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Question
Use the mirror equation to deduce that an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
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Solution
For a concave mirror, the focal length (f) is negative.
∴ f < 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u < 0
It is placed between the focus (f) and the pole.
∴ f > u > 0
`1/"f" < 1/"u" < 0`
`1/"f" - 1/"u" < 0`
For image distance v, we have the mirror formula:
`1/"v" + 1/"u" = 1/"f"`
`1/"v" = 1/"f" - 1/"u"`
∴ `1/"v" < 0`
∴ v > 0
The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write:
`1/"u" > 1/"v"`
v > u
Magnification, m = `"v"/"u" > 1`
Hence, the formed image is enlarged.
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