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Refraction of Light by a Converging Lens

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Edgar Miranda

on 28 March 2011

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Transcript of Refraction of Light by a Converging Lens

Aim: What are the Refraction Rules for Converging Lenses? What are the object-image relationships for Converging Lenses? Refraction is the bending of a wave (e.g.light wave) when it enters another medium where the speed (v) of the wave is different. Refraction is most commonly observed when an electromagnetic wave (i.e., light wave) enters another medium at any angle other than 90 degrees or zero degree (i.e. oblique angle). Although refraction of light is the most commonly observed example, any wave can undergo refraction whenever the wave interacts with another medium. A lens is a carefully ground or molded piece of transparent material that refracts light rays in such as way as to form an image. Lenses serve to refract light at each boundary.Two types of lenses are shown below. This lesson is about the refraction on light through a type of converging lens- the biconvex lens.


As a ray of light enters a lens, the light ray is refracted; and as the same ray of light exits the lens, it is refracted again. The overall effect of the refraction of light at these two boundaries is that the light ray has changed directions. Because of the special geometric shape of a lens, the light rays are refracted such that they form images.

A convex lens is a converging lens which works much like a concave mirror. This kind of lens is thicker in the middle and thinner towards the edges, like the lens in a magnifying glass. The image is changed by the position of the object in relation to the focal length and the radius of curvature. Refraction Rules for a Converging Lens

1. Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point (F) on the opposite side of the lens.

Converging Lens Image Formation

Converging lenses can produce both real and virtual images while diverging images can only produce virtual images. The process by which images are formed for lenses is the same as the process by which images are formed for plane and curved mirrors. Images are formed at locations where any observer is sighting as they view the image of the object through the lens. So if the path of several light rays through a lens is traced, each of these light rays will intersect at a point upon refraction through the lens. Each observer must sight in the direction of this point in order to view the image of the object. While different observers will sight along different lines of sight, each line of sight intersects at the image location. The diagram below shows several incident rays emanating from an object - a light bulb. Three of these incident rays correspond to our three strategic and predictable light rays. Each incident ray will refract through the lens and be detected by a different observer (represented by the eyes). The location where the refracted rays are intersecting is the image location.

The image in the above diagram is a real image since the light rays are actually passing through the image location. To each observer, it appears as though light is coming from this location. Converging Lenses - Ray Diagrams

We see an object because light from the object travels to our eyes as we sight along a line at the object. We see an image of an object because light from the object reflects off a mirror or refracts through a transparent material and travel to our eyes as we sight at the image location of the object. The image location is the location in space where light appears to diverge from. Because light emanating from the object converges or appears to diverge from this location, a replica or likeness of the object is created at that location. For both reflection and refraction scenarios, ray diagrams have been a valuable tool for determining the path of light from the object to our eyes. Step-by-Step Method for Drawing Ray Diagrams

The method of drawing ray diagrams for double convex lens (biconvex) is described below. The description is applied to the task of drawing a ray diagram for an object located beyond the 2F point of a double convex lens.
1. Pick a point on the top of the object and draw three incident rays traveling towards the lens.

Using a ruler, accurately draw one ray so that it passes exactly through the focal point (F) on the way to the lens. Draw the second ray such that it travels exactly parallel to the principal axis. Draw the third incident ray such that it travels directly to the exact center of the lens. Place arrowheads upon the rays to indicate their direction of travel. 2. Once these incident rays strike the lens, refract them according to the three rules of refraction for converging lenses.

The ray that passes through the focal point (F) on the way to the lens will refract and travel parallel to the principal axis. Use a straight edge to accurately draw its path. The ray that traveled parallel to the principal axis on the way to the lens will refract and travel through the focal point. And the ray that traveled to the exact center of the lens will continue in the same direction. Place arrowheads upon the rays to indicate their direction of travel. Extend the rays past their point of intersection. 3. Mark the image of the top of the object.

The image point of the top of the object is the point where the three refracted rays intersect. All three rays should intersect at exactly the same point. This point is merely the point where all light from the top of the object would intersect upon refracting through the lens. Of course, the rest of the object has an image as well and it can be found by applying the same three steps to another chosen point. 4. Repeat the process for the bottom of the object.

One goal of a ray diagram is to determine the location, size, orientation, and type of image that is formed by the double convex lens. Typically, this requires determining where the image of the upper and lower extreme of the object is located and then tracing the entire image. After completing the first three steps, only the image location of the top extreme of the object has been found. Thus, the process must be repeated for the point on the bottom of the object. If the bottom of the object lies upon the principal axis (as it does in this example), then the image of this point will also lie upon the principal axis and be the same distance from the mirror as the image of the top of the object. At this point the entire image can be filled in.
Some students have difficulty understanding how the entire image of an object can be deduced once a single point on the image has been determined. If the object is merely a vertical object (such as the arrow object used in the example below), then the process is easy. The image is merely a vertical line. In theory, it would be necessary to pick each point on the object and draw a separate ray diagram to determine the location of the image of that point. That would require a lot of ray diagrams as illustrated in the diagram below.


Converging Lenses - Object-Image Relations
Case 1: The object is located beyond 2F

When the object is located at a location beyond the 2F point, the image will always be located somewhere in between the 2F point and the focal point (F) on the other side of the lens. Regardless of exactly where the object is located, the image will be located in this specified region. In this case, the image will be an inverted image. That is to say, if the object is right side up, then the image is upside down. In this case, the image is reduced in size; in other words, the image dimensions are smaller than the object dimensions. If the object is a six-foot tall person, then the image is less than six feet tall. Magnification (M) is the ratio of the height of the object to the height of the image. In this case, the magnification is a number with an absolute value less than 1. Finally, the image is a real image. Light rays actually converge at the image location. If a sheet of paper were placed at the image location, the actual replica or likeness of the object would appear projected upon the sheet of paper. Case 2: The object is located at 2F

When the object is located at the 2F point, the image will also be located at the 2F point on the other side of the lens. In this case, the image will be inverted (i.e., a right side up object results in an upside-down image). The image dimensions are equal to the object dimensions. A six-foot tall person would have an image that is six feet tall; the absolute value of the magnification is exactly 1. Finally, the image is a real image. Light rays actually converge at the image location. As such, the image of the object could be projected upon a sheet of paper.

Case 3: The object is located between 2F and F

When the object is located in front of the 2F point, the image will be located beyond the 2F point on the other side of the lens. Regardless of exactly where the object is located between C and F, the image will be located in the specified region. In this case, the image will be inverted (i.e., a right side up object results in an upside-down image). The image dimensions are larger than the object dimensions. A six-foot tall person would have an image that is larger than six feet tall. The absolute value of the magnification is greater than 1. Finally, the image is a real image. Light rays actually converge at the image location. As such, the image of the object could be projected upon a sheet of paper.
Case 4: The object is located at F

When the object is located at the focal point, no image is formed. As discussed earlier in Lesson 5, the refracted rays neither converge nor diverge. After refracting, the light rays are traveling parallel to each other and cannot produce an image.
Case 5: The object is located in front of F

When the object is located at a location in front of the focal point, the image will always be located somewhere on the same side of the lens as the object. Regardless of exactly where in front of F the object is located, the image will always be located on the object's side of the lens and somewhere further from the lens. The image is located behind the object. In this case, the image will be an upright image. That is to say, if the object is right side up, then the image will also be right side up. In this case, the image is enlarged; in other words, the image dimensions are greater than the object dimensions. A six-foot tall person would have an image that is larger than six feet tall. The magnification is greater than 1. Finally, the image is a virtual image. Light rays diverge upon refraction; for this reason, the image location can only be found by extending the refracted rays backwards on the object's side the lens. The point of their intersection is the virtual image location. It would appear to any observer as though light from the object were diverging from this location. Any attempt to project such an image upon a sheet of paper would fail since light does not actually pass through the image location

It might be noted from the above descriptions that there is a relationship between the object distance and object size and the image distance and image size. Starting from a large value, as the object distance decreases (i.e., the object is moved closer to the lens), the image distance increases; meanwhile, the image height increases. At the 2F point, the object distance equals the image distance and the object height equals the image height. As the object distance approaches one focal length, the image distance and image height approaches infinity. Finally, when the object distance is equal to exactly one focal length, there is no image. Then altering the object distance to values less than one focal length produces images that are upright, virtual and located on the same side of the lens as the object. Finally, if the object distance approaches 0, the image distance approaches 0 and the image height ultimately becomes equal to the object height. These patterns are depicted in the diagram below. Eight different object locations are drawn in red and labeled with a number; the corresponding image locations are drawn in blue and labeled with the identical number.


2.Any incident ray traveling through the focal point (F) on the way to the lens will refract through the lens and travel parallel to the principal axis. 3.An incident ray that passes through the center of the lens (converging and diverging lenses) will continue in the same direction that it had when it entered the lens. Note: References for this HW were the Physics Classroom,the HyperPhysics, and the YouTube websites. The material used for this presentatation were copied mostly from the physics classroom website and pasted here.
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