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Propagation of Light Study Guide (page 2)

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Updated on Sep 27, 2011

Example 2

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.

Solution 2

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.

Images in Curved Mirrors

A few characteristics of curved mirrors follow.

  • The curvature, focal distance, and radius are related by the following expression:
  • CV = R = 2 · f

    R = 2 · f

  • The mirror equation connects the characteristic distances:
  • Magnification of the image is
  • 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).
  • Images in Curved Mirrors

  • 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).

Images in Curved Mirrors

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.

Example

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.

Solution

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
  • Upright

Images in Curved Mirrors

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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