26.11

Angular Magnification and the Magnifying Glass

If you hold a penny at arm’s length, the penny looks larger than the moon. The reason is that the penny, being so close, forms a larger image on the retina of the eye than does the more distant moon. The brain interprets the larger image of the penny as arising from a larger object. The size of the image on the retina determines how large an object appears to be. However, the size of the image on the retina is difficult to measure. Alternatively, the angle subtended by the image can be used as an indication of the image size. Figure 26-38 shows this alternative, which has the advantage that is also the angle subtended by the object and, hence, can be measured more easily. The angle is called the angular size of both the image and the object. The larger the angular size, the larger the image on the retina, and the larger the object appears to be.

Figure 26-38 The angle is the angular size of both the image and the object.

According to Equation 8.1, the angle (measured in radians) is the length of the circular arc that is subtended by the angle divided by the radius of the arc, as Figure 26-39a indicates. Part b of the drawing shows the situation for an object of height ho viewed at a distance do from the eye. When is small, ho is approximately equal to the arc length and do is nearly equal to the radius, so that

This approximation is good to within one percent for angles of 9° or smaller. In the next example the angular size of a penny is compared with that of the moon.

Figure 26-39 (a) The angle , measured in radians, is the arc length divided by the radius. (b) For small angles (less than 9°), is approximately equal to ho/do, where ho and do are the object height and distance.

| Example 14 | | | A Penny and the Moon

Compare the angular size of a penny held at arm’s length with that of the moon (, and ). Reasoning The angular size of an object is given approximately by its height ho divided by its distance do from the eye, , provided that the angle involved is less than roughly 9°; this approximation applies here. The “heights” of the penny and the moon are their diameters.
Solution The angular sizes of the penny and moon are

The penny thus appears to be about three times as large as the moon.

An optical instrument, such as a magnifying glass, allows us to view small or distant objects because it produces a larger image on the retina than would be possible otherwise. In other words, an optical instrument magnifies the angular size of the object. The angular magnification (or magnifying power) M is the angular size of the final image produced by the instrument divided by a reference angular size . The reference angular size is the angular size of the object when seen without the instrument.

| (26.9)

The physics of a magnifying glass. A magnifying glass is the simplest device that provides angular magnification. In this case, the reference angular size is chosen to be the angular size of the object when placed at the near point of the eye and seen without the magnifying glass. Since an object cannot be brought closer than the near point and still produce a sharp image on the retina, represents the largest angular size obtainable without the magnifying glass. Figure 26-40a indicates that the reference angular size is , where N is the distance from the eye to the near point. To compute , recall from Section 26.7 and Figure 26-28 that a magnifying glass is usually a single converging lens, with the object located inside the focal point. In this situation, Figure 26-40b indicates that the lens produces a virtual image that is enlarged and upright with respect to the object. Assuming the eye is next to the magnifying glass, the angular size seen by the eye is , where do is the object distance. The angular magnification is

According to the thin-lens equation, do is related to the image distance di and the focal length f of the lens by

Substituting this expression for 1/do into the previous expression for M leads to the following result:

| (26.10)

Two special cases of this result are of interest, depending on whether the image is located as close to the eye as possible or as far away as possible. To be seen clearly, the closest the image can be relative to the eye is at the near point, or . The minus sign indicates that the image lies to the left of the lens and is virtual. In this event, Equation 26.10 becomes . The farthest the image can be from the eye is at infinity ; this occurs when the object is placed at the focal point of the lens. When the image is at infinity, Equation 26.10 simplifies to . Clearly, the angular magnification is greater when the image is at the near point of the eye rather than at infinity. In either case, however, the greatest magnification is achieved by using a magnifying glass with the shortest possible focal length. Example 15 illustrates how to determine the angular magnification of a magnifying glass that is used in these two ways.

Figure 26-40 (a) Without a magnifying glass, the largest angular size occurs when the object is placed at the near point, a distance N from the eye. (b) A magnifying glass produces an enlarged, virtual image of an object placed inside the focal point F of the lens. The angular size of both the image and the object is .

| Example 15 | | | Examining a Diamond with a Magnifying Glass

A jeweler, whose near point is 40.0 cm from his eye and whose far point is at infinity, is using a small magnifying glass (called a loupe) to examine a diamond. The lens of the magnifying glass has a focal length of 5.00 cm, and the image of the gem is −185 cm from the lens. The image distance is negative because the image is virtual and is formed on the same side of the lens as the object. (a) Determine the angular magnification of the magnifying glass. (b) Where should the image be located so the jeweler’s eye is fully relaxed and has the least strain? What is the angular magnification under this “least strain” condition?

An employee of the department for counterfeit money in Munich, Germany, uses a magnifying glass to examine a 200-euro note. (Peter Kneffel/ dpa/Landov)

Reasoning The angular magnification of the magnifying glass can be determined from Equation 26.10. In part (a) the image distance is −185 cm. In part (b) the ciliary muscle of the jeweler’s eye is fully relaxed, so the image must be located infinitely far from the eye, at its far point, as Section 26.10 discusses.
Solution

a. | With f = 5.00 cm, , and N = 40.0 cm, the angular magnification is

b. | With f = 5.00 cm, , and N = 40.0 cm, the angular magnification is

Jewelers often prefer to minimize eyestrain when viewing objects, even though it means a slight reduction in angular magnification.

Check Your Understanding 5

A person who has a near point of 25.0 cm is looking with unaided eyes at an object that is located at the near point.The object has an angular size of 0.012 rad.Then, holding a magnifying glass (f = 10.0 cm) next to her eye, she views the image of this object, the image being located at the near point. What is the angular size of the image? (The answer is given at the end of the book.)
Background: The angular magnification of a magnifying glass is the angular size of the image seen with the magnifying glass divided by the reference angular size.The reference angular size is the angular size of the object when placed at the near point and seen without the magnifying glass.
For similar questions (including conceptual counterparts), consult Self-Assessment Test 26.2, which is described at the end of Section 26.14.

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.