EXERCISE 8
The Lens Equation
8.1. Introduction
One of the simplest optical device in common usage is no doubt the thin lens. The
development of optical devices using lenses dates to the sixteenth and seventeenth
centuries and includes eyeglasses, cameras, magnifying glasses, telescopes, binoculars,
microscopes, and many specialized instruments. A thin lens is usually circular in cross
section, and its two faces are portions of a sphere. (Although cylindrical surfaces are
also possible, we will concentrate on spherical.) The two faces can be concave, convex
or plane.
Consider the rays parallel to the axis of double convex lens which is shown in cross
section in Fig. 8.1. We assume the lens is made of glass or transparent plastic, so its
index of refraction is greater than that of air outside. The axis of lens is a straight line
passing through the center of the lens and perpendicular to its two surfaces (Fig. 8.1).
From Snell’s law, we can see that each ray in Fig. 8.1 is bent toward the axis at both
lens surfaces. If rays fall parallel to the axis of a thin lens, they will be focused to a
point called the focal point denoted with F. The focal point existence is not precisely
true for lens with spherical surfaces. It will be nearly true only for lense, which
diameter will be small, compared to the radii of curvature of the two lens surfaces and
it focus parallel rays to a tiny region that is nearly a point. This criterion is satisfied by
a thin lens, one that is very thin compared to its diameter, and we consider only thin
lenses here.
F
Axis
f
Fig. 8.1 Parallel rays are brought to a focus by a converging thin lens.
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THE LENS EQUATION
The most important parameter of a lens is a focal length f. Once f is known, the image
position can be found for any given object. Finding the image point by drawing rays
would be difficult if we would have to determine all the refractive angles. Instead, we
can do it in a simpler manner by making use of a certain facts we already know, such
as that a ray parallel to the axis of the lens passes (after refraction) through the focal
point. In fact, to find an image point, we need consider only three rays indicated in Fig.
8.2, which shows an arrow as the object and a converging lens forming an image to the
right.
Object
1
1
F’
F
Object
(a) Ray 1
1
1
F’
Image
F
2
(b) Ray 2
2
Object
1
3
1
F’
Image
F
2
(c) Ray 3
2
Fig.8.2 Finding the image by ray tracing for a converging lens. Rays leave each point on the object.
The three most useful rays are shown , leaving the tip of the object, for determining where the image
of that point is formed.
These rays, emanating from a single point on the object, are drawn as if the lens were
infinitely thin, and we show only a single sharp bend within the lens instead of the
refractions at each surface. These three rays are drawn as follows:
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THE LENS EQUATION
Ray 1 is drawn parallel to the axis; therefore it is refracted by the lens so that it
passes along a line through the focal point F, (Fig. 8.2.a).
Ray 2 is drawn on a line passing through the other focal point F’ (front side of
lens in Fig. 2) and emerges from the lens parallel to the axis, (Fig. 8.2.b).
Ray 3 is directed toward the center of the lens, where the two surfaces are
essentially parallel to each other; this ray therefore passes through the lens
without bending; the ray would be displaced slightly to one side, but since we
assume the lens is thin, we draw ray 3 straight through as shown (Fig. 8.2.c).
Actually, any two of these rays will suffice to locate the image point, which is the point
where they intersect. Drawing the third can serve as a check.
In this way we can find the image point for one point of the object (the top of arrow in
Fig. 8.2). The image points on the object can be found similarly to determine the
complete image of the object. Because the rays actually pass through the image for the
case shown in Fig. 8.2, it is a real image.
Now, we can derive an equation that relates the image distance to the object distance
and the focal length of the lens. This will make the determination of image position
quicker and more accurate than doing ray tracing. Let d0 be the distance of the object
from the center of the lens, and di be the distance of the image from the center of the
lens; let also h0 and hi refer to the heights of the object and image.
O’
B
ho
F
O
F’
I
A
hi
I’
f
do
di
Fig. 8.3 Deriving the lens equation for a converging lense.
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THE LENS EQUATION
Consider the two rays shown in Fig. 8.3 for a converging lens (assumed to be very
thin). The triangles FI’I and FBA (Fig. 8.3) are similar because angle AFB equals
angle IFI’; so
hi d i  f

h0
f
(8.1)
since length AB=h0. Triangles OAO’ and IAI’ are similar. Therefore,
hi d i

h0 d 0
(8.2)
We equate the right sides of these two equations, divide by di, and rearrange to obtain
1
1 1
 
d0 di
f
(8.3)
This is called the lens equation. It relates the image distance di to the object distance
d0 and the focal length f. It is the most useful equation in geometric optics.
The lateral magnification m, of a lens is defined as the ratio of image height to object
height, m=hi/h0. Thus according to the conventions stated in Fig. 8.3 we have
m
hi
d
 i
h0
d0
(8.4)
8.2. Measurements
do
So
di
f
Crossed arrow target
f
Si
lens
Viewing screen
Fig. 8.4.Experimental setup.
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THE LENS EQUATION
Set up the equipment as shown in Fig. 8.4. Turn on the Light Source and slide the lens
toward or away from the Crossed Arrow Target, to focus the image of the Target onto
the Viewing Screen. Now set d0 to the values (in milimeters) listed in the table below.
At each setting, locate the image and measure di. Also measure hi, the height of image
(h0 is the height of the arrow on the crossed arrow target). Measurements should be
repeated 6 times.
Using the data you have collected, perform the calculations shown in the table 8.1.
Are your results in complete agreement with the Fundamental Lens Equation? If not,
what is the source of the discrepancies?
8.3. Results, calculation and uncertainty
The data should be collected in a table 8.1.
Table 8.1
Data
d0 [mm]
di [mm]
Calculations
hi [mm]
1/di + 1/d0
f [mm]
hi/h0
-di/d0
500
450
400
350
300
250
200
150
100
75
50
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THE LENS EQUATION
Calculate the focal length of the lens and lateral magnification using eq. 8.1, 8.3 and
8.4. Estimate the uncertainty of the measured values by calculating the standard
deviation.
The final results read:
f  f  Sf
(8.5)
m  m  Sm
(8.6)
8.4. Questions
1. Based on the Fundamental Lens Equation, explain what would happen to di if you
increased d0 even further? What would happened to di if d0 were very, very large?
2. What value of d0 will make you unable to focus an image onto the screen? Use the
Fudamental Lens Equation to explain why.
3. What kind of Fundamental Lens Equation do you know?
4. What kind of parameters decide about the kind of obtained image?
5. What kind of image can you observe in a microscope, cinema and in the human eye?
Explain it using the Fundamental Lens Equation.
6. What types of lenses do you know ?
7. What happens to the image as you change the distance between an object and lens.
8. What is an aberration ?
9. Explain vision correction by glasses.
10. Derive law of refraction from Fermat’s principle.
8.5. References
1. Dryński T., Ćwiczenia laboratoryjne z fizyki, PWN, Warszawa, 1959
2. Resnick R., Halliday D., Fizyka, Tom 2, PWN, Warszawa, 1989
3. Szydłowski H., Pracownia fizyczna, PWN, Warszawa, 1994
4. Young H.D., Freedman R.A., University Physics with Modern Physics,
Addison-Wesley Publishing Company, 2000
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