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A
tall object is located to the left of a lens. The image formed by the lens is tall and located to the left of the lens. Is this a converging or diverging lens, and
what is the focal length of the lens? |
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- sign indicates diverging lens! |
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So Correct Answer is B ! |
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A
object is to the left of a lens. The image is found to be tall and upright. What is the focal length of the lens, and
it is a diverging or converging lens? |
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So image distance is |
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Now find focal length |
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- sign indicates diverging lens! |
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So Correct Answer is A ! |
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An
object is located to the left of a converging lens with a
focal length of . A diverging lens with a focal length of is placed to the right of the converging lens. Where is the final image of the two lenses
formed, relative to the diverging lens?
Is this image upright or inverted? |
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The first lens creates an image
found from |
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This is to the right of the
converging lens. This is a real image
and would therefore be inverted. So,
with the second lens 30 cm to right of converging lens, this image acts as
the object of the second lens. |
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Note: the – sign indicates a virtual
object. Now find the second image. |
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This is positive so it is a real
image 4.99 cm to right of diverging lens.
Now since the object was virtual the image retains the same
orientation, so the virtual object was inverted since the first image was
real and inverted. Thus, the final
image is also inverted. |
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Also consider the magnifications |
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So is negative, so is negative and image is inverted! |
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So Correct Answer is C ! |
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A
person has a far point (distance which they can see clearly without
corrective lenses) of . They would like to be able see objects up
to away.
What type of contact lenses (converging or diverging) and with what
focal length should the lens have to correct their vision? |
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So the image distance is the far
point but remember this is a virtual image so, , while the object distance is the
distance you want to now make out or .
Use the lens equation |
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negative focal length, so diverging
lens! |
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So the Correct Answer is D ! |
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A
person has a near point of . What should be the focal length of a lens,
which sits from their eye, which would allow them to
read at the usual near point of ? |
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So the image distance is the near
point but remember first you need to
subtract off the distance to the eye, and remember this is a virtual image so, , while the object distance is the
distance you want to now make out or .
Use the lens equation |
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So Correct answer is B ! |
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