Question

Magnification of mirror equals $-\frac{v}{u}$why the minus sign is here?

Answer 1

Step 1: Magnification

1. Magnification produced by a lens is defined as the ratio pf the size of the image formed by the lens to the size of the object.
2. If size of the image formed by the lens is $h_2$ and the original size of the image is $h_1$ then the magnification is given as,

$m=\frac{h_2}{h_1}$

Step 2: Magnification of concave mirror

The figures show the ray diagram for the formation of image $A\text{'}B\text{'}$ of a finite object $AB$ by a concave mirror.

Now, $△APB$ is similar to $∆A\text{'}PB\text{'}$

$\therefore \frac{A\text{'}B\text{'}}{AB}=\frac{PB\text{'}}{PB}$

Upon applying the sign convention we get,

$A\text{'}B\text{'}=-h_2$ (Downward image height)

$AB=+h_1$ (Upward object height)

$PB=-u$ (Object distance on left)

$PB\text{'}=-v$ (Image distance on left)

$\therefore \frac{-h_2}{+h_1}=\frac{-v}{-u}$

$\therefore$Magnification, $m=\frac{h_2}{h_1}=-\frac{v}{u}$

Step 3: Magnification of convex mirror

Figure shows the formation of image $A\text{'}B\text{'}$ of a finite object $AB$ by a convex mirror.

Now, $△APB$ is similar to $∆A\text{'}PB\text{'}$

$\therefore \frac{A\text{'}B\text{'}}{AB}=\frac{PB\text{'}}{PB}$

Upon applying the sign convention we get,

$A\text{'}B\text{'}=+h_2$ (Downward image height)

$AB=+h_1$ (Upward object height)

$PB=-u$ (Object distance on left)

$PB\text{'}=+v$ (Image distance on left)

$\therefore \frac{+h_2}{+h_1}=\frac{+v}{-u}$

$\therefore$Magnification, $m=\frac{h_2}{h_1}=-\frac{v}{u}$

Hence, the minus sign in magnification indicates the use of a spherical mirror for the formation of the image of an object.