Image Formation by Lenses
Student Extras
Visit The Physics Classroom's Flickr Galleries and enjoy a photo overview of the topic of refraction and lenses.
Flickr PhysicsVisit The Physics Classroom's Flickr Galleries and enjoy the terrific display of photos showing the refraction of light by dew drops.
Shockwave StudiosInteract, view and manipulate variables with the Lenses activity from the Shockwave Studios.
PhET Simulation: Geometric Optics
Explore the refraction of light and the formation of images by lenses with this Java applet from PhET.
Teacher's Guide
The Lenses activity from the Shockwave Studios is an excellent accompaniment to this reading.
The LaboratoryLooking for a lab that coordinates with this page? Try the Exploring Lenses Lab from The Laboratory.
The LaboratoryLooking for a lab that coordinates with this page? Try The L•O•S•T Art of Image Description Lab from The Laboratory.
Curriculum CornerPractice makes perfect with this diagramming activity from The Curriculum Corner.
Workshop Tutorials for Physics: Waves and Optics - Introductory
These workshop-style tutorials and accompanying activities guide students through various wave topics - including lenses.
Treasures from TPFNeed ideas? Need help? Explore The Physics Front's treasure box of catalogued resources on ray optics, including the topic of lenses.
Converging Lenses - Object-Image Relations
Previously in Lesson 5, ray
diagrams were constructed in order to determine the general
location, size, orientation, and type of image formed by
double convex lenses. Perhaps you noticed that there is a
definite relationship between the image characteristics and
the location where an object placed in front of a double
convex lens. The purpose of this portion of the lesson is to
summarize these object-image relationships. The 
best
means of summarizing this relationship is to divide the
possible object locations into five general areas or
points:
- Case 1: the object is located beyond the 2F point
- Case 2: the object is located at the 2F point
- Case 3: the object is located between the 2F point and the focal point (F)
- Case 4: the object is located at the focal point (F)
- Case 5: the object is located in front of the
focal point (F)
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. Earlier in Unit 13, the term magnification
was introduced; the
magnification 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 2F 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.
Check
Your Understanding
1. Identify the means by which you can use a converging lens to form a real image.
2. Identify the means by which you can use a converging lens to form a virtual image.
3. A converging lens is sometimes used as a magnifying glass. Explain how this works; specifically, identify the general region where the object must be placed in order to produce the magnified effect.