Question

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

Answer 1

Hint: This question can be solved by the application of the mirror formula. When the image distance is negative, the image is behind the mirror.
Formula Used: The formula used in the solution is given here.
${\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$ where f is the focal length, $u$ is the object distance, $v$ is the image distance.

Complete Step by Step Solution: Spherical mirrors can be thought of as a portion of a sphere that was sliced away and then silvered on one of the sides to form a reflecting surface. Concave mirrors are silvered on the inside of the sphere. The point on the mirror's surface where the principal axis meets the mirror is known as the vertex and the distance from the vertex to the center of curvature is known as the radius of curvature. The radius of curvature is the radius of the sphere from which the mirror was cut. Since the focal point is the midpoint of the line segment adjoining the vertex and the center of curvature, the focal length would be one-half the radius of curvature.
It is an equation relating object distance and image distance with focal length is known as a mirror equation. It is also known as a mirror formula.
${\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$ where f is the focal length, $u$ is the object distance, $v$ is the image distance.
All distances are measured from the mirror as the origin. A concave mirror has a positive focal length while a convex mirror has a negative focal length.
It has been given that an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
Thus, for a concave mirror, $f>0$ and $u<0$.
As object lies between f and 2f, it can be written that, $u=-f$
By, the mirror formula,
${\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{f}}+{\displaystyle \frac{1}{f}}={\displaystyle \frac{2}{f}}$
$\Rightarrow v=\mathrm{\infty }$
Now, for $u=-2f$
From the mirror formula,
${\displaystyle \frac{1}{v}}=-{\displaystyle \frac{1}{f}}+{\displaystyle \frac{1}{2f}}=-{\displaystyle \frac{1}{2f}}$
$\Rightarrow v=-2f$
Thus, image distance,
$v⩾-2f$.
Since image distance is negative, image is formed at the same side of mirror as the object at a distance greater than 2f.
Magnification is given by: $m={\displaystyle \frac{-v}{u}}<0$.

Hence, the image formed is real.

Note: Alternatively, the mirror formula states that, ${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}$
For a concave mirror, $f<0,u<0$.
Thus, we can write that,
$2f<u<f$
$\Rightarrow {\displaystyle \frac{1}{2f}}>{\displaystyle \frac{1}{u}}>{\displaystyle \frac{1}{f}}$
Subtracting the inverse of focal length from each of the values in the equation,
${\displaystyle \frac{1}{2f}}-{\displaystyle \frac{1}{f}}>{\displaystyle \frac{1}{u}}-{\displaystyle \frac{1}{f}}>{\displaystyle \frac{1}{f}}-{\displaystyle \frac{1}{f}}=0$
$\Rightarrow -{\displaystyle \frac{1}{2f}}>{\displaystyle \frac{1}{v}}>0$ because, we already know that, ${\displaystyle \frac{1}{u}}>{\displaystyle \frac{1}{f}}>{\displaystyle \frac{1}{v}}$.
Simplifying the equation further, we can write,
${\displaystyle \frac{1}{2f}}<{\displaystyle \frac{1}{v}}<0$
$\Rightarrow v<0$
Thus, the image formed is real.