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dorthod
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Can someone get me started on this, I know how to work a diverging lens and a concave mirror, just not together. The problem:
A concave mirror with a radius of curvature of 20.0 cm is placed 25.0 cm from a diverging lens with a focal length of 16.7 cm. An object is placed midway between the lens and the mirror. Image you are looking through the lens and considering only the light that leaves the object and travels first to the mirror:
a. Locate the image formed by the mirror(which becomes a virtual object for the lens) by using a ray diagram, and then verify that image distance with the lens equation.
b. Locate the final image formed by the lens/mirror system, using a ray diagram, and then verify that image distance with the lens equation.
c. Is this image real or virtual? Explain why this is not such a simple answer.
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Here's what I can do:
a)1/10 = 1/Si + 1/12.5(not sure though about 12.5)
50cm = Si
b)1/16.7 = 1/Si - 1/50(assuming part a is correct)
12.52 = Si
c) No clue
Image I cooked up:
http://home.earthlink.net/~emerson6/lensmirror.gif
A concave mirror with a radius of curvature of 20.0 cm is placed 25.0 cm from a diverging lens with a focal length of 16.7 cm. An object is placed midway between the lens and the mirror. Image you are looking through the lens and considering only the light that leaves the object and travels first to the mirror:
a. Locate the image formed by the mirror(which becomes a virtual object for the lens) by using a ray diagram, and then verify that image distance with the lens equation.
b. Locate the final image formed by the lens/mirror system, using a ray diagram, and then verify that image distance with the lens equation.
c. Is this image real or virtual? Explain why this is not such a simple answer.
----------------------------
Here's what I can do:
a)1/10 = 1/Si + 1/12.5(not sure though about 12.5)
50cm = Si
b)1/16.7 = 1/Si - 1/50(assuming part a is correct)
12.52 = Si
c) No clue
Image I cooked up:
http://home.earthlink.net/~emerson6/lensmirror.gif
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