Use the mirror equation to show that an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

#### Solution

According to the mirror equation, we have

`1/"u" + 1/"v" = 1/"f"`

where

u = Distance of object from the mirror

v = Distance of image from the mirror

f = Focal length of the mirror

From mirror equation, we have

`"v" = ("uf")/("u" - "f")` ...........(i)

Applying new cartesian sign convention, we get

f = −ve and u = −ve

Given: f < u < 2f

v = −ve .....................[From (i)]

Magnification is given by

`"m" = -(-"v")/(-"u")` = −ve

Hence, the image formed is real.

From mirror formula, when u = −2f,

`1/(-2"f") + 1/"v" = 1/-"f"`

`1/"v" = 1/(2"f") - 1/"f" = -1/(2"f")`

Therefore, when the object is at 2f, the image is formed at 2f

Now, when u = f

`1/-"f" + 1/"v" = -1/"f"`

`1/"v" = 1/0`

v = ∞

Therefore, when the object is at f, the image is formed at infinity.

This shows that when f < u < 2f, ∞ < v < 2f.