Compound Optics Prelab - SCIPP

Compound Optics Prelab
1 Astronomical Telescope
An astronomical telescope, intended for visual observation, is constructed with an objective
lens with focal length f 1 = +150 mm and an eyepiece lens with focal length f 2 = +75 mm.
The telescope is used for viewing distant objects (s 1 >> f 1 ). Make a scale drawing of the
telescope, showing the position of the intermediate image. What is the magnification of
the telescope?
The astronomical telescope is a system consisting of two lenses. The way to make calculations is to
divide this system into two smaller subsystems, each consisting of one lens.
The first system consists of the object, the objective lens, and the intermediate image, which is the
image produced by the objective lens. The object is at a distance s 1 from the objective lens. The objective
lens has focal length f 1 . The distance between the objective lens and the (real) intermediate image
is s ′ 1. (Since the intermediate image is real, do you expect it to be on the same side of the objective lens
as the object, or do you expect it to be on the other side of the lens?)
The second system consists of the intermediate image, which acts as the object for the eyepiece lens, the
eyepiece lens, and the final image, viewed by the human eye. The distance between the intermediate
image and the eyepiece lens is s 2 . The focal length of the eyepiece lens is f 2 . The distance between the
eyepiece lens and the (virtual) final image is s ′ 2.
If you look at the diagram of the telescope in your lab manual, you will see that the total distance
between the objective lens and the eyepiece lens is the distance from the objective lens to the intermediate
image plus the distance from the intermediate image to the eyepiece lens. In other words,
D = s ′ 1 + s 2 .
In order to make a scale drawing of the telescope, we need to find the two distances s ′ 1 and s 2 .
order to find these two distances, we will have to make some assumptions about s 1 and s ′ 2.
In
1.1 Intermediate Image
The focal length of the objective lens is f 1 . We know that
f 1 = 150 mm
Let s 1 be the distance between the object and the objective lens. We are told that the telescope is
used to view distant objects. This means that the distance s 1 is very, very large,
s 1 → ∞
Define the distance from the objective lens to the intermediate image to be s ′ 1. We need to find s ′ 1.
Since we know the focal length f 1 and the distance s 1 , we can use the lens equation to find s ′ 1. The lens
equation is
1
+ 1 s 1 s ′ = 1
1 f 1
1

So
1
s ′ 1
= 1 f 1
− 1 s 1
We have made an assumption about the object distance s 1 . Use this assumption to calculate the
inverse of the object distance,
Remember from your calculus class that
lim
x→∞
1
s 1
1
x → 0
The lens equation should simplify to include just two terms. Now it is easy to take the inverse of the
lens equation.
What is the distance s ′ 1 from the object lens to the intermediate image in millimeters?
1.2 Final Image
“The astronomical telescope is intended for visual observation.” What does this statement
mean?
The first assumption which you should make is that your eye is as close to the eyepiece lens as possible.
For the sake of convenience, define the distance between the eye and the eyepiece lens to be zero.
The second assumption which you should make is that the magnitude of the distance between the
eye and the intermediate image is extremely large. (In this case we say that the eye is “relaxed”.)
Since your eye is very close to the eyepiece lens, the magnitude of the distance s ′ 2 between the eyepiece
lens and the final image is also extremely large,
|s ′ 2| → ∞
(The final image is virtual, so the image distance s ′ 2 is negative.)
The lens equation is
Consider what
implies about the term
1
+ 1 s 2 s ′ = 1
2 f 2
|s ′ 2| → ∞
1
s ′ 2
The lens equation should once again simplify to contain two terms. Take the inverse of these two
terms. What is the distance s 2 from the intermediate image to the eyepiece lens, as a multiple of the
focal length f 2 ?
2

1.3 Drawing
You now know enough to make a very accurate diagram of the telescope. Your drawing does not need
to be a principal ray diagram, because it is difficult to include objects and images which are at infinity
in a ray diagram. However,the finite distances depicted in your drawing must be to scale, and it must be
drawn with a ruler.
Start with the “distant object” on the left. The distance from object to objective lens is s 1 . The
distance from the objective lens to the intermediate image is s ′ 1. Indicate the focal length f 1 of the
objective lens in your diagram.
You calculated the distance s 2 from the intermediate image to the eyepiece lens.
length f 2 of the eyepiece lens in the diagram.
Include the focal
You also know the distance s ′ 2 between the final image and the eyepiece lens. Note that the final
image is virtual. Will it be on the same side of the eyepiece lens as the intermediate image, or will it
be on the opposite side of the lens? Remember that we are assuming the eye is positioned very close to
the eyepiece lens, and the image needs to be in front of the eye.
1.4 Magnification
The magnification provided by the objective lens is
( ) s
′
M 1 = − 1
s 1
The magnification provided by the eyepiece lens is
( ) s
′
M 2 = − 2
s 2
The total magnification is the product of these two quantities,
M = M 1 · M 2
( ) ( )
s
′
M = 1 s
′
· 2
s 1 s 2
Note that s 1 → ∞ because we are viewing a distant object, and s ′ 2 → −∞ because the final image
is virtual and the eye is relaxed. So we can say that s 1 ≈ −s ′ 2 and simplify the expression for the
magnification,
( ) s
′
M ≈ − 1
s 2
You calculated the distance s ′ 1 of the intermediate image as a multiple of the focal length f 1 of the
objective lens. You calculated the object distance s 2 as a multiple of the the focal length f 2 of the
eyepiece lens. If you insert these quantities into the above expression for the magnification, you should
obtain equation 33 in your lab manual.
2 Optical Microscope
An optical microscope is intended for visual observation. The objective lens has a focal
length f 1 = +75 mm. The eyepiece lens has a focal length f 2 = +150 mm. The microscope is
used for viewing an object at a distance of 125 mm from the objective lens. Determine the
position of the intermediate image, and make a scale drawing of the microscope, showing
3

the position of the intermediate image. Calculate the optimum magnification of the microscope.
Like the telescope, the microscope is a system consisting of two lenses.
this system into two smaller subsystems, each consisting of one lens.
Once again we will divide
The first system consists of the object, the objective lens, and the intermediate image. The object
is at a distance s 1 = 125 mm from the objective lens. The objective lens has focal length f 1 . The distance
between the objective lens and the (real) intermediate image is s ′ 1.
The second system consists of the intermediate image, which acts as the object for the eyepiece lens, the
eyepiece lens, and the final image, viewed by the human eye. The distance between the intermediate
image and the eyepiece lens is s 2 . The focal length of the eyepiece lens is f 2 . The distance between the
eyepiece lens and the (virtual) final image is s ′ 2.
We need to find the two distances s ′ 1 and s 2 .
make some assumptions about s 1 and s ′ 2.
In order to find these two distances, we will have to
2.1 Intermediate Image
Determine the position of the intermediate image.
The focal length of the objective lens to be f 1 . The problem states that
f 1 = 75 mm
Define the distance between the object which is being viewed and the objective lens to be s 1 . The
problem states that the object is at a distance of 125 mm from the objective lens, so
s 1 = 125 mm
Define the distance between the objective lens and the intermediate image to be s ′ 1. You need
to find s ′ 1.
The lens equation relates the focal length of the lens to the distances between the object, the image,
and the lens,
1
+ 1 s 1 s ′ = 1
1 f 1
Since you know the focal length f 1 and the distance s 1 , you can find the distance s ′ 1,
1
s ′ 1
= 1 f 1
− 1 s 1
s ′ 1
1 = ( )
1
f 1
− 1 s 1
How far (in millimeters) is the intermediate image from the objective lens?
Is this distance positive? In other words, is this a real image?
Calculate the magnification M 1 ,
( ) s
′
M 1 = − 1
s 1
Is the intermediate image magnified? What does the negative sign mean?
4

2.2 Final Image
“The microscope is intended for visual observation.” In figure 14 your lab manual says “the
optimum configuration is for |s ′ 2| ≈ 250 mm”. What does this mean?
If you have ever used a microscope, you will remember that you had to lean down until your eye was
very close to the eyepiece lens. The shortest allowed distance between the eye (and thus the eyepiece
lens) and the final image is 250 mm. (If the image is any closer than this, a human eye cannot focus on
it.) In order to obtain the best possible magnification, the final image must be at exactly this position.
For this reason, we define the image distance
|s ′ 2| = 250 mm
Since the final image is virtual, s ′ 2 will be negative.
Define f 2 to be the focal length of the eyepiece lens. We know that the focal length of the eyepiece
lens is
f 2 = 150 mm
The intermediate image is the image which was produced by the object lens. This image serves
as the object for the eyepiece lens.
Define the distance between the intermediate image and the eyepiece lens to be s 2 .
to find s 2 .
We need
Use the lens equation to solve for the distance s 2 between the intermediate image and the eyepiece
lens,
1
+ 1 s 2 s ′ = 1
2 f 2
What is s 2 ?
1
s 2
= 1 f 2
− 1 s ′ 2
How much is the final image magnified with respect to the intermediate image?
( ) s
′
M 2 = − 2
s 2
Is the final image inverted with respect to the intermediate image?
2.3 Magnification
We know that the intermediate image is magnified by a factor of M 1 with respect to the object.
The final image is magnified by a factor of M 2 with respect to the intermediate image. The total
magnification M is just the product of these two intermediate steps,
M = M 1 · M 2
M 1 is negative, since the objective lens produces a real, inverted intermediate image. M 2 is positive,
since the eyepiece lens produces a virtual final image which is upright with respect to the intermediate
image. The total magnification M is therefore negative, and the final image is inverted with respect to
the original object.
(Alternatively, you could have used equation 36 in your lab manual to calculate the total magnification
M.)
5

2.4 Drawing
You now know enough to make a very accurate principal ray diagram of the microscope. Your ray
diagram must be to scale and it must be drawn with a ruler.
Start with the object on the left hand side. The distance between the object and the objective lens
is s 1 . Include the focal length f 1 of the objective lens. Draw two or three rays as shown in your textbook
(figure 31.17).
Use your ray diagram to predict the image distance s ′ 1 of the intermediate image.
with your calculation? Is the intermediate image real and inverted?
Does this agree
You calculated the distance s 2 between the intermediate image and the eyepiece lens. Include the
focal length f 2 of the eyepiece lens. Draw two or three rays, and use your ray diagram to predict the
position s ′ 2 of the virtual final image.
Don’t be surprised if the final image is between the objective lens and the intermediate image.
Does the distance s ′ 2 agree with our initial definition of this distance as being the closest position where
the human eye can still focus?
Is the final image virtual? Is the final image in your diagram M times the size of the initial object?
You will need your diagram during the lab.
6