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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.

A light bulb at d sub o equals 0.75 m is placed in front of a convex lens of f equals 0.50 meter. The convex lens produces a real, inverted, and enlarged image on a screen at d sub I equals 1.50 meters.
A light bulb placed 0.750 m from a lens having a 0.500 m focal length produces a real image on a poster board as discussed in the example above. Ray tracing predicts the image location and size.

Strategy and Concept

Since the object is placed farther away from a converging lens than the focal length of the lens, this situation is analogous to those illustrated in [link] and [link] . Ray tracing to scale should produce similar results for d i . Numerical solutions for d i and m can be obtained using the thin lens equations, noting that d o = 0.750 m and f = 0.500 m .

Solutions (Ray tracing)

The ray tracing to scale in [link] shows two rays from a point on the bulb’s filament crossing about 1.50 m on the far side of the lens. Thus the image distance d i is about 1.50 m. Similarly, the image height based on ray tracing is greater than the object height by about a factor of 2, and the image is inverted. Thus m is about –2. The minus sign indicates that the image is inverted.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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cm
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Can you compute that for me. Ty
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Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Source:  OpenStax, Concepts of physics. OpenStax CNX. Aug 25, 2015 Download for free at https://legacy.cnx.org/content/col11738/1.5
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