Propagation of Light Study Guide (page 2)

Introduction

The knowledge of geometrical optics is applied in this lesson to the work of some optical tools such as plane mirrors, curved mirrors, and convergent and divergent lenses, We will determine if an image formed with these tools is real or virtual, straight or inverted, or smaller or larger than the object.

Images in Plane Mirrors

In the previous lesson, we were already exposed to the concept that makes plane mirrors work: reflections from a flat surface. In this case, we will use a plane mirror, investigate the incident and reflected rays, the normal, and their relative positions. A plane mirror is an opaque, flat, shiny object that reflects light that propagates toward it. Think of your morning look in the mirror: What do you see? The image of your face staring back at you as if you were copied at an equal distance from the real you into the mirror. But is that an image that is real? In other words, is that an image that you can project on a screen? If you put a screen at the back of the mirror, you will not get an image: The mirror is opaque and light does not propagate through. The image that you see is called a virtual image. Figure 19.1 shows how the virtual image is formed. In order to form a point on the image, we need at least two rays (or geometrical projections of two rays) to intersect. The horizontal axis on which the object and the mirror sit will be called the optical axis. The quantities in Figure 19.1 are shown in Table 19.1.

We define lateral magnification to be the ratio of the image height to the object height:

where

m > 0 if the image is upright

m < 0 if the image is reversed

m > 1 if the image is larger than the object

m < 1 if the image is smaller than the object

The image in the mirror forms at the intersection of two extensions of light rays, and therefore, the image is virtual.

Right triangles ABO and BCO are equilateral triangles having a common side and all angles are equal. Accordingly, we can determine that the two distances, d_o and d_i, are equal and the height of the object and of the image is the same.

Also, due to light propagation in a straight line, the top of the object corresponds to the top of the image, and object points on the optical axis will have a corresponding image also on the axis.

In conclusion, the image in the previous plane mirror is

• Straight (m > 0)
• Virtual (image formed by ray extensions)
• The same size as the object (m = 1)
• At the same distance from the mirror as the object (d_o = d_i)

Example 1

A pendulum is suspended 10 cm in front of a plane mirror and above ground at a height of 30 cm. Characterize the properties of the image formed by the mirror.

Solution 1

We will draw a diagram of the image, and using the previous geometrical analysis, we will find the solution of the problem (see Figure 19.2).

As before, right triangles ABO and BCO are equilateral due to all angles being equal and one common side. The height of the image with respect to the optical axis is the same for both the object and the image, and the distance from the object to the mirror is the same as the distance of the image to the mirror.

Note

The light ray propagating perpendicular to the mirror is reflected and follows the same path as the incident light ray.

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.

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.

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.

d_o = +30 cm = +0.30 m

f = + 10 cm = +0.1 m

d_i = ?

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):

• 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):

• 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.

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