The thin lens equation and the paraxial equation say what happens to light as it travels through
one lens or one refracting surface, but there are times when we want to know what happens when light travels through
many lenses or refracting surfaces.
For example, look at Figure 1. There are two lenses here: the first lens has a power of
\( +2{\text{D}} \) and the second lens has a power of
\( +3{\text{D}} \), and they are \( 0.5{\text{m}} \)
apart. There is an object \( 0.75{\text{m}} \) to the left of the first lens. An image forms
\( 0.25\text{m} \) to the right of the second lens. The question we will be answering in this chapter is how to calculate that image distance.
Figure 1. Light travelling through two positive lenses separated by a gap of
\( 0.5\text{m} \). The lens powers are written above the lenses. The object is placed
\( 0.75\text{m} \) to the left of the first lens and the image forms
\( 0.25\text{m} \) to the right of the second. Note that in reality the light doesn't stop where it converges, as shown here. Instead, it continues until it hits something (click the check box to see).
Both lenses obey the thin lens equation. The equation for the first lens is
\[V_{in(1)}+F_{(1)}=V_{out(1)}\tag{lens 1}\]
where the \( (1) \) in the subscripts indicate that these quantities refer to the first lens. The object is
\( 0.75\text{m} \) to the left of the first lens, so
\( V_{in(1)}=1/(-0.75)=-1.333\text{D} \). Since \( F_{(1)}=+2\text{D}
\), this means \( V_{out(1)}=0.667\text{D} \).
The equation for the second lens is
\[V_{in(2)}+F_{(2)}=V_{out(2)}\tag{lens 2}\]
where the \( (2) \) indicates that these quantities refer to the second lens. To use this equation, we need to know what
\( V_{in(2)} \) is, but what is it?
The light hitting the second lens is not coming directly from an object, so we can't calculate
\( V_{in(2)} \) as \( 1/(\text{object distance})
\), like we did for the first lens. You might guess that
\( V_{in(2)} \) is just the same as \( V_{out(1)} \)
, but that would be wrong because the vergence of the light changes
as the light travels through the gap between the lenses.
We need to work out how the vergence changes as the light travels from the first lens to the second lens, and we can do this with a bit of trickery. Let's (temporarily) replace the second lens with a flat lens, which has no power (
\( F_{(2)}=0\text{D} \)). This is shown in
Figure 2. There are two things to note:
\( V_{in(2)} \) for the flat lens must be the same as for the original lens, because neither lens has done anything
to the light at that point.
Since the power of the flat lens is \( 0 \), the thin lens equation for the flat lens is just
\( V_{in(2)}=V_{out(2)} \).
So if we can work out \( V_{out(2)} \) for the flat lens, we know
\( V_{in(2)} \) for the flat lens, and so we know it for the original lens.
Figure 2. In this diagram, the second lens of
Figure 1 has been replaced by a lens with zero power.
The substituted flat lens has no power, so it doesn't change the vergence of the light, and the light converges to where it would have done if the lens wasn't there. The light leaves the first lens with a vergence of
\( V_{out(1)}=0.667\text{D} \), so it converges to a point which is
\( 1.5\text{m} \) from the first lens. This is only
\( 1\text{m} \) from the flat lens (\( 1.5-0.5=1 \)
, so \( V_{out(2)}=1\text{D} \).
Thus, \( V_{in(2)}=1\text{D} \). This doesn't depend on the power of the lens (point 1 above), so we've worked out
\( V_{in(2)} \) for the original lens, not just our substituted flat lens. Now we can return to the situation in
Figure 1.
Using the thin lens equation \( V_{in(2)}+F_{(2)}=V_{out(2)}
\), with \( V_{in(2)}=1\text{D} \), we can (finally) work out that
\( V_{out(2)}=1+F = 1+3 = 4\text{D} \). Thus the light converges to a point which is
\( 1/V_{out(2)}=0.25\text{m} \) from the second lens.
The Step-Along Procedure
While the above trick might be good for explaining why \( V_{in(2)}
\) is what it is, it's quite complicated to replace the lens with one of zero power, and then work out
\( V_{in(2)} \). To make life a little easier, we can just pull out the math steps from the above procedure to work out how vergence changes as light travels from one lens to another. The procedure is as follows:
For any two lenses separated by a gap, work out \( V_{out(1)}
\) for the first lens.
Work out the image distance \( d = 1/V_{out(1)} \) for the first lens (ignoring the second lens).
Subtract the gap width from the image distance \( d \)
. The gap width is always positive, but \( (d-\text{gap width})
\) might not be, either because \( d \) is negative or because the gap is bigger than
\( d \).
\( V_{in(2)} \) for the second lens is one over the result of step 3; that is
\( V_{in(2)} = 1/(d-\text{gap width}) \).
This is called the step-along procedure, because it tells you how vergence changes as we step-along from one lens to another in a sequence. For the two lenses in
Figure 1, this procedure works out as follows:
\( V_{out(1)} \) for the first lens is
\( 0.667\text{D} \)
The image distance \( d \) for the first lens is
\( d=1/0.667 = 1.5\text{m} \).
Subtracting the gap width \( 0.5\text{m} \) from the image distance gives us
\( 1.5-0.5 = 1\text{m} \)
\( V_{in(2)} \) for the second lens is one over the result of step 3, which is
\( V_{in(2)}=1/1 = 1\text{D} \).
Once we have \( V_{in(2)} \), we can work out
\( V_{out(2)} \) using the thin lens equation.
An Example.
Look at the pair of lenses in Figure 3. The first lens has a power of
\( F_{(1)}=+10{\text{D}} \). The second lens has a power of
\( F_{(2)}=+4{\text{D}} \). The lenses are separated by a gap of
\( 0.2{\text{m}} \). An object is placed
\( 0.5{\text{m}} \) to the left of the first lens. The vergence of the light entering the first lens is
\( V_{in(1)}=1/(-0.5)=-2{\text{D}} \).
Figure 3. Another example of light passing through two lenses. In this case, the light is converged to a real image between the two lenses, before diverging again to strike the second lens.
Now use the step-along procedure:
\( V_{out(1)} \) for the first lens is
\( V_{in(1)}+F_{(1)}=1/(-0.5)+10 = 8{\text{D}} \).
The image distance \( d \) for the first lens is
\( d=1/V_{out(1)} = 0.125\text{m} \).
Subtracting the gap width \( 0.2\text{m} \) from the image distance gives us
\( 0.125-0.2 = -0.075\text{m} \). (Notice that this is a negative distance, and it means that the light converged to a point to the left of the second lens, and then started diverging again.)
\( V_{in(2)} \) for the second lens is one over the result of step 3, which is
\( 1/(-0.075) = -13.33\text{D} \).
Now that we have \( V_{in(2)} \) for the second lens, we can apply the thin lens equation to get
\( V_{out(2)} \):
This light is diverging, and forms a virtual image which is
\( 1/(-9.33)=-0.107{\text{m}} \) to the left of the second lens.
The Stepalong Formula.
The above procedure can be reduced to a single formula. It is no more or less work than the above procedure, but is
occasionally useful. The stepalong formula lets you calculate \(
V_{in(2)} \) from \( V_{out(1)} \) and the gap width
\( g \):
You can construct this from the stepalong procedure as follows:
Start with \( V_{out(1)} \) for the first lens.
Work out the image distance \( d = 1/V_{out(1)} \) for the first lens (ignoring the second lens).
Subtract the gap width \( g \) from the image distance
\( d \) to give \( 1/V_{out(1)}-g \)
.
\( V_{in(2)} \) for the second lens is one over the result of step 3; that is
\( V_{in(2)} = 1/(1/V_{out(1)}-g) \).
When you multiply the top and bottom of the fraction \( 1/(1/V_{out(1)}-g)
\) by \( V_{out(1)} \) you get the stepalong formula.
Although the stepalong formula is used to work out the vergence
\( V_{in(2)} \) entering the second lens at distance \( g
\), there really doesn't need to be a lens there at all for the formula to work, and it works for just a single lens too.
That is, if you have a lens whose vergence is \( V_{out} \)
, then the vergence \( V \) of light at a distance
\( g \) from it is
\[
V = \frac{V_{out}}{1-gV_{out}}
\]
(provided of course there is no other lens in that distance
\( g \))
Many Refracting Surfaces
We can trace vergence through two (or more) refracting surfaces in the same way as we traced light through lenses. All we have to do is use the paraxial equation instead of the thin lens equation. Apart from the added complication of the refractive indices, the task isn’t any different.
Figure 4. Two refracting surfaces as described in the text. The scale of the drawing means that the object and image points are off the sides of the image. The different colours show where the refractive indices are different. Light blue to the left is water (
\( n=1.33 \)), dark blue in the middle is glass (
\( n=1.5 \)), and grey to the right is air (\( n=1 \)
).
Look at Figure 4. This shows a glass lens (with a refractive index of
\( 1.5 \)). On the left of the glass lens is water (refractive index 1.33), and on the right is air. This could be say a lens used in prescription swimming goggles. The distance between the two surfaces is
\( 0.05{\text{m}} \), which is fairly unrealistic - few lenses are
\( 5\text{cm} \) thick - but helps us to see what is going on with the calculations. Both surfaces obey the paraxial equation. The paraxial equation for the first surface is
The power of the first surface is \( F_{surface(1)}=+6.8D \)
. (This depends on \( n_{in(1)} \) and
\( n_{out(1)} \).) An object is placed \( 0.8{\text{m}} \)
to the left of the first surface. The vergence of the light hitting the first surface is thus
\( V_{in(1)}=1/(-0.8) = -1.25{\text{D}} \). Taking the paraxial equation for the first surface, and filling in all the known quantities, gives
Here \( n_{in(2)}=1.5 \) (the same as
\( n_{out(1)} \)) and \( n_{out(2)}=1 \). The surface power for the second surface is
\( F_{surface(2)}=-2.5D \). The vergence of the light entering the second surface is
\( V_{in(2)}=4.13{\text{D}} \), so taking the paraxial equation for the second surface, and filling in all the known quantities, gives
from which \( V_{out(2)}=+3.70{\text{D}} \). The image thus forms at a distance of
\( 1/3.7=0.270{\text{m}} \) to the right of the second surface
Reduced Distance
There is a clever little trick to make calculations with refracting surfaces quite a lot easier, which depends on the following definitions:
The reduced distance is defined as the actual distance divided by the refractive index.
The actual distance is the reduced distance multiplied by the refractive index.
When all distances are given as reduced distances, we can use the thin lens equation instead of the paraxial equation, because it turns out that the thin lens equation using reduced distances gives exactly the same result as the paraxial equation with actual distances.
Let's try this trick with the problem shown in Figure 4
. An object is placed \( 0.8{\text{m}} \) to the left of the first surface. The reduced distance is
\( 0.8/n_{in(1)} = 0.8/1.33=0.602{\text{m}} \). The gap between the surfaces is
\( 0.05{\text{m}} \), which as a reduced distance is
\( 0.05/1.5=0.0333{\text{m}} \). The surface powers are unchanged.
The thin lens equation for the first surface says
\[ V_{in(1)}+F_{surface(1)}=V_{out(1)} \]
Using the reduced distance for the object gives us
\[ 1/(-0.602)+6.8=V_{out(1)}=5.13{\text{D}} \]
Now use the step-along procedure:
\( V_{out(1)} \) for the first surface is
\( 5.13\text{D} \).
The reduced image distance \( 1/V_{out(1)} \) for the first surface is
\( 1/5.13 = 0.195\text{m} \).
Subtracting the reduced gap \( 0.0333\text{m} \) from the reduced image distance gives us
\( 0.195-0.0333 = 0.162\text{m} \)
\( V_{in(2)} \) for the second surface is one over the result of step 3, which is
\( 1/0.162 = 6.17\text{D} \).
The thin lens equation for the second surface says
\[ V_{in(2)}+F_{surface(2)}=V_{out(2)} \]
Thus \( V_{out(2)} \) is \( 6.17+(-2.5)=V_{out}=3.67{\text{D}}
\) The image thus forms at a (reduced) distance of \( 1/3.67=0.272{\text{m}}
\) from the second surface. The actual distance is this reduced distance multiplied by the refractive index to the right of the second lens, which is just 1, so the actual distance here is the same as the reduced distance. There is a slight difference between this result and the result earlier due to rounding.
If you are using the stepalong formula, you use the reduced distance for the gap; that is, the stepalong formula becomes
In Chapter 1 the power of a thin lens is defined using parallel light in two ways:
When \( V_{in}=0 \), the power is \(
F=V_{out} \)
When \( V_{out}=0 \), the power is
\( F=-V_{in} \).
These two definitions for a single thin lens can be used to define the power of a set of lenses or surfaces. When we have two (or more) lenses or surfaces in a line, the
Back Vertex is the last lens or surface in the series and the
Front Vertex is the first lens or surface in the series. Such a system of lenses or surfaces has two powers:
The Back Vertex Power (or BVP) is the vergence of light leaving the Back Vertex when parallel light hits the Front Vertex.
The Front Vertex Power (or FVP) is the negative of the vergence of light hitting the Front Vertex when parallel light leaves the Back Vertex.
For a thin lens, the back vertex power and front vertex power are both equal to the power
\( F \), but for two or more lenses they can be quite different. The corresponding focal lengths are, as before, just one over the powers.
The Front Vertex and Back Vertex Power of Two Lenses.
Consider a pair of thin lenses separated by a distance \( d \)
, as shown in the top of Figure 5. The first lens has power
\( F_1 \) and the second lens has power
\( F_2 \). The first lens is the Front Vertex, and the second lens is the Back Vertex. In
Figure 5, the lenses are both positive, but they need not be.
\( F_1 \)\( F_2 \)\( F_1 \)\( F_2 \)
Figure 5. Back and front vertex powers BVP and FVP. The top half of this figure shows two positive lenses separated by a distance
\( d \). Parallel light enters the first lens and is converged a little. It continues until it hits the second lens where it is converged further. It leaves the second lens and converges to a point, which is a kind of focal point (since parallel light was sent into the two-lens system). The distance from the second lens to the focal point is the back vertex focal length, which is one over the back vertex power (BVP). The bottom half of this figure shows an object which is placed to cause parallel light to leave the second lens. The distance from the first lens to the object is the front vertex focal length, which is one over the front vertex power (FVP).
We can work out a formula for the Back Vertex Power of two lenses using the step-along procedure and a bit of algebra:
Step
Statement
Reason
1
\( V_{in(1)}+F_1=V_{out(1)} \)
Thin Lens equation for lens 1
2
\( F_1=V_{out(1)} \)
Substitute \( V_{in(1)}=0 \)
(parallel light enters the front vertex)
3
\( \text{image distance} = 1/V_{out(1)}=1/F_1
\)
Step-along procedure step 1: image distance from first lens
4
\( 1/F_1-d \)
Step-along procedure step 2: subtract the gap from image distance
5
\( V_{in(2)}=\dfrac{1}{1/F_1-d} \)
Step-along procedure step 3:
\( V_{in(2)} \) is reciprocal of step 4.
6
\( V_{in(2)}=\dfrac{F_1}{1-dF_1}
\)
Multiply top and bottom of fraction by
\( F_1 \) (just tidying up)
7
\( V_{in(2)}+F_2=V_{out(2)} \)
Thin Lens equation for lens 2
8
\( \dfrac{F_1}{1-dF_1}+F_2=V_{out(2)}
\)
Substitute step 6 for \( V_{in(2)}
\)
9
\( \dfrac{F_1}{1-dF_1}+F_2=\text{BVP}
\)
BVP definition: the vergence
\( V_{out(2)} \) leaving the last lens when \( V_{in(1)}=0
\)
The formula for the front vertex power looks the same as the one for back vertex power, just with the two lens powers swapped:
\[ \text{FVP} = \frac{F_2}{1-d F_2}+F_1 \]
When two thin lenses get very close together, the distance \(
d \) between them is essentially zero. In this case, the Back Vertex Power is simply
\( \text{BVP}=F_1+F_2 \), and the Front Vertex Power is also
\( \text{FVP}=F_1+F_2 \). In this case, we can treat the two lenses as a single lens with a power equal to the sum of the powers of the individual lenses. This is why you can usually put sets of lenses in an optometrist's trial frame to get any particular power you want - the lenses are close enough together that their powers add together into a single power.
The Front Vertex and Back Vertex Power of a Thick Lens.
Consider two refracting surfaces in air, separated by a distance
\( d \), as shown in Figure 6. The index of refraction between the two surfaces is
\( n \). This is a thick lens. When light travels through a thick lens, the vergence of the light is changed by the power of the first refracting surface; then the vergence of the light changes as it travels through the lens; and finally the vergence is changed again by the power of the second refracting surface.
We can define the Front Vertex Power and Back Vertex Power of a thick lens just as we did for a pair of thin lenses. Thanks to the idea of a reduced distance, a thick lens with thickness
\( d \) and surface powers \( F_{surface(1)}
\) and \( F_{surface(2)} \) is exactly the same as two thin lenses with the same powers, separated by a reduced distance
\( d/n \).
Figure 6. A thick lens is two refracting surfaces separated by a distance
\( d \). In this image parallel light enters the first surface and is refracted. It continues through the lens and is then refracted again as it leaves. The light converges on a point, called the back vertex focal point.
For a thick lens, with surface powers \( F_{surface(1)} \)
and \( F_{surface(2)} \), we can write down the BVP and FVP as:
These are the same as the two-lens formulae, except we are using surface powers and the reduced distance between the surfaces.
Thin Lenses.
If the thickness of a lens \( d \) becomes very small (near zero), we have a
thin lens. If we put \( d=0 \) into the FVP and BVP formulas for a thick lens, we get
That is, Front and Back Vertex Powers are equal to the sum of the surface powers.
Thus, if we want a thin lens with power \( F \), we can choose any front surface power
\( F_{surface(1)} \) and back surface power
\( F_{surface(2)} \) we like, just so long as they add up to the power
\( F \). This gives lens designers some freedom in the lens shapes they can make to give the same lens power. Some lens shapes are shown in
Figure 7.
Figure 7. Typical lens cross-sections. The three lenses in the top row are all positive, the three on the bottom row all negative. On the top row, the first lens is a plano-convex lens. One surface is flat (the “plano” surface, with zero power) and one surface has positive power. The second lens is biconvex. Both surfaces are positive. The third lens is a positive meniscus lens. The front surface is positive and the back surface is negative, but since the front surface is more powerful, the total power is positive. On the bottom row, the first lens is plano-concave. The second is biconcave, with both surfaces negative. The last lens is a negative meniscus, but this time the back surface is more powerful than the front, so the total power is negative.
A Look Ahead.
The above calculations can be used for three, four, or more lenses or surfaces. The step-along procedure just needs the vergence leaving the previous lens to work out the vergence entering the next lens in the series. However, the calculations quickly become intimidating, and they have to be done again and again every time a different vergence
\( V_{in(1)} \) hits the first lens in the set.
It turns out that any system of lenses or surfaces, no matter how complicated, can be described by just
four numbers. Once these four numbers are calculated (and they only have to be calculated once), they can be used again and again to compute the change in vergence of light passing through the entire system. This method of calculation is described in Chapters 17 to 19.
Self-Test
Now that you've read this chapter, you can do a self-test
