Which of the following is the correct magnification?

 

Magnification is also equal to the image-to-object distance ratio. Because the object is always above the principal axis, its height is always positive. However, the sign for image height may vary depending on the type of image formed.

$\text{M}=\frac{\text{V}}{\text{u}}$

Because the object is always above the principal axis, its height is always positive. However, the sign for image height may vary depending on the type of image formed. The height of virtual images should be assumed to be positive, whereas the height of real images should be assumed to be negative.

Using mirror formula we get $\frac{1}{\text{f}}=\frac{1}{\text{u}}+\frac{1}{\text{v}}$

$\frac{1}{\text{f}}-\frac{1}{\text{u}}=\frac{1}{\text{v}}$

$\frac{1}{\text{v}}=\frac{\text{u-f}}{\text{uf}}$

$\text{v=}\frac{\text{uf}}{\text{u-f}}$

$\text{M}=\frac{\frac{{\text{u}}^f}{\text{u}-\text{f}}}{\text{u}}=\frac{\text{f}}{\text{u}-\text{f}}$