Notice
that in Mirrors we get a virtual image to the right of the surface. Thus for
mirrors we say that
s
i
s
i
is positive to the left of the mirror. This allows to retain
correspondence between
s
i
s
i
being negative and an image being virtual.
Again we use the small angle approximation. By inspection of the figure we see
that
2
θ
i
=
α
+
β
2
θ
i
=
α
+
β
and
θ
i
=
α
+
γ
.
θ
i
=
α
+
γ
.
Now
we multiply the second equation by two and subtract the first equation from it
and we get:
2
α
+
2
γ
−
α
−
β
=
0
2
α
+
2
γ
−
α
−
β
=
0
or
α
−
β
=
−
2
γ
.
α
−
β
=
−
2
γ
.
Using
the small angle approximation we see that this is
h
s
o
+
h
s
i
=
−
2
h
r
h
s
o
+
h
s
i
=
−
2
h
r
where
I have used the fact that
s
i
s
i
is negative to the right of the mirror. So I can write the mirror equation as
1
s
o
+
1
s
i
=
−
2
R
1
s
o
+
1
s
i
=
−
2
R
or
1
s
o
+
1
s
i
=
1
f
1
s
o
+
1
s
i
=
1
f
where
for a mirror
1
/
f
=
−
2
/
R
1
/
f
=
−
2
/
R
