Question

What are conjugate foci? Deduce the following relationship between the focal length $f$, of a spherical mirror, distance of object of u and distance of image v. $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$.

Answer 1

The sets of points on the principal axis of a lens where the position of the object and the image can be interchanged are called conjugate foci.

For concave mirror:
Let MPN=Concave mirror
P=Pole
F=Principle focus
C=Centre of curvature
PC=Principle axis
AB=Principle acis
A;B'=Real image of object AB formed by concave mirror
DE=Perpendicular drawn from D on the principle axis
In $\Delta$ABC and $\Delta A^{\prime }{B}^{\prime }{C}^{\prime }$
$\angle$ABC$=\angle$A'B'C'(each ${90}^o$)
$\angle$ACB$=\angle$A'CB'(vertically opposite angles)
Hence, $\frac{AB}{A^{\prime }{B}^{\prime }}=\frac{CB}{B^{\prime }C}$ ........(1)
In $\Delta$DEF and $\Delta$A'B'F
$\angle$DEF$=$A'B'F(each ${90}^o$)
$\angle$DFE$=$A'FB'(vertically opposite angles)
Hence, $\frac{DE}{A^{\prime }{B}^{\prime }}=\frac{EF}{F{B}^{\prime }}$
But; DE$=$AB
$\therefore \frac{AB}{A^{\prime }{B}^{\prime }}=\frac{EF}{F{B}^{\prime }}$ ......(2)
From equation (1) and (2)
$\frac{CB}{B^{\prime }C}=\frac{EF}{F^{\prime }{B}^{\prime }}$
If the aperture of mirror is very small i.e., point E is very near to point P.
Then, $EF=PF$
$\therefore \frac{CB}{B^{\prime }C}=\frac{PF}{F{B}^{\prime }}$
$\frac{PB-PC}{PC-P{B}^{\prime }}=\frac{PF}{P{B}^{\prime }-PF}$ ......(3)
Applying sign convention, $PB=-U,PC=-R=-2f$
$P{B}^{\prime }=-y,PF=-f$
Put in equation (3)
$\frac{-U+2f}{-2f+v}=\frac{-f}{-v-(-f)}$
$\frac{-U+2f}{-2f+v}=\frac{-f}{-v+f}$
$(-u+2f)(-v+f)=-f(-2f+v)$
$uv-uf-2fv+2f^2=2f^2-fv$
$uv=uf+2fv-fv$
$uv=uf+fv$
Dividing uvf, we get
$\frac{uv}{uvf}=\frac{uf}{uvf}+\frac{fv}{uvf}$

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

This is the mirror formula for concave mirror.
(2) For convex mirror:
Let, MPN$=$convex mirror
P$=$Pole
F$=$Principle focus
C$=$Centre of curvature
PC$=$Principle axis
AB$=$Object
A'B'$=$Image of object AB formed by convex mirror
DE$=$Perpendicular drawn from point D on the principle axis.
In $\Delta$ABC and $\Delta$A'B'C
$\angle$ABC$=\angle$A'B'C(each ${90}^o$)
$\angle$ACB$=\angle$A'CB'(Common angles)
Hence,
$\frac{AB}{A^{\prime }{B}^{\prime }}=\frac{BC}{B^{\prime }C}$ .....(1)
In $\Delta$DEF and $\Delta$A'B'F
$\angle$DEF$=\angle$A'B'F(each ${90}^o$)
$\angle$DEF$=\angle$A'FB'(Common angles)
$\frac{DE}{A^{\prime }{B}^{\prime }}=\frac{EF}{B^{\prime }F}$ ......(2)
From equation (1) and (2)
$\frac{BC}{B^{\prime }C}=\frac{PF}{B^{\prime }F}$
If the aperture of mirror is very small i.e., point E is very near to point P then EF$=$PF(approx.)
$\frac{BC}{B^{\prime }C}=\frac{PF}{B^{\prime }F}$
$\frac{PB+PC}{PC-P{B}^{\prime }}=\frac{PF}{PF-P{B}^{\prime }}$ ........(3)
Applying sign convention,
$PB=-u,PC=+R=+2f$
$P{B}^{\prime }=+v,PF=+f$
Put in equation (3),
$\frac{-U+2f}{2f-v}=\frac{f}{f-v}$
$(-u+2f)(f-v)=f(2f-v)$
$-uf+uv+2f^2=2f^2-fv$
$uv=uf+2fv-fv$
$uv=uf+fv$
Dividing uvf, we get
$\frac{uv}{uvf}=\frac{uf}{uvf}+\frac{fv}{uvf}$

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

This is the mirror formula for convex mirror.