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Ray diagrams In Figure 18-11, rays are traced from an object located far from a convex lens For the purpose of locating the image, you only need to use two rays Ray 1 is parallel to the principal axis It refracts and passes through F on the other side of the lens Ray 2 passes through F on its way to the lens After refraction, its path is parallel to the principal axis The two rays intersect at a point beyond F and locate the image Rays selected from other points on the object converge at corresponding points to form the complete image Note that this is a real image that is inverted and smaller compared to the object You can use Figure 18-11 to locate the image of an object that is closer to the lens than the object in the figure If a refracted ray is reversed in direction, it will follow its original path in the reverse direction This means that the image and object may be interchanged by changing the direction of the rays Imagine that the path of light through the lens in Figure 18-11 is reversed and the object is at a distance of 15 cm from the right side of the lens The new image, located 30 cm from the left side of the lens, is a real image that is inverted and larger compared to the object If the object is placed at twice the focal length from the lens at the point 2F, as shown in Figure 18-12, the image also is found at 2F Because of symmetry, the image and object have the same size Thus, you can conclude that if an object is more than twice the focal length from the lens, the image is smaller than the object If the object is between F and 2F, then the image is larger than the object.



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9-8 Longwire (greater than 2 ) antenna is terminated in a resistance (typically 400 to 800 ) Radials under the resistor improve the grounding of the antenna

Figure 18-12 When an object is placed at a distance equal to twice the focal length from the lens, the image is the same size as the object





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What happens when you mask, or cover, part of a lens Does this cause only part of a real image to be formed by the lens 1 Stick the edge of a convex lens into a ball of clay and place the lens on a tabletop CAUTION: Lenses have sharp edges Handle carefully 2 Use a small lamp on one side and a screen on the other side to get a sharp image of the lamp s lightbulb CAUTION: Lamps get hot and can burn skin 3 Predict what will happen to the image if you place your hand over the top half of the lens This is called masking 4 Observe the effects of masking more of the lens and masking less of the lens Analyze and Conclude 5 How much of the lens is needed for a complete image 6 What is the effect of masking the lens

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Answer (for Objective 51): E is correct The getNetworkPrinterID() method implies a functionality that could be used by many different classes, not just the RobotDog class A is incorrect because the code is legal B s change would improve encapsulation, not cohesion C and D s changes would have no effect on cohesion (it s okay for a class to print its own state)

An Image Formed by a Convex Lens An object is placed 320 cm from a convex lens that has a focal length of 80 cm a Where is the image b If the object is 30 cm high, how tall is the image c What is the orientation of the image

Sketch the situation, locating the object and the lens Draw the two principal rays Known:

Unknown:

much longer than 5 , the four main lobes begin to converge along the length of the wire, and the antenna becomes bidirectional This effect occurs at physical lengths greater than about 20 In general, the following rules apply to longwire antennas: On each side of the antenna, there is at least one lobe, minor or major, for each half-wavelength of the wire element For the overall element, there is one lobe for every quarter-wavelength If there are an even number of lobes on either side of the antenna wire, then half of the total number of lobes are tilted backward, and half are tilted forward; symmetry is maintained If there are an odd number of lobes on either side of the wire, then one lobe on either side will be perpendicular to the wire, with the other lobes distributed either side of the perpendicular lobe

11 cm (11 cm away from the lens on the side opposite the object) b Use the magnification equation and solve for image height m hi

50 Given:

Are the units correct All are in centimeters Do the signs make sense Image position is positive (real image) and image height is negative (inverted compared to the object), which make sense for a convex lens

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