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Image formation by thin lenses

We use ray tracing to investigate different types of images that can be created by a lens. In some circumstances, a lens forms a real image, such as when a movie projector casts an image onto a screen. In other cases, the image is a virtual image, which cannot be projected onto a screen. Where, for example, is the image formed by eyeglasses? We use ray tracing for thin lenses to illustrate how they form images, and then we develop equations to analyze quantitatively the properties of thin lenses.

Consider an object some distance away from a converging lens, as shown in [link] . To find the location and size of the image, we trace the paths of selected light rays originating from one point on the object, in this case, the tip of the arrow. The figure shows three rays from many rays that emanate from the tip of the arrow. These three rays can be traced by using the ray-tracing rules given above.

  • Ray 1 enters the lens parallel to the optical axis and passes through the focal point on the opposite side (rule 1).
  • Ray 2 passes through the center of the lens and is not deviated (rule 2).
  • Ray 3 passes through the focal point on its way to the lens and exits the lens parallel to the optical axis (rule 3).

The three rays cross at a single point on the opposite side of the lens. Thus, the image of the tip of the arrow is located at this point. All rays that come from the tip of the arrow and enter the lens are refracted and cross at the point shown.

After locating the image of the tip of the arrow, we need another point of the image to orient the entire image of the arrow. We chose to locate the image base of the arrow, which is on the optical axis. As explained in the section on spherical mirrors, the base will be on the optical axis just above the image of the tip of the arrow (due to the top-bottom symmetry of the lens). Thus, the image spans the optical axis to the (negative) height shown. Rays from another point on the arrow, such as the middle of the arrow, cross at another common point, thus filling in the rest of the image.

Although three rays are traced in this figure, only two are necessary to locate a point of the image. It is best to trace rays for which there are simple ray-tracing rules.

Figure shows the cross section of a bi-convex lens. Three rays originate from the top of an object and enter the lens. Ray 1 is parallel to the optical axis. Ray 2 crosses the center of the lens. Ray 3 crosses the focal point in front of the lens. All rays converge on the other side at the top of an inverted image. Ray 1 crosses the focal point behind the lens. Ray 2 is undeviated. Ray 3 becomes parallel to the optical axis. The object distance and image distances are d subscript o and d subscript I respectively. The object and image heights are h subscript o and h subscript I respectively. The focal length is f.
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.

Several important distances appear in the figure. As for a mirror, we define d o to be the object distance, or the distance of an object from the center of a lens. The 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 the height of the image are indicated by h o and h i , respectively. Images that appear upright relative to the object have positive heights, and those that are inverted have negative heights. By 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.

Questions & Answers

For the question about the scuba instructor's head above the pool, how did you arrive at this answer? What is the process?
Evan Reply
as a free falling object increases speed what is happening to the acceleration
Success Reply
of course g is constant
Alwielland
acceleration also inc
Usman
which paper will be subjective and which one objective
jay
normal distributiin of errors report
Dennis
normal distribution of errors
Dennis
photo electrons doesn't emmit when electrons are free to move on surface of metal why?
Rafi Reply
What would be the minimum work function of a metal have to be for visible light(400-700)nm to ejected photoelectrons?
Mohammed Reply
give any fix value to wave length
Rafi
40 cm into change mm
Arhaan Reply
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
Prema
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
Prema
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
Prema
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
Prema
this msg is out of mistake. sorry friends​.
Prema
what is physics?
sisay Reply
why we have physics
Anil Reply
because is the study of mater and natural world
John
because physics is nature. it explains the laws of nature. some laws already discovered. some laws yet to be discovered.
Yoblaze
is this a physics forum
Physics Reply
explain l-s coupling
Depk Reply
how can we say dirac equation is also called a relativistic equation in one word
preeti Reply
what is the electronic configration of Al
usman Reply
what's the signeficance of dirac equetion.?
Sibghat Reply
what is the effect of heat on refractive index
Nepal Reply
As refractive index depend on other factors also but if we supply heat on any system or media its refractive index decrease. i.e. it is inversely proportional to the heat.
ganesh
you are correct
Priyojit
law of multiple
Wahid
if we heated the ice then the refractive index be change from natural water
Nepal
can someone explain normalization condition
Priyojit Reply
please tell
Swati
yes
Chemist
1 millimeter is How many metres
Darling Reply
1millimeter =0.001metre
Gitanjali
Practice Key Terms 5

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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