Question

Derive the relationship between the focal length and radius of curvature of spherical mirror.

Answer 1

MPN$=$ Concave spherical mirror
P$=$Pole
F$=$ Principle focus
C$=$Centre of curvature
PC$=$Principle axis
f$=$ Focal length of mirror
R$=$ Radius of curvature of mirror
AD$=$ incident ray
EF$=$reflected ray
DC$=$ Normal at point D
Angle of incidence$=\angle$ADC
Angle of reflection$=\angle$CDF
$\angle$ADC$=\angle$ CDF$=\theta$(let)(By second law of reflection)
$\angle$ADC$=\angle$DCF$=\theta$(alternate angles)
In $\Delta$DCF,
$\angle$CDF$=\angle$DCF
$\Delta$DCF is a isocsleles triangle
$\therefore DF=PC\left[\begin{array}{c}sidesopp.toequal.\\ anglesinatriangleare\\ equal\end{array}\right]$
If the aperture of mirror is very small i.e., point D is very near to point P then.
$DF=PF$(approx)
Hence, $PF=FC$
$PE+PF=PF+FC$
$2PF=PC$ ...(i)
Applying sign convention,
$PF=-f,PF=R$
Put in eq. (1)
$2(-f)=-R$

$R=2f$

OR

$f=\frac{R}{2}$

Hence, focal length of concave mirror is half of its radius of curvature.
For convex mirror:
Let, MPN$=$convex spherical mirror
P$=$Pole
F$=$ Principle focus
C$=$ Centure of curvature
PC$=$ Principle axis
f$=$ Focal length of mirror
R$=$ Radius of curvature of mirror
AD$=$ Incident ray
DB$=$Reflected ray
CE$=$ Normal at point D
Angle of incidence$=\angle$ADE
Angle of reflection$=\angle$EDB
$\angle$ADE$=\angle$EDB$=\theta$(let) (By $2^{nd}$ law of reflection)
$\angle$EDB$=\angle$ FDC$=\theta$(vertically opposite angles)
$\angle$ADE$=\angle$FCD$=\theta$(corresponding angles)
$\Delta$DCF is a isosceles triangle
$\therefore DF=FC\left[\begin{array}{c}Sidesopp.toequal\\ anglesinatriangleare\\ equal\end{array}\right]$
If the aperture of mirror is very small i.e., point D is very near to point P then.
$DF=PF$
Hence, $PF=FC$
$PF+PF=PF+FC$ .......(1)
Applying sign convetion, $
$PF=+f,PC=+R$
Put in eq.(1)

$R=2f$

OR

$f=\frac{R}{2}$

Hence focal length of concave mirror is half of its radius of curvature