PHYS471
Lab 2. Thin lenses and optical instruments
J. James Jun
2007 January 25
Goal
To observe the operation of thin lenses
To gain experience with the placement and alignment of optical components
To examine and measure real and virtual images in simple optical systems
To measure the focal lengths of double convex and double concave lenses
To understand the operation of simple optical instruments: microscope and telescope
Materials and equipments
1 meter optical bench with 5 moveable mounts
lens holders (3)
short focal length double convex lenses (L1, E1) (2)
long focal length double convex lens (L4)
diopter gauge
Ruler
*The default quantity is (1)
Lighted test object
Screen
Medium focal length double convex lenses (L2,
L3)(2)
Double concave lens (L5)
Meter stick
Flashlight
Procedures
The laboratory lights were dimmed. The focal lengths of converging lenses were determined by the direct method. The
lighted object is placed at the far end of the laboratory to approximate an infinite distance and the distance from a lens to a screen
was measured when a lens focused an image to a point. This distance was taken as the focal distance of the convex lens. The
lenses were labeled from L1 to L5 sorted by the shortest to the longest.focal length. After the direct measurement method, a
diopter guage was used to determine the focal distance. After this, the displacement method was used to measure the focal length
of the lens L2. The screen was located at position 0cm on the optical bench scale. The lighted object was located at 110cm. The
lens was moved between the object and the screen and two positions were located where a real image was formed on the screen. a
and b were recorded where “b” is the distance from the object and the screen and “a” is the distance from the 1st position and the
2nd position. And then the lighted object is fixed at 0cm and the object distance was changed by moving the lens. Lens L3 was
used for this part of the experiment. The image distance was determined by moving the screen until a focused image was formed
on the screen. This procedure was repeated for 5 different object distances. 1/ i vs 1/ o was plotted. Then the displacement
method was used to find the focal length of the diverging lens. The screen was located at 0cm and the lighted object was located at
115cm. Lens L5 was positioned at 100cm and lens L3 was positioned between lens L5 and the screen. The object distance is
measured by moving L3 until a focused image appeared on the screen. Then the focal length of the same diverging lens L5 was
determined using a different indirect method. Then another displacement method was used to determine the focal distance of the
same diverging lens L5. The lighted object was located at 115cm and the screen was located at 0cm. A converging lens L1 was
located at 110.51cm to create an image and the displacement of the image created by L1 was recorded. Then lens L5 was located
at 92.6cm to create an image of the image created by lens L1. The displacement of the image created by L5 was recorded. Then
Data & calculations
Requirement 1
Table 1: Focal lengths of the convex lenses determined by the direct method
Lens #
Focal Distance
[cm]
L1
2.8
L2
5.3
L3
9.9
L4
22.2
L5
n/a
*Error in focal distance: ±2mm
Requirement 2
Table 2: Focal lengths of the convex lenses determined by the direct method
Dleft
Dright
f
Δf
Lens #
⎡⎣ m −1 ⎤⎦
[ m]
[ m]
1.73E+01
2.E+01 3.06E-02
9.9E+00 9.9E+00 5.07E-02
5.0E+00 5.0E+00 1.00E-01
2.3E+00 2.3E+00 2.17E-01
-5.3E+00 -5.3E+00 -9.5E-02
has an uncertainty of 1/ 8 ⎡⎣ m −1 ⎤⎦
⎡⎣ m −1 ⎤⎦
L1
L2
L3
L4
L5
* Dleft , Dright
2.E-04
5.E-04
2.E-03
8.E-03
2.E-03
* f : focal distance
Requirement 3
⎧O1 = ( 5.5 ± .2 ) cm
⎪
⎪O2 = (104.0 ± .2 ) cm
⎨
⎪a = O2 − O1 = ( 98.5 ± .4 ) cm
⎪b = 110.0 ± .4 cm
(
)
⎩
Requirement 4
Lens L3 is used ( f = (10.0 ± .2 ) [ cm ] )
Δ (1/ o )
Data #
do
di
o
i
1/ o
unit
[cm]
[cm]
[ m]
[ m]
⎡⎣ m ⎤⎦
⎡⎣ m
3.32E+00
3.60E+00
4.16E+00
4.56E+00
5.00E+00
30.1
45.0
1
27.8
43.6
2
24.0
41.0
3
21.9
40.6
4
20.0
39.6
5
*screen fixed at the zero distance
di : image location on the bench
0.301
0.278
0.240
0.219
0.200
d o : lens location on the bench
di , d o has an uncertainty of 2mm
Graph 1: 1/i vs 1/o to confirm the thin lens equation
−1
0.149
0.158
0.170
0.187
0.196
⎤⎦
2.E-02
3.E-02
3.E-02
4.E-02
5.E-02
−1
Δ (1/ i )
1/ i
⎡⎣ m ⎤⎦
⎡⎣ m −1 ⎤⎦
6.73E+00
9.E-02
6.33E+00
8.E-02
5.88E+00
7.E-02
5.35E+00
6.E-02
5.11E+00
5.E-02
−1
Graph 1. 1/i vs 1/o to confirm the thin lens equation
7.00E+00
6.80E+00
6.60E+00
1/i [m^-1]
6.40E+00
6.20E+00
6.00E+00
5.80E+00
y = -0.9702x + 9.8878
R2 = 0.9851
5.60E+00
5.40E+00
5.20E+00
5.00E+00
3.00E+00
3.50E+00
4.00E+00
4.50E+00
5.00E+00
5.50E+00
1/o [m ^-1]
mobs = ( −.97 ± .07 ) , bobs = ( 9.9 ± .3) ⎡⎣ m −1 ⎤⎦
Let’s find an expected slope and the expected y-intersect from the thin lens formula:
1 1 1
+ =
where o: object distance, i: image distance, f: focus distance
o i f
By solving this equation for a dependent variable 1 , we get
i
1 =− 1 + 1
i
o
f
Thus the expected slope and intersects are
mexp = −1
( )
⎞
1 ⎛
1
−1
=⎜
⎟⎟ = 10.0 ± .2 ⎣⎡ m ⎦⎤
⎜
f ⎝ (.100 ± .002 ) [ m ] ⎠
Comparing the expected and observed values, they agree within the calculated uncertainties. Thus the thin lens formula is
experimentally verified to be correct.
bexp =
Requirement 5
Let’s first calculate the focal distance of the concave lens L5 for procedure #5.
Table 3: Distance measurements of the procedure #5
[m]
x1
0.000
x2
0.893
x3
1.000
x4
n/a
x5
1.150
* x1 ~ x5 all have uncertainty of .002 [ m ]
Figure 1: Procedure #5 illustration
The thin lens formula needs to be applied twice iteratively.
First let’s apply it to the real object and the virtual image created by L5
1
1 1
⎪⎧o = ( x5 − x3 ) = (.150 ± .004 ) [ m ]
= + , where ⎨ 1
(1.1)
i1 = − ( x4 − x3 )
f L 5 o1 i1
⎪⎩
Since we do not know x4 , which is the displacement of the virtual image, we need to find it from L2.
We can again apply the thin lens formula for the optical system created by the virtual image, L3, and the real image (at screen).
o2 = x4 − x2
⎧
1
1 1
= + , where ⎨
f L 3 o2 i2
⎩i2 = x2 − x1 = (.893 ± .004 ) [ m ]
Since we know f L 3 , this equation can be solved for x4
−1
−1
⎛
⎞
⎛ 1
1⎞
1
1
o2 = ⎜
− ⎟ =⎜
−
⎟⎟ = (.113 ± .003) [ m ]
⎜
⎝ f L 3 i2 ⎠
⎝ (.100 ± .002 ) [ m ] (.893 ± .004 ) [ m ] ⎠
Thus x4 = o2 + x2 = (.113 ± .003) [ m ] + (.893 ± .002 ) [ m ] = (1.006 ± .005 ) [ m ]
Now let’s find f L 5 using (1.1)
f L5
−1
⎛
⎞
⎛ 1 1⎞
1
1
=⎜ + ⎟ =⎜
+
⎟
⎜
⎟
⎝ o1 i1 ⎠
⎝ (.150 ± .004 ) [ m ] − ( (1.006 ± .005 ) [ m ] − (1.000 ± .002 ) [ m ]) ⎠
−1
−1
⎛
⎞
1
1
=⎜
+
= ( −.006 ± .007 ) [ m ]
⎜ (.150 ± .004 ) [ m ] − (.006 ± .007 ) [ m ] ⎟⎟
⎝
⎠
This focus distance does not agree with the focus distance calculated from the diopter meter, which is − (.095 ± .002 ) [ m ]
The image was very blurry when it was created on the screen and for this reason I do not trust the location of L3.
Now let’s calculate f L 5 using the procedure #6 method.
Table 4: Distance measurements of the procedure #6
[m]
x1
0.878
x2
0.898
x3
0.926
x4
1.105
x5
1.150
* x1 ~ x5 all have uncertainty of .002 [ m ]
Figure 2: Procedure #6 illustration
Since we know the displacement of the virtual object by direct measurement, we can apply the thin lens formula only once with
the virtual object and the real image created by L5.
−1
⎛ 1 1⎞
1
1 1
= + ⇒ f L5 = ⎜ + ⎟
f L 5 o2 i2
⎝ o2 i2 ⎠
⎧⎪o = − ( x3 − x2 ) = − (.028 ± .004 ) [ m ]
where ⎨ 2
⎪⎩ i2 = ( x3 − x1 ) = (.048 ± .004 ) [ m ]
Thus f L 5 = − (.07 ± .02 ) [ m ]
This focal length determined by the displacement method does not agree with the diopter meter measurement value of
− (.095 ± .002 ) [ m ] but considering huge amount of error introduced by difficulty of determining the optimally focused image, the
real uncertainty might be greater than the nominal uncertainty of .02 [ m ] of the displacement method. If the error were greater by
10%, the focal distance determined by the diopter meter would agree with the focal distance determined by the displacement
method.
Requirement 6
Derivation of eqn(7):
From thin lens formula,
1 1 1
ss'
⎛1 1 ⎞
= + ⇒ f = 1/ ⎜ + ⎟ =
(1.2)
f s s'
⎝ s s'⎠ s + s'
From the definition, b = s + s ', a = s '− s assuming s ' > s
b2 − a 2
(1.3)
4
ss'
b2 − a 2 1 b2 − a2
=
=
QED
Plug (1.3) to (1.2) and we get f =
s + s'
4 b
4b
Thus b 2 − a 2 = ( s '+ s ) − ( s '− s ) = 4s ' s ⇒ s ' s =
2
2
Derivation of eqn(8):
There exists two configurations. Let o1 be the object distance of the first configuration and i1 be the image distance of the first
configuration. The magnification of the first configuration is therefore M 1 =
i1
o1
Since these two configuration is completely symmetric, the second configuration has o2 = i1 , i2 = o1 . Thus the magnification of the
second configuration is M 2 =
i2 o1
1
. Therefore M 1 M 2 = 1
= =
o2 i1 M 1
QED
Requirement 7
The observed angular magnification by the direct method is
( 5.0 ± .5)
= 2.5 ± .7
( 2.0 ± .5 )
The eqn(10) predicts the angular magnification ( M α ) to be
25cm
.25
=
= 4.93 ± .05
fL2
.0507 ± .0005
These two magnifications do not agree within the calculated uncertainties. The formula of angular magnification (eqn(10))
assumes a human eye to have a near focus of 25cm but this varies individually. I personally measured this and I have myopia
meaning that My vision has a near-focus less than 25cm. The effect of this is decrease in angular magnification thus the calculated
angular magnification will be shifted toward the observed angular magnification. I performed the direct measurement with my left
17.5cm
.175
=
= 2.66 ± .03 and
eye and my left eye has a near focus of 13.5cm. Thus the eqn(10) is modified to M a =
fL2
.0507 ± .0005
this agrees with the direct measurement within the calculated uncertainties.
Ma =
Requirement 8
The distance measurements
⎧ x1 = (.999 ± .002 ) m
⎪
⎨ x3 = (1.094 ± .002 ) m
⎪
⎩ x4 = (1.150 ± .002 ) m
The observed magnification is M obs =
6 ± .5
= 3.0 ± .8
2 ± .5
Now let’s calculate the expected angular magnification M exp using eqn(11)
M =
si 13.5cm* si .135m
×
= ×
so
fe
so
fe
(1.4)
* f e : Focus distance of the L2, f e = (.0507 ± .0005 ) [ m ]
so = x3 − x4 = (.056 ± .004 ) [ m ]
Let’s find si using thin lens formula
−1
−1
⎛ 1
1⎞
1
1
⎛
⎞
si = ⎜
− ⎟ =⎜
−
⎟ [ m ] = (.067 ± .002 ) [ m ]
⎝ .100 ± .002 .056 ± .004 ⎠
⎝ f L1 so ⎠
Now let’s substitute all these values to (1.4) to evaluate M exp
M =
si 13.5cm* (.067 ± .002 )
.135
×
=
×
= 3.2 ± .2
so
fe
(.056 ± .004 ) (.0507 ± .0005 )
*Note that I used the near focus distance of my near-sighted left eye, which is 13.5cm.
By comparing M exp to M obs , they agree within the calculated uncertainties.
Requirement 9
The distance measurements
⎧⎪ x1 = (.112 ± .002 ) m
⎨
⎪⎩ x3 = (1.094 ± .002 ) m
4 ± .5
= .8 ± .1
5 ± .5
Now let’s use the eqn(12) to calculate the expected magnification M exp
The observed magnification is M obs =
M exp =
fo
f
(.100 ± .002 )
= L3 =
= 3.26 ± .03
fe
f L1 (.0306 ± .0007 )
The observed magnification does not agree with the expected magnification.
Requirement 10
I have observed chromatic aberration and distortion for the duration of this experiment. The effect of distortion became significant
as an object distance decreased and as a focus distance decreased. The shape of the distortion was spherical.
Discussions and Conclusion
The thin lens formula and lensmaker’s formula is experimentally verified. A compound optical system is studied and both linear
and angular magnifications are calculated and experimentally verified. Thin lens formula is used to determine the focal distance of
lenses using displacement method. The focus distance are measured in three different methods: direct, diopter meter, and
displacement method and the focus distances using different methods generally agree. A microscope and telescope is constructed
using two convex lens system but the theoretic magnification did not agree with the experimental value for the telescope. The
angular magnification formula is modified since the observer’s eye (author’s) has smaller near focus distance.
-end-