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.
