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First of four images shows an incident ray 1 coming from an object (a girl ) placed on the axis. After refraction, the ray passes through F on other side of the lens. Second of four images shows an incident ray 2 passing through the center without any deviation. Third of four images shows an incident ray passing through F, which after refraction goes parallel to the axis. Fourth image shows a combination of all three rays, 1, 2, and 3, incident on a convex lens; after refraction, they converge or cross at a point below the axis at some distance from F. Here the height of the object h sub o is the height of the girl above the axis and h sub i is the height of the image below the axis. The distance from the center to point F is small f. The distance from the center to the girl is d sub o and that to the image is d sub i.
Ray tracing is used to locate the image formed by a lens. Rays originating from the same point on the object are traced—the three chosen rays each follow one of the rules for ray tracing, so that their paths are easy to determine. The image is located at the point where the rays cross. In this case, a real image—one that can be projected on a screen—is formed.

The image formed in [link] is a real image    , meaning that it can be projected. That is, light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye, for example. [link] shows how such an image would be projected onto film by a camera lens. This figure also shows how a real image is projected onto the retina by the lens of an eye. Note that the image is there whether it is projected onto a screen or not.

Real image

The image in which light rays from one point on the object actually cross at the location of the image and can be projected onto a screen, a piece of film, or the retina of an eye is called a real image.

Figure (a) shows incident rays coming from an object (a girl) and falling on a convex lens in a camera. The rays after refraction produce an inverted, real, and diminished image on the film of the camera. Figure (b) shows the same object in front of a human eye. The rays from the object fall on the convex lens and on refraction produce a real, inverted, and diminished image on the retina of the eyeball.
Real images can be projected. (a) A real image of the person is projected onto film. (b) The converging nature of the multiple surfaces that make up the eye result in the projection of a real image on the retina.

Several important distances appear in [link] . We define d o to be the object distance, the distance of an object from the center of a lens. Image distance d i is defined to be the distance of the image from the center of a lens. The height of the object and height of the image are given the symbols h o and h i , respectively. Images that appear upright relative to the object have heights that are positive and those that are inverted have negative heights. Using the rules of ray tracing and making a scale drawing with paper and pencil, like that in [link] , we can accurately describe the location and size of an image. But the real benefit of ray tracing is in visualizing how images are formed in a variety of situations. To obtain numerical information, we use a pair of equations that can be derived from a geometric analysis of ray tracing for thin lenses. The thin lens equations are

1 d o + 1 d i = 1 f

and

h i h o = d i d o = m.

We define the ratio of image height to object height ( h i / h o size 12{h rSub { size 8{i} } /h rSub { size 8{o} } } {} ) to be the magnification     m size 12{m} {} . (The minus sign in the equation above will be discussed shortly.) The thin lens equations are broadly applicable to all situations involving thin lenses (and “thin” mirrors, as we will see later). We will explore many features of image formation in the following worked examples.

Image distance

The distance of the image from the center of the lens is called image distance.

Thin lens equations and magnification

1 d o + 1 d i = 1 f
h i h o = d i d o = m

Finding the image of a light bulb filament by ray tracing and by the thin lens equations

A clear glass light bulb is placed 0.750 m from a convex lens having a 0.500 m focal length, as shown in [link] . Use ray tracing to get an approximate location for the image. Then use the thin lens equations to calculate (a) the location of the image and (b) its magnification. Verify that ray tracing and the thin lens equations produce consistent results.

Practice Key Terms 8

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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