Now we can turn to the case of lenses. A lens can be considered the
combination of two spherical interfaces. To solve an optical problem using
multiple interfaces or lenses, one considers each one by one. For example one
finds the image created by the first surface and then uses it as the object of
the second surface.
Consider
a lens of thickness
d
d
as shown in the drawing. At interface 1 (coloured red in the drawing) we
have
n
1
s
o
1
+
n
2
s
i
1
=
n
2
−
n
1
R
1
n
1
s
o
1
+
n
2
s
i
1
=
n
2
−
n
1
R
1
For
surface 2 (coloured blue) the image of the the first surface becomes the object
of the second. Note the sign of
s
o
2
s
o
2
is negative so that
s
o
2
=
d
−
s
i
1
s
o
2
=
d
−
s
i
1
Thus
n
2
s
o
2
+
n
3
s
i
2
=
n
3
−
n
2
R
2
n
2
s
o
2
+
n
3
s
i
2
=
n
3
−
n
2
R
2
becomes
n
2
d
−
s
i
1
+
n
3
s
i
2
=
n
3
−
n
2
R
2
n
2
d
−
s
i
1
+
n
3
s
i
2
=
n
3
−
n
2
R
2
Now
add the equations
n
1
s
o
1
+
n
2
s
i
1
+
n
2
d
−
s
i
1
+
n
3
s
i
2
=
n
2
−
n
1
R
1
+
n
3
−
n
2
R
2
n
1
s
o
1
+
n
2
s
i
1
+
n
2
d
−
s
i
1
+
n
3
s
i
2
=
n
2
−
n
1
R
1
+
n
3
−
n
2
R
2
Now
we take the thin lens case,that is
d
→
0
d
→
0
n
1
s
o
+
n
3
s
i
=
n
2
−
n
1
R
1
+
n
3
−
n
2
R
2
n
1
s
o
+
n
3
s
i
=
n
2
−
n
1
R
1
+
n
3
−
n
2
R
2
That
equation would work for making prescription swim goggles for example, however
most of the time
n
1
=
n
3
n
1
=
n
3
(namely air for eyeglasses). So making that the case we get
n
1
s
o
+
n
1
s
i
=
(
n
2
−
n
1
)
[
1
R
1
−
1
R
2
]
n
1
s
o
+
n
1
s
i
=
(
n
2
−
n
1
)
[
1
R
1
−
1
R
2
]
which
in the case of air (n=1) is
1
s
o
+
1
s
i
=
(
n
l
−
1
)
[
1
R
1
−
1
R
2
]
.
1
s
o
+
1
s
i
=
(
n
l
−
1
)
[
1
R
1
−
1
R
2
]
.
That
is The lensmaker's formula(where
n
l
n
l
is the index of refraction of the lens material)Now we can find
the foci
f
i
=
lim
s
i
→
∞
f
i
=
lim
s
i
→
∞
f
o
=
lim
s
o
→
∞
f
o
=
lim
s
o
→
∞
Which
we see from the lensmaker's formula must be the same so lets drop the
subscripts:
1
f
=
(
n
l
−
1
)
[
1
R
1
−
1
R
2
]
1
f
=
(
n
l
−
1
)
[
1
R
1
−
1
R
2
]
and
1
s
o
+
1
s
i
=
1
f
1
s
o
+
1
s
i
=
1
f
which
is the Gaussian Lens Formula
A convex lens will have a positive focal length
f
f
.
We can use the same equation as before for the magnification,
m
=
h
i
h
o
=
−
n
1
s
i
n
2
s
o
.
m
=
h
i
h
o
=
−
n
1
s
i
n
2
s
o
.
but
now note that
n
1
=
n
2
n
1
=
n
2
so the equation becomes
m
=
h
i
h
o
=
−
s
i
s
o
.
m
=
h
i
h
o
=
−
s
i
s
o
.
You
can examine the figure above to verify that this is true.
We can also consider a concave lens which has a negative focal length. Notice
that in this case the image is upright and virtual. Notice that in this case
s
i
s
i
is
negative.
