Question

Use a mirror equation to show that an object placed between $f$ and $2f$ of a concave mirror forms an image beyond $2f$.

Answer 1

Hint: To solve this problem, first we have to know what is the mirror equation, and the sign of the focus value of the concave mirror. The focus value of the concave mirror is negative because the concave mirror reduces the distance of the focus of the object.

Useful formula
The mirror equation is given by,
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Where, $f$ is the focus of the object, $v$ is the distance of the image from the mirror and $u$ is the distance of the object from the mirror.

Complete step by step solution
Given that,
The object is placed between $f$ and $2f$ of a concave mirror.
The mirror equation is given by,
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\,...................\left( 1 \right)$
When the object lies between the $f$ and $2f$, then
At $u = f$, in the mirror equation, then
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{f}$
By rearranging the terms, then the above equation is written as,
$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{f}$
By subtracting the above equation, then the above equation is written as,
$\dfrac{1}{v} = \dfrac{0}{f}$
By taking reciprocal then, the above equation is written as
$v = \dfrac{f}{0}$
On dividing the above equation, then the above equation is written as,
$v = \infty $
At $u = 2f$, in the mirror equation, then
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{{2f}}$
By rearranging the terms, then the above equation is written as,
$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{{2f}}$
By cross multiplying the terms, then the above equation is written as,
$\dfrac{1}{v} = \dfrac{{2f - f}}{{2{f^2}}}$
On further simplification, then the above equation is written as,
$\dfrac{1}{v} = \dfrac{f}{{2{f^2}}}$
On dividing the terms, then the above equation is written as,
$\dfrac{1}{v} = \dfrac{1}{{2f}}$
By taking reciprocal, then
$v = 2f$
The above equation is also written as, $v \geqslant 2f$

Thus, the object is real and formed beyond the $2f$.

Note: The alternative method is, for concave mirrors the focus of the object is less than zero and the distance of the image is also less than zero. By using this condition, the distance of the object is greater than $2f$, then the solution is the image is formed beyond $2f$.