GateWay CC
PHY101 Physics Lab:
OPTICS
Purpose: To introduce the basic ideas of geometrical optics
Part I – To determine the index of refraction
Part II – To determine the image distance based on Ray tracing method
Part III – To determine focal length of a thin lens
Part I: Index of refraction
Please go to: www.gwc.maricopa.edu/class/phy101/simulation/optics/index1.htm
The idea of geometrical optics is used to understand the light behavior. This concept is based on
the fact that light travel along the straight line until it get reflected (if it hits a non-transparent
material), refracted (if it hits a transparent material) or absorbed (if it hits a reflective material).
If you place a straight stick half-submerged into water, you will notice that the stick appears
bent at the point it enters the water. This optical effect is due to refraction of light. As light
passes from one transparent medium to another (air to water), it changes speed, and bends.
How much slower light travels, and how much light bends it depends on the index of
refraction of the mediums (air – water) and the angle of incidence (see figure 1).
Figure 1. Light refraction
Light waves propagate through air with a constant speed c = 3x108 m/s. As the waves travel
through transparent material (water, glass, etc) it travels with speed v which is smaller than c.
The ratio between these two velocities is called a index of refraction:
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n
c
v
(1)
In this simulation you will determine the index of refraction of various materials, using the
following equation known as Snell’s Law:
n1  angleofincidence  n2  angleofrefraction
n1  angleofinc idence
 n2 
angleofref raction
(2)
Follow the directions in the simulation to complete the table below.
Medium
Air
Ice
Water
Glass
Sapphire
Berisha:optics.doc
Angle of
Incidence
20
30
40
50
60
70
80
20
30
30
40
50
60
70
80
20
30
40
50
60
70
80
20
30
40
50
60
70
80
20
30
40
50
60
70
80
Angle of
Refraction
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Index of
Refraction
Speed of light in
Medium
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Part II: Image Location using Ray Tracing Model
Please go to: www.gwc.maricopa.edu/class/phy101/simulation/optics/index2.htm
The Ray Tracing Model is used to predict the image location and its orientation for a converging
or diverging lens. In this experiment we will study the propagation of light through a lens,
determine the focal length of the lens, and verify lens equation:
1
1
1


f do di
(3)
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 d i. The distance from
the center of the lens to focal point is called focal length f.
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 focal length of a lens is inversely related to the focusing
power of the lens: a shorter focal length has a stronger focusing power.
The optical axis of the lens is indicated by the dotted line in figures bellow.
The three principle light rays are indicated by the arrows, and are labeled 1,2, and 3, in figures
1, 2, and 3. Note that the three principle rays come from one point on the object – top of an
object.
Ray 1.
The ray parallel to the optical axis is bent-refracted by the lens pass through f2:
Figure 2. 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 3. 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:
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Figure 4. Ray 3 goes through the focal point
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.
SIGN CONVENTION
Start with a real object to the left of the lens, with light going from left to right as shown figures
4 in this manual.
do > 0 - object distance is positive when object is on the left side of the lens (this will be real
object)
do < 0 – object distance is negative when object is on the right side of the lens (this will be
virtual object)
f > 0 - for a converging lens (positive lens)
f < 0 - for a diverging lens (negative lens)
di > 0 – image distance is positive when image is on the right side of the lens (this will be a
real image)
di < 0 – image distance is negative when image is on the left side of the lens (this will be a
virtual image)
In the simulation you will change the object distance according to table below. At the end of
each run, computer will display image distance and other data related to image. Record these
data on your table.
Focal Length
Object
distance
cm
10
10
10
10
10
Image
distance
Calculated
focal length
Absolute
difference
Relative
difference
25
20
15
10
5
Calculate the absolute and relative difference between Focal length and calculated focal length
using the following formulas:
Absolute Difference  Focal Length  Calculated Focal Length
Relative Difference 
Berisha:optics.doc
Focal Length  Calculated Focal Length
x100%
Calculated Focal Length
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Part II: Image Location using Optical Bench
Please go to: www.gwc.maricopa.edu/class/phy101/simulation/optics/index3.htm
Simulation for part III, is designed based on figure 5.
Figure 5. Optical bench
Start the simulation by selecting the largest object distance of 30 cm. At this point image screen
will start moving through the optical bench. You can control the screen by pressing STOP and
GO buttons. The goal is to stop the screen at the point when the image is the sharpest. Read the
image distance in cm from the optical bench. Calculate the focal length of the lens using
equation 3. Repeat the procedure by selecting the object distance according to table below.
Focal
Length
cm
10
10
10
10
10
Object
distance
cm
25
20
15
10
5
Image
distance
cm
Calculated
focal length
cm
Absolute
difference
cm
Relative
difference
%
Magnification
Image
Description
Magnification is the ratio between the size (height) of an image to the size (height) of an object.
For this example the magnification of the image is:
M
hi
d
 i
ho
do
When the magnification is equal to one, the size of an image is equal to size of an object. When
the magnification is negative, the image is inverted. When the magnification is positive, the
image is erect. When the magnification is larger than 1, image is larger than object. When the
magnification is less than 1, image is smaller than object.
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