Monday, May 22, 2006

16

Ray Tracings – Mirrors


When doing ray tracings, there are three rays that follow simple known paths before and after their refraction by the mirror, just as there were with lenses. Use two of the three construction rays to find the image.

  • Draw a line from the object to the mirror, parallel with the direction of light. At the lens, draw the line through the primary focal point of the mirror.
  • Draw a line from the object, through the center of curvature of the mirror, then to the mirror.
  • Draw a line from the object to F and then to the mirror. At the mirror, draw the line back parallel to the axis of the mirror.



Click on image to enlarge.



Click on image to enlarge.

Where the lines cross is where the image will be formed. By doing this, you can estimate the approximate location of the image, tell whether the image is erect or inverted as well as real or virtual.



Figure 23 shows the construction of mirror images produced by a concave mirror:

  • Object farther from mirror than C, image real and inverted
  • Object at C, image also at C, real and inverted
  • Object at F, image at infinity
  • Object closer to mirror than F, image virtual and erect



Click on image to enlarge.

Figure 24 shows the construction of mirror images produced by a convex mirror:

  • Object real, image virtual, erect and minified.



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Plane Mirror: (Figure 25)

This mirror has a power of zero. Therefore U = V and m = +1. This indicates that any real object has a virtual erect image of the same size and any virtual object has a real, erect image of the same size. The virtual image of a real object will be located as far behind the mirror as the real image is in front of the mirror.



Click on image to enlarge.


Question: An object is placed 1m to the left of a concave mirror with a radius of curvature of 20cm. Where will the image be focused? Will the image be real or virtual; magnified or minified; erect or inverted? Additionally, do a line drawing to show these results. (Figure 26)



Answer:

The focal length of the mirror (f) = r/2 = 0.2/2 = 0.10m.

The power of the mirror (Dm) = 1/f = 1/0.10 = +10.00D.


The position of the image is:

u = object distance = -1m
U = object vergence = 1/u = 1/(-1) = -1.00D
Dm = reflecting power of the mirror = +10.00D

V = image vergence = U + D m = -1.00D + (+10.00D) = +9.00D

Image position is v = 100/V = 100/+9.00 = +11.11cm.

Magnification = U/V = -1.00D/+9.00D = -0.11.

Therefore, the image will be real, inverted and minified.


Click on image to enlarge.



Question: (Figure 27) An object is placed 0.5m to the left of a cornea with a radius of curvature of 10mm. Where will the image be focused? Will the image be real or virtual; magnified or minified; erect or inverted? Additionally, do a line drawing to show these results.

Answer:

The focal length of the cornea (f) = -(r/2) = -0.01/2 = -0.005m.

The reflective power of this cornea is (Dm) = 1/f = 1/-0.005 = -200.00D.

U = object vergence = 1/u = 1/(-0.5) = -2.00

V = image vergence = U + Dm = -2.00D + (-200.00D) = -202D

Image position = v = 100/V = 100/-202 = -4.95mm.

Magnification = M = U/V = -2/-202 = 0.0099.

Therefore, the image will be virtual, erect and minified.


Click on image to enlarge.

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