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Numerical

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.

Concept: Ray Optics - Mirror Formula

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