Geometric Optics

13.2  Geometric Optics

13.2.1  Definitions for Lenses

One extremely useful application of the properties of light propagation through materials is the formation of an image of an object through the reflection or refraction of light. We form the following definitions:

object
Anything from which light rays emanate. The object can be luminous in which case it is the source of light or can be reflective of light from some other source. Objects can be

point objects
having no physical extent
extended objects
having a length, width, and breadth

image
A reproduction of an object formed from light. Images can be

real images
formed on a surface such as a screen
virtual images
exist only within the brain but are perceived to be at a particular location. An example is shown in figure  13.9.


Figure 13.9: A toy for showing virtual images.

A clear material, e.g. glass, which reflects or refracts light can, for particular curve shapes, cause parallel rays of light to converge at a point. Reflecting surfaces, curved or not, are referred to as mirrors in optics. Mirrors have one focal point to go with their one curved surface. A refracting material with two curved surfaces is called a lens. Since a lens has two curved surfaces, it has two focal points. If the curved surfaces are close enough together that we can neglect the distance between the surfaces, we refer to it as a thin lens. A lens can be one of two types:

converging
a lens in which parallel rays of light passing through the lens are brought together at the focal point. Rays of light which come from a point object placed at one of the focal points and which pass through the lens are converted into parallel rays (see figure  13.10).
diverging
a lens in which parallel rays of light diverge after passing through the lens. The focal length of a diverging lens is defined as a negative quantity (see figure  13.11).


Figure 13.10: First and second focal points of a converging thin lens.


Figure 13.11: First and second focal points of a diverging thin lens and the negative focal length.

We can use a lens to image an object. In the case of a thin lens, we define the object distance, s, as the distance of the object from the center of the thin lens. The image distance, s¢, is the distance of the image formed from the center of the thin lens, and we usually term the focal distance, f, as the distance of the focal point from the center of the thin lens (see figure  13.12).


Figure 13.12: Definition of image, object, and focal lengths for a thin lens.

The object, image, and focal lengths are related by the formula
1
s
+ 1
s¢
= 1
f
(13.2.1.17)
Furthermore, the size of the image in the plane of the image, object, and lens, which we depict as y¢, is related to the size of the object (call it y) by the magnification. The magnification is
m º y¢
y
= - s¢
s
(13.2.1.18)
We define images which are on the same side of a converging lens as the object as virtual. Note that in such cases s¢ < 0 and the magnification is positive. For real images, the image is inverted compared to the object, i.e. y¢ and y have opposite signs. Hence a positive magnification corresponds to an erect, virtual image while a negative magnification corresponds to an inverted, real image. Let's consider an example.

Problem 1:
A converging lens with a focal length of 7.00 cm forms a 1.30 cm tall image of a 4.00 mm tall real object that is to the left of the lens. The image is erect. Find the locations of the object and the image and determine whether the image is real or virtual.
Solution:
Since we have the sizes of the images, we can find the magnification.
m = y¢
y
= 1.30 cm
0.400 cm
= +3.25.
(13.2.1.19)
Notice that since the image is erect, y¢ > 0 and the image is virtual. The magnification also implies
m = - s¢
s
Þ s¢ = -3.25s
(13.2.1.20)
Since s is positive (the object is real), the image distance is negative so it is located to the left of the lens as the object is. We now find the distances for the object and the image.
1
s
+ 1
s¢
=
1
f
1
s
- 1
3.25s
=
1
f
3.25 - 1
3.25s
=
1
f
Þ
s
=
2.25f
3.25
=
2.25(7.00 cm)
3.25
s
=
4.85 cm Þ s¢ = -15.8 cm
(13.2.1.21)
So the image is located 15.8 cm to the left of the lens and the object is located 4.85 cm to the left of the lens.


Figure 13.13: A converging lens producing a magnified virtual image.



Send comments to larryg@upenn5.hep.upenn.edu.
This page was last modified on 04/11/2003 at 09:44:03 (EDT).
Current date/time is Saturday, 07-Nov-2009 23:40:07 EST

15442 hits since