# Relation Between Focal Length And Radius of Curvature

### For Concave Mirror

​Let the concave spherical mirror has focal length $f$ and radius of curvature $R$. Let $P$, $F$ and $C$ be its pole, focus and centre of curvature respectively. Consider a ray of light $OA$ is incident on the mirror which is parallel to the principal axis of the mirror. Join $AC$. Then, $AC$ is normal to the mirror at point $A$. Let the ray of light gets reflected along $AB$ cutting the pricipal axis at point $F$. Let $i$ be the angle of incidence and $r$ be the angle of reflection.

​Then, $\text{angle of incidence} = \text{angle of reflection}$ $∠OAC=∠CAF\text{ __(1)}$ also, $∠OAC= ∠ACF\text{ __(2) [Being alternate angles]}$

From $(1)$ and $(2)$, $∠CAF=∠ACF$ Thus, $ΔACF$ is an isosceles triangle. $∴CF=AF\text{ __(3)}$ Now, $CP=CF+FP\text{ __(4)}$

From $(3)$ and $(4)$, $CP=AF+FP\text{ __(5)}$ The points $A$ and $P$ lies very close to each other because the aperture of the mirror is small. Thus, $\text{when }A\to P\text{,}$ $AF ≈FP$

Therefore, equation $(4)$ becomes, $CP=FP+FP$ $CP=2FP$ $FP=\frac{1}{2}CP$ $∴f=\frac{1}{2}R$ This is the relation between focal length and radius of curvature for concave mirror.

[Also see: Lateral Shift]

### For Convex Mirror

​In the similar way, let the convex spherical mirror has focal length $f$ and radius of curvature $R$. Let $P$, $F$ and $C$ be its pole, focus and centre of curvature respectively. Consider a ray of light $OA$ is incident on the mirror which is parallel to the principal axis of the mirror. Then, the ray of light gets reflected along $AB$. Join $AC$ and produce outward. Then, $AC$ is normal to the mirror at point $A$. Let $i$ be the angle of incidence and $r$ be the angle of reflection.

​Then, $\text{angle of incidence} = \text{angle of reflection}$ $∠OAN= ∠BAN\text{ __(a)}$

Now, $∠FAC=∠BAN\text{ __(b) [Being vertically opp. angles]}$ $∠FCA=∠OAN\text{ __(c) [Being corresponding angles]}$

From $(a)$, $(b)$ and $(c)$, $∠FAC=∠FCA$ Thus, $ΔACF$ is an isosceles triangle. $∴CF=AF\text{ __(3)}$ Now, $CP=CF+FP\text{ __(4)}$ From $(3)$ and $(4)$, $CP=AF+FP\text{ __(5)}$

Since the aperture of the mirror is small, $\text{when }A\to P\text{,}$ $AF ≈FP$ Therefore, equation $(4)$ becomes, $CP=FP+FP$ $CP=2FP$ $FP=\frac{1}{2}CP$ $∴f=\frac{1}{2}R$ This is the relation between focal length and radius of curvature for convex mirror.

[Also See: Dispersion]