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# Propagation of Light Study Guide

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Updated on Sep 27, 2011

## 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, do and di, 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 (do = di)

#### 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.

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