Question

The linear magnification for a mirror is the ratio of the size of the image to the size of the object, and is denoted by m. then m is equal to (symbols have their usual meanings)
$\begin{array}{rl}& \text{A}\text{. }{\displaystyle \frac{uf}{u-f}}\\ & \text{B}\text{. }{\displaystyle \frac{u+f}{u-f}}\\ & \text{C}\text{. }{\displaystyle \frac{f}{f-u}}\\ & \text{D}\text{. }{\displaystyle \frac{f+u}{u-f}}\end{array}$

Answer 1

Hint: Magnification of mirror is expressed as the ratio of the image height to the object height. Express it in terms of the image distance and the object distance. Obtain the expression for the mirror equation for spherical mirrors. Make the required modification to express the linear magnification

Complete step-by-step answer:
Linear magnification of a spherical mirror can be defined as the ratio of the height of the image produced by the lens to the height of the object. We can also define the linear magnification in terms of the object distance and the image distance as the negative of the ratio of the distance of the image from the lens to the distance of the object from the lens. Mathematically we can express the linear magnification as,
$m={\displaystyle \frac{h^{\prime }}{h}}=-{\displaystyle \frac{v}{u}}$
Where, m is the linear magnification of the mirror, h is the height of the object, $h^{\prime }$ is the height of the image, u is the object distance and v is the image distance.
Now, we can express the linear magnification in terms of the focal length of the mirror and object distance with the help of the mirror equation.
The mirror equation is mathematically expressed as,
${\displaystyle \frac{1}{u}}+{\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{f}}$
Where, f is the focal length of the mirror.
Multiply both of the above equations by the object distance u.
$\begin{array}{rl}& {\displaystyle \frac{u}{u}}+{\displaystyle \frac{u}{v}}={\displaystyle \frac{u}{f}}\\ & 1+{\displaystyle \frac{u}{v}}={\displaystyle \frac{u}{f}}\\ & {\displaystyle \frac{u}{v}}={\displaystyle \frac{u}{f}}-1\\ & {\displaystyle \frac{u}{v}}={\displaystyle \frac{u-f}{f}}\\ & {\displaystyle \frac{v}{u}}={\displaystyle \frac{f}{u-f}}\end{array}$
Now, linear magnification,
$\begin{array}{rl}& m=-{\displaystyle \frac{v}{u}}=-{\displaystyle \frac{f}{u-f}}\\ & m={\displaystyle \frac{f}{f-u}}\end{array}$

So, the correct answer is “Option C”.

Note: If the magnitude is higher than 1, then the image formed is bigger than the object. If the magnitude is less than 1, then the image is smaller than the object. A negative magnification gives us that the image is real and inverted. A positive magnification tells us that the image is virtual and erect.