Worksheet-Lec15

# Worksheet-Lec15 - “object” created by the second lens...

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Physics 123 Interactive Lecture Monday, May 4, 2009 Multiple Lenses A 7-cm-height luminous object is placed 30 cm to the left of a converging lens of focal length 12 cm as shown. A diverging lens of focal length -14 cm is 48 cm further to the right. A person looks at this setup from the far right as shown. Our goal is to locate the final image seen by the person and determine its height, whether it is real or virtual, and whether it is upright or upside down. 1. If only the first (converging) lens were present, where would it produce an image? Compute an answer and draw in the ray diagram (lightly) showing the image on the figure above. Consult with your neighbors as needed. 2. Assume that the image you found above is now the object. Determine its distance from the second (diverging) lens. Now compute the location of the image of this

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Unformatted text preview: “object” created by the second lens. Again, draw in the ray diagram (lightly) on the figure above. Consult with your neighbors. The figure is repeated on the back of this page for your convenience. page 1 of 2 48 cm 30 cm x x • 12 cm-14 cm • Physics 123 Interactive Lecture Monday, May 4, 2009 3. The magnification equation works exactly like the lens/mirror equation. That is, find the magnification of the image created by the first lens, then repeat for the mirror, etc., until you reach the final image. The overall magnification is the product of your individual answers. 4. You should now have the answers to all parts of the original goal. page 2 of 2 48 cm 30 cm x x • 12 cm-14 cm •...
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Worksheet-Lec15 - “object” created by the second lens...

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