experience total internal reflection.

The property of refraction is what allows lenses to be used in eyeglasses to correct vision problems. It is also what allows lenses to be used to magnify images in telescopes, binoculars, and microscopes. Lenses can be ground into many different types of shapes, but we will limit our discussion here to two major types of lenses: convex and concave.

Concave lenses are called diverging lenses because light that passes through such a lens will refract away from the focal point. This type of lens would not help you start a fire, but it can be used to correct some types of vision problems.

Figure 8.7

Convex lenses are called converging lenses, because incident rays that are parallel to the primary axis will be refracted through the focal point (f). Incident rays that pass through the secondary focus (f”) will be refracted in such a way that they end up parallel to the primary axis. The result is that rays of light that pass through the lens are made to converge at a point. This type of lens can be used as a magnifying glass and to start fires.

The following ray diagram (Figure 8.8) illustrates why convex lenses are called converging lenses. Notice that the diagram includes a notable simplification. In reality, the ray of light will refract twice. Once when it enters the lens, and once when it leaves the lens. Most ray diagrams are simplified to show only one refraction, but you should always remember that this represents the net effect of two refractions.

Figure 8.8

You can see from the ray diagram that we use many of the same concepts for lenses that we use for mirrors. We still talk about the primary axis, vertex, center of curvature, focal length, and focal point. In fact, as you will see shortly, we even use some of the same equations for lenses that we used for mirrors in our previous lesson.

Although the equations are identical, we must pay careful attention to another set of sign conventions when dealing with lenses.

Concave (Diverging) Lens

The focal length (f) is always negative.

The distance of the object (d_o) is always positive.

The distance of the image (d_i) is always negative, because only a virtual image can be formed by this type of lens.

The height of the image (h_i) will always be positive, because all images formed by this lens will be upright.

Convex (Converging) Lens

The focal length (f) is always positive.

The distance of the object (d_o) is positive if the object is placed in front of the lens, and negative if the object is placed behind the lens.

The distance of the image (d_i) is positive for a real image, which is formed on the opposite side of the lens as the object, and negative for a virtual image, which is formed on the same side of the lens as the object.

The height of the image (h_i) will be positive if the image is upright, and negative if it is inverted.

Example 7

A bulb with a height of 6.00 cm is placed 24.0 cm in front of a convex lens with a focal length of 16.0 cm. Calculate the height of the image formed and the distance the image will be formed from the lens.

Checking the sign conventions for convex lenses reveals that we don”t need to list any of our given values with negative signs.

Given: h_o = 6.00 cm d_o = 24.0 cm f = 16.0 cm

Find: d_i and h_i

The fact that the value for the distance of the image is positive means that it forms in front of the lens. It is a real image.

The fact that the height of the image has a negative value means that the image is inverted. It is also larger than the object.

Example 8

A candle with a height of 4.00 cm is placed 14.0 cm in front of a concave lens with a focal length of 18.0 cm. Calculate the height of the image formed and the distance the image will formed from the lens.

Checking the sign conventions for concave lenses reveals that our focal length should be given a negative sign.

Given: h_o = 6.00 cm d_o = 14.0 cm f = −18.0 cm

Find: d_i and h_i

As is always the case with concave lenses, the distance of the image is negative, placing the image on the same side of the lens as the object.

The height of the image is positive, because concave lenses always form upright images. The image is also reduced.

Lesson 8–6 Review

1. A diamond has an index of refraction of 2.42. Calculate the speed of light in the diamond.

2. A ray of light passes through air and strikes a smooth body of water at an angle of 22.0° from the normal. If the index of refraction for the water is 1.33, what will be the angle of refraction?

3. An object with a height of 5.00 cm is placed 22.0 cm in front of a concave lens with a focal length of 11.0 cm. Calculate the height of the image formed and the distance the image will formed from the lens.