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Propagation of Light Study Guide (page 3)

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Updated on Sep 27, 2011

Characteristics of Thin Convergent and Divergent Lenses

Now that we have a complex understanding of the light propagation because of opaque objects, let us study some that are transparent to light that we call lenses. Although a mirror works mostly through reflection, lenses having light transmitted through their material also deal with refraction. Two types of lenses exist: convergent, which gathers together an incoming bunch of light rays (see Figure 19.12), and divergent, which spreads apart a bunch of incoming light rays (see Figure 19.13). Both types of lenses are transparent. We consider the lenses thin so that expressions connecting lens characteristics do not depend on the thickness of the lens.

Characteristics of Thin Convergent and Divergent Lenses

Characteristics of Thin Convergent and Divergent Lenses

The lens equation is similar to the mirror equation and is

But in this case, the focal length is

where R1 and R2 are the radii of the two curvatures (left and right) and n is the index of refraction of the material the lens is made of.

Example 1

A slab of glass has a plane on both sides. Find the "extreme" focal distance of such a lens and the position of an image formed by the lens.

Solution 1

We will start with the definition of the focal length and then solve for f.

If the "lens" is a glass slab, then both R1 and R2 are very large and the focal distance is infinity.

The magnification equation is identical to the one we defined for mirrors:

and the convention of sign is shown in Table 19.2.

Table 19.2 Convention of Sign

Because lenses are transparent and light can travel from both sides, each lens has two corresponding focal points.

Example 2

Considering the sign conventions and using the lens equation, find the position of an image and its magnification if the object is placed in front of a convergent lens beyond its focal point at a distance of 30 cm. The focal distance is 10 cm.

Solution 2

The object's distance and focal distance of the lens will be converted first in meters. Then we can determine the image position by setting up the lens equation with considerations on the sign.

do = +30 cm = +0.30 m

f = + 10 cm = +0.1 m

di = ?

The image is in the back of the lens and it is real.

And the image is inverted.

Images in Thin Convergent and Divergent Lenses

As we have drawn diagrams for the plane and curved mirrors, we will check our understanding of image formation by finding the way to construct an image in a thin lens. The same important rays will become handy in this process with the main difference being that, although mirrors are opaque, lenses are transparent, so instead of reflection phenomena, we deal with refraction.

Considering a convergent lens, let's study the light propagation (I will use lens symbols; see Figure 19.14):

Images in Thin Convergent and Divergent Lenses

  • All rays parallel to the optical axis converge beyond the lens and propagate through the focal point.
  • All rays propagating along the optical axis continue to propagate in the same direction beyond the lens.
  • A light ray passing through the tip of the lens will propagate in the same direction beyond the lens.
  • Incoming rays tend to be bent toward the thicker part of the convergent lens.

The following are the characteristics of a divergent lens (see Figure 19.15):

Images in Thin Convergent and Divergent Lenses

  • All rays parallel to the optical axis diverge beyond the lens and their extension passes through the virtual focal point.
  • All rays propagating along the optical axis continue to propagate in the same direction beyond the lens.
  • A light ray passing through the tip of the lens will propagate in the same direction beyond the lens.
  • Incoming rays tend to be bent toward the thicker part of the divergent lens top of the lens.

Depending on the relative position of the object to the focal point and the type of lens, image characteristics are different.

Example

In a previous example, you considered the sign conventions and the lens equation in order to find the position of an image and its size if the object is placed in front of a convergent lens beyond its focal point at a distance of 30 cm with a focal distance of 10 cm. Draw the diagram corresponding to this situation and compare it with the quantitative results and conclusions.

Solution

As we see from Figure 19.16, the image is reversed (m < 0), real, and at the back of the lens, and the image is also smaller than the object. In addition, the image is beyond the focal point, so it should be positive and larger than 0.10 m (which it is since the result is +0.15 m). These conclusions based on the image are the same as the ones we determined based on the lens equation and magnification in the previous example.

Images in Thin Convergent and Divergent Lenses

 

Practice problems of this concept can be found at:Propagation of Light Practice Questions

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