26.15

Concepts & Calculations

One important phenomenon discussed in this chapter is how a ray of light is refracted when it goes from one medium into another. Example 18 reviews some of the important aspects of refraction, including Snell’s law, the concept of a critical angle, and the notion of index matching.

| Concepts & Calculations Example 18 | | | Refraction

A ray of light is incident on a glass–water interface at the critical angle , as Figure 26-46 illustrates. The reflected light then passes through a liquid (immiscible with water) and into air. The indices of refraction for the four substances are given in the drawing. Determine the angle of refraction for the ray as it passes into the air.

Figure 26-46 A ray of light strikes the glass–water interface at the critical angle . The reflected ray passes through a liquid and then into air.

Concept Questions and Answers What determines the critical angle when the ray strikes the glass–water interface?

> Answer According to Equation 26.4, the critical angle is determined by the indices of refraction nglass and nwater: .

When light is incident at the glass–water interface at the critical angle, what is the angle of refraction, and how is the angle of reflection related to the critical angle?

> Answer When light is incident at the critical angle, the angle of refraction is 90°, so the refracted ray skims along the surface. (For simplicity, this ray is not shown in the drawing.) We know from our work with mirrors that the angle of reflection is equal to the angle of incidence. Therefore, the angle of reflection is equal to the critical angle, so .

When the reflected ray strikes the glass–liquid interface, how is the angle of refraction related to the angle of incidence ? Note that the two materials have the same indices of refraction.

> Answer According to Snell’s law, Equation 26.2, the angle of refraction is related to the angle of incidence by . Since the two indices of refraction are equal, a condition known as “index matching,” , and the two angles are equal. Therefore, the ray crosses the glass–liquid interface without being refracted.

When the ray passes from the liquid into the air, is the ray refracted?

> Answer Yes. Because the two materials have different indices of refraction, the angle of refraction is different from the angle of incidence . The angle of refraction can be found from Snell’s law.

Solution Let us follow the ray of light as it progresses from left to right in Figure 26-46. It strikes the glass–water interface at the critical angle of

The angle of reflection from the glass–water interface is the same as the critical angle, so .
The ray then strikes the glass–liquid interface with an angle of incidence of . Note, as usual, that the angle of incidence is measured relative to the normal. At the glass–liquid interface, the angle of refraction is the same as the angle of incidence , since the indices of refraction of the two materials are the same . Thus, .
At the liquid–air interface we use Snell’s law, Equation 26.2, to determine the angle at which the refracted ray enters the air: . Note from the drawing that the angle of incidence is the same as , so . The angle of refraction is

One of the most important uses of refraction is in lenses, the behavior of which is governed by the thin-lens and magnification equations. Example 19 discusses how these equations are applied to a two-lens system and reviews the all-important sign conventions that must be followed (see the Reasoning Strategy in Section 26.8). In particular, the example shows how to account for a virtual object, in which the object lies to the right (rather than to the left) of a lens.

| Concepts &Calculations Example 19 | | | A Two-Lens System

In Figure 26-47 a converging lens and a diverging lens are separated by a distance of 10.0 cm. An object with a height of ho1 = 5.00 mm is placed do1 = 45.0 cm to the left of the first (converging) lens. What is (a) the image distance di1 and (b) the height hi1 of the image produced by the first lens? (c) What is the object distance for the second (diverging) lens? Find (d) the image distance di2 and (e) the height hi2 of the image produced by the second lens.

Figure 26-47 The image produced by the first lens, the “first image,” falls to the right of the second lens, so the image becomes a virtual object for the second lens.

Concept Questions and Answers Is the image produced by the first (converging) lens real or virtual?

> Answer A converging lens can form either a real or a virtual image, depending on where the object is located relative to the focal point. If the object is to the left of the focal point, as it is in this example, the image is real and falls to the right of the lens. (Had the object been located between the focal point and the lens, the image would have been a virtual image located to the left of the converging lens.)

As far as the second lens is concerned, what role does the image produced by the first lens play?

> Answer The image produced by the first lens acts as the object for the second lens.

Note in Figure 26-47 that the image produced by the first lens is called the “first image,” and it falls to the right of the second lens. This image acts as the object for the second lens. Normally, however, an object would lie to the left of a lens. How do we take into account that this object lies to the right of the diverging lens?

> Answer According to the Reasoning Strategy in Section 26.8, an object that is located to the right of a lens is called a virtual object and is assigned a negative object distance.

How do we find the location of the image produced by the second lens when its object is a virtual object?

> Answer Once the object distance for the second lens has been assigned a negative number, we can use the thin-lens equation in the usual manner to find the image distance.

Solution

a. | The distance di1 of the image from the first lens can be found from the thin-lens equation, Equation 26.6. The focal length, f1 = +20.0 cm, is positive because the lens is a converging lens, and the object distance, do1 = +45.0 cm, is positive because it lies to the left of the lens:

This image distance is positive, indicating that the image is real.
b. | The height hi1 of the image produced by the first lens can be obtained from the magnification equation, Equation 26.7:

The minus sign indicates that the first image is inverted with respect to the object (see Figure 26-47).
c. | The first image falls 36.0 cm to the right of the first lens. However, the second lens is located 10.0 cm to the right of the first lens, and therefore the first image is located 36.0 cm − 10.0 cm = 26.0 cm to the right of the second lens. This image acts as the object for the second lens. Since the object for the second lens lies to the right of it, the object is a virtual object and is assigned a negative number: .
d. | The distance di2 of the image from the second lens can be found from the thin-lens equation. The focal length, f2 = −15.0 cm, is negative because the lens is a diverging lens, and the object distance, do2 = −26.0 cm, is negative because it lies to the right of the lens:

or

The negative sign for di2 means that the image is formed to the left of the diverging lens and, hence, is a virtual image.
e. | The image height hi1 produced by the first lens becomes the object height ho2 for the second lens: . The height hi2 of the image produced by the second lens follows from the magnification equation:

The positive value for hi2 means that the final image has the same orientation as the first image—that is, inverted with respect to the original object.

At the end of the problem set for this chapter, you will find homework problems that contain both conceptual and quantitative parts. These problems are grouped under the heading Concepts & Calculations.

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