Image Formation by Lenses

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PHY101 Physics Lab:
BY LENSES
IMAGE
FORMATION
RAY TRACING MODEL FOR IMAGE FORMATION BY LENSES
The Ray Tracing Model is used to predict the image location and its orientation for a converging
or diverging lens. In the figures below, f1 or f2 are the focal points of the lens. The focal length of
the lens is the distance from the center of the lens to f1 or f2. If the lens is symmetrical (most
often is) then the focal length on both sides is equal. The optical axis of the lens is indicated by
the dotted line.
The three principle light rays are indicated by the arrows, and are labeled 1,2, and 3. Note that
the three principle rays come from one point on the object – top of an object.
CONVERGING LENS – CONVEX LENSES
Ray 1.
The ray parallel to the optical axis is bent-refracted by the lens pass through f2:
Figure 1. Ray 1 parallel to optical axis
Ray 2.
The ray through the center of the lens continues in a straight line with no change in direction:
Figure 2. Ray 2 goes through the center of the lens
Ray 3.
The ray through f1 is bent-refracted by the lens to travel in a path parallel to the optical axis:
Figure 3. Ray 3 goes through the focal point
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The location of the image is at the point where three rays intercept. This example shows that
when object is located to the left of focal point f1, the image is INVERTED.
The distance from the center of the lens to the object is called object distance d o. The distance
from the center of the lens to the image location is called image distance di. The distance from
the center of the lens to focal point is called focal length f.
SIGN CONVENTIONS
Start with a real object to the left of the lens, with light going from left to right as shown figures
in this manual.
do &gt; 0 for a real object.
do &lt; 0 for a virtual object.
f &gt; 0 for a converging lens
f &lt; 0 for a diverging lens
di &gt; 0 for an image on the right side of the lens (this will be a real image)
di &lt; 0 for an image on the left side of the lens (this will be a virtual image)
When the object is located at the point exactly 2F (object distance equal to two focal length),
the image is also located at the distance of 2F from the lens.
Figure 4. Object is located at 2F or between F and 2F. In both cases image is INVERTED
Example 1: An object is placed a distance of do = 20 cm from a converging lens having a focal
length of f = 10 cm. Determine the image distance and the image magnification.
1
1
1


f do di
1
1
1


10 20 d i
1
1
1
1



d i 10 20 20
d i  20cm
Because di &gt; 0, image is located to the right of the lens
Magnification of the image is:
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h
d
M  i  i
ho
do
d
20
M  i 
 1
do
20
Image is INVERTED and has the same size as the object.
Example 2: An object is placed a distance of do = 15 cm from a converging lens having a focal
length of f = 10 cm. Object distance is between f and 2f (f &lt; d0 &lt; 2f). Determine the image
distance and the image magnification.
1
1
1


f do di
1
1
1
 
10 15 d i
1
1 1
1

 
d i 10 15 30
d i  30cm
Because di &gt; 0, image is located to the right of the lens
Magnification of the image is:
h
d
M  i  i
ho
do
d
30
M  i 
 2
do
15
Image is INVERTED. REAL, TWICE THE SIZE OF THE OBJECT and located to the right of the lens.
What happened when object is located at focal point?
Figure 5. Object is located at focal point
When object is located at the focal point, refracted rays do not intercept. As a consequence
image is located at infinity. See Figure 5 and example 3.
Example 3: An object is placed a distance of do = 10 cm from a converging lens having a focal
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length of f = 10 cm. Determine the image distance and the image size.
1
1
1


f do di
1
1
1
 
10 10 d i
1
1 1
  0
d i 10 10
di  
When object is located at focal point (do = f), image is located at infinity and it is small in size.
Magnification of the image is:
M
hi
d
 i
ho
do
M 

 
10
What happened now when object is located between the lens and focal point (d o &lt; f)?
Figure 6. Object is located between the lens and object
In this case image is image is VIRTUAL and located in front of the lens
Example 4: An object is placed a distance of do = 5 cm from a converging lens having a focal
length of f = 10 cm. Determine the image distance and the image size.
1
1
1


f do di
1 1 1
 
10 5 d i
1
1 1
1

 
d i 10 5
10
d i  10cm
Because di &lt; 0, image is located to the left of the lens
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Magnification of the image is:
h
d
M  i  i
ho
do
M 
 10
2
5
Image is virtual, erected, twice the size of the object and located in front of the lens.
DIVERGING LENS - CONVEX LENS
The method of three rays is used to determine image location for Diverging lens too.
Ray 1.
The ray parallel to the optical axis is bent-refracted by the lens, and appears to come from f2:
Figure 7. Ray 1 parallel to optical axis
Ray 2.
The ray through the center of the lens continues in a straight line with no change in direction:
Figure 8. Ray 2 goes through the center of the lens
Ray 3.
The ray headed toward f1 is bent-refracted by the lens to travel in a path parallel to the optical
axis:
Figure 9. Ray 3 is headed toward the focal point
The image location will be at the intersection of the principle rays. If the rays diverge after
passing through the lens, then the point where the extensions of these rays intersect determines
the location of the image.
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The location and the size of the image for other object locations are illustrated in the figure 10.
Figure 10. Image location for diverging lens
LAB ACTIVITY I: Determine the characteristics of Image for Converging and Diverging
Lens
Type of
lens
Object
location
Image location
relative to lens
Image
RIGHT or
LEFT
CONV
  do  f
CONV
do  2 f
CONV
do  f
CONV
0  do  f
Object
location
Image location
relative to lens
RIGHT or
LEFT
DIV
- Same, Larger, or
Smaller
Image
REAL or
VIRTUAL
Image
ERECT or
INVERTED
Size of the image
relative to object
- Same, Larger, or
Smaller
do  
DIV
  do  f
DIV
do  2 f
DIV
do  f
DIV
0  do  f
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ERECT or
INVERTED
Size of the image
relative to object
do  
CONV
Type of
lens
REAL or
VIRTUAL
Image
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LAB ACTIVITY II: Finding the focal length
The equipments for this experiment is called “optical bench”. The apparatus consist of a light
source, a screen in which the image is displayed, and number of different lenses. Lenses can be
positioned along the straight line that is scaled (meter stick) for reading the position of an object
as well as image.
Figure 11. Optical Bench
The object is a light source which is a bright square with two dark, crossed arrows. These arrows
actually represent the object. The screen is a white square of cardboard on which a real image of
an object will be displayed – when located at a proper location.
A) FINDING THE FOCAL LENGTH OF A CONVERGING LENS
Based on the lens equation:
1
1
1


f do di
one can conclude that when do is very large then f = di. If an object distance is very large, we
can very good approximation of focal length, by simply measuring the image distance. In this
experiment we will place an object (light) and an image (screen) at opposite ends of optical
bench, and move the converging lens until a sharp image is formed on the screen. The distance
between lens and the screen is approximate value of the focal length of the lens.
Repeat the procedure for at least two different lenses.
Lens
Object
Distance
Image
Distance
Focal
Length
m
m
m
1
2
B) DETERMINE THE IMAGE LOCATION, SIZE AND OTHER CHARACTERISTICS OF AN IMAGE
FORMED BY CONVERGING LENS
The lens equation:
1
1
1


f do di
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is going to be used to determine the focal length of a lens. When we measure do and di solution
to lens equation will give the focal length.
Experiment procedure:
Position the lens anywhere on the scaled optical bench. Position the light (object) anywhere on
the scaled optical bench. Adjust the screen (image) to get a clear image.
Determine the object distance and image distance, do and di, from the scale on the optical
bench. Use lens equation to calculate the focal length.
On you chart describe the image: VIRTUAL or REAL, ERECT or INVERTED, SMALLER or LARGER
then object.
Repeat the procedure for several object locations.
Calculate the average value for focal length.
Tria
l
Object
Distance
Image
Distance
Focal
Length
m
m
m
Image
Description
1
2
3
4
Average focal length:
B) DETERMINE THE IMAGE LOCATION, SIZE AND OTHER CHARACTERISTICS OF AN IMAGE
FORMED BY DIVVERGING LENS
The image formed by diverging lens is always a virtual image. In order to determine the focal
length of a diverging lens we need to use two lenses:
- one converging lens that will create a virtual object for diverging lens, and
- on diverging lens whose image will be real
First we make a real image with converging lens. Then we place a diverging lens between the
converging lens and its image. This image now becomes an object (virtual) for diverging lens.
The final image is now formed by diverging lens and is going to be real image. The
arrangements of two lenses are shown in figure 12.
Figure 12. Determine the focal length of diverging lens
First we determine the image distance for converging lens only d iC by adjusting the screen until
image is sharp. After this we place the diverging lens behind the converging lens, as shown in
figure 12. Determine the image distance diD formed by diverging lens.
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The object distance for the virtual object of the diverging lens is:
doD = d – diC
Focal length of diverging lens is:
f 
d oD  d iD
d oD  d iD
Total magnification is:
d
d
M  iC  iD
d oC d oD
#
Units
Image Distance
Object distance for DIV lens
m
m
Image distance for DIV lens
Magnification for DIV lens
Lens Separation
Object distance for DIV lens
Image distance for DIV lens
Focal length of DIV lens
Magnification of DIV lens
Total Magnification
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