Propagation of Light Study Guide (page 2)
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:
- 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)
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.
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.
The light ray propagating perpendicular to the mirror is reflected and follows the same path as the incident light ray.
Consider two mirrors at 90° with respect to each other and an image in front of them as in the figure. Find the number and position of the images with respect to the mirrors.
Again, we will start with a diagram of the setup and a geometrical analysis. Image 1 and Image 2 are formed similarly to the images in the previous examples. They will be virtual images, as large as the object, and the object distance and the image distance are the same relative to their respective mirrors.
Image 3 is more complex because it is formed after a double reflection (once on each of the two mirrors). Again, the image is virtual, as tall as the object, and at the same distance from the tip of the two mirrors (see Figure 19.3)
Images in Curved Mirrors
A curved mirror forms an image differently than a plane mirror because the surface is no longer a plane, and every point on the mirror will have a different normal. A curved mirror has a few characteristics, such as curvature point C, radius R, focal point F, focal distance f, the tip of the mirror, and the type of curvature. We will analyze each of them on a diagram.
Figure 19.7 shows two types of mirrors. The one on the left is called concave, and the one on the right is called convex. A concave mirror has the reflecting side on the interior of the curvature, whereas convex mirrors have the reflecting side on the outside of the mirror.
A few characteristics of curved mirrors follow.
- The curvature, focal distance, and radius are related by the following expression:
- The mirror equation connects the characteristic distances:
- Magnification of the image is
CV = R = 2 · f
R = 2 · f
- ho and hi are greater than 0 if the object/image is upright and are less than 0 if inverted.
- di is greater than 0 if the image is at the left of the mirror and is less than 0 if at right.
- f is greater than 0 if the focal point is real and is less than 0 if the focal point is virtual.
- m is greater than 0 when the image is upright and is less than 0 when the image is inverted.
- All parallel light rays passing through the curvature C will be reflected by the mirror on the same path as the incident light (Ray 1 in Figure 19.8).
- Light passing through the focal point will be reflected by the mirror on a direction parallel to the optical axis (Ray 2). The reverse path is also possible; light parallel to the optical axis is reflected through the focal point.
- All light rays directed along the optical axis toward the mirror will be reflected back on the same path as the incident light ray (Ray 3).
- The focal point is real for the concave mirrors and virtual for the convex mirrors (see Figure 19.9 as only the extension of the ray passes through the focal point because the mirror is opaque).
Let us use the previous information in an example. Remember that to obtain an image point, you need to intersect at least two rays or extensions of rays.
An object 3 cm high is placed in front of a concave mirror at a distance of 5 cm, and the radius of the mirror is 12 cm. Find the characteristics of the image obtained in the mirror.
We will convert all measurements in meters, draw the geometric diagram of the experiment, and set up our equations.
ho = 3 cm = 0.03 m
do = 5 cm = 0.05 m
R = 12 cm = 0.12 m
Image characteristics = ?
f = R/2 = O. 06 cm
All three rays were drawn as in the previous example. Without any calculation, we can draw qualitative conclusions on the image (see Figure 19.10):
- Virtual (extension of rays)
- Larger than the object
The equations defined before will have to agree with the geometry of the setup.
We solve for di:
Because di is negative, it means that the image is at the right of the mirror, and the object will be virtual. Because the mirror is opaque, the light rays cannot pass through it.
The magnification m = 6 (m > 0 and m > 1) means the image is larger than the object and upright.
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