Question

What is Mirror formula? Derive the relation $\frac{1}{u}+\frac{1}{v}=\frac{2}{R}$ for concave mirror

Answer 1

The formula which gives the relation between object distance $\left(u\right)$, the image distance $\left(v\right)$ and the focal length $\left(f\right)$ of the mirror is known as mirror formula.

In the figure shown above, an object $AB$ is placed at a distance $u$ from the pole of the concave mirror of small aperture, just beyond the center of curvature. Hence, its real,inverted and diminished image $AB$ is formed at a distance $v$ in front of the mirror.

According to Cartesian sign convention,

Object distance $\left(PB\right)=-u$

Image distance $\left(PB\right)=-v$

Focal length $\left(PF\right)=-f$

Radius of curvature $\left(PC\right)=-R$

It is clear from the geometry of the figure, right angled $△$ ABP and $△$ABP are similar.

$\therefore \frac{A^{\prime }{B}^{\prime }}{AB}=\frac{P^{\prime }{B}^{\prime }}{PB}=\frac{-v}{u}$

$\therefore \frac{A^{\prime }{B}^{\prime }}{AB}=\frac{v}{u}$ ........(i)

Similarly, $△$ $ABC$ and $△$ $A^{\prime }{B}^{\prime }C$' are similar.

$\therefore \frac{A^{\prime }{B}^{\prime }}{AB}=\frac{C^{\prime }{B}^{\prime }}{CB}$ .......(ii)

From figure,

$C{B}^{\prime }P{C}^{\prime }P{B}^{\prime }=-R-(-v)=-R+v$

and $CB=PB-PC=-u-(-R)=-u+R$

From (ii),

$\frac{A^{\prime }{B}^{\prime }}{AB}=\frac{-R+V}{-u+R}$ ...... (iii)

Comparing (i) and (iii),

$\frac{v}{u}=\frac{-R+v}{-u+R}$

$\therefore -uv+Rv=-Ru+vu$

or, $R(u+v)=2uv$

$\therefore \frac{1}{v}+\frac{1}{u}=\frac{2}{R}$ [Dividing both sides by $Ruv$.]

Hence, Proved.