TP3 polarisation

advertisement
PHYSICS 332 LABORATORY REPORT
OPTICS Laboratory
Name and ID:
1. ………………………………………………………………………………………………
2. ……………………………………………………………………………………………….
Date performed: ……………………..… Time section meets:………………………
Title: Polarization of light
Grading Checklist
LABORATORY JOURNAL:
PREDICTIONS
(individual predictions and warm-up completed in journal before each lab
session)
LAB PROCEDURE
(measurement plan recorded in journal, tables and graphs made in journal as
data is collected, observations written in journal)
BONUS POINTS FOR TEAMWORK
LABORATORY REPORT:
ORGANIZATION
(clear and readable; logical progression from problem statement through
conclusions; pictures provided where necessary; correct grammar and
spelling; section headings
provided; physics stated correctly)
DATA AND DATA TABLES
(clear and readable; units and assigned uncertainties clearly stated)
RESULTS
(results clearly indicated; correct, logical, and well-organized calculations
with uncertainties indicated; scales, labels and uncertainties on graphs;
physics stated correctly)
CONCLUSIONS
(comparison to prediction & theory discussed with physics stated correctly ;
possible sources of uncertainties identified; attention called to experimental
problems)
TOTAL
1 ‫الصفحة‬
Polarisation
Points
I. Objects of the experiments
The purpose of this laboratory activity is to determine the relationship between the intensity
of the transmitted light through two polarizers and the angle, Ø, of the axes of the two
polarizers.
II. Principles and Theory
A polarizer only allows light which is
vibrating in a particular plane to pass
through it. This plane forms the “axis” of
polarization. Unpolarized light vibrates in
all planes perpendicular to the direction of
propagation.
If unpolarized light is incident upon an
“ideal” polarizer, only half will be
transmitted through the polarizer. Since in
reality no polarizer is “ideal”, less than half
the light will be transmitted.
The transmitted light is polarized in one
plane. If this polarized light is incident upon
a second polarizer, the axis of which is oriented such that it is perpendicular to the plane of
polarization of the incident light, no light will be transmitted through the second polarizer.
However, if the second polarizer is oriented at an angle so that it is not perpendicular to the
first polarizer, there will be some component of the electric field of the polarized light that lies
in the same direction as the axis of the second polarizer, thus some light will be transmitted
through the second polarizer (see the bottom figure).
The component, E, of the polarized electric field is found by:
𝐸 = 𝐸0 cos ∅
Since the intensity of the light varies as the square of the electric field, the light intensity
transmitted through the second filter is given by:
𝐼 = 𝐼0 𝑐𝑜𝑠 2 ∅
where 𝐼0 is the intensity of the light passing through the first filter and Ø is the angle between
the polarization axes of the two filters.
Consider the two extreme cases illustrated by this equation:
• If Ø is zero, the second polarizer is aligned with the first polarizer, and the value of 𝑐𝑜𝑠 2 Ø is
one. Thus the intensity transmitted by the second filter is equal to the light intensity that
passes through the first filter. This case will allow maximum intensity to pass through.
• If Ø is 90º, the second polarizer is oriented perpendicular to the plane of polarization of the
first filter, and the 𝑐𝑜𝑠 2 Ø gives zero. Thus no light is transmitted through the second filter.
This case will allow minimum intensity to pass through.
2 ‫الصفحة‬
Polarisation
• These results assume that the only absorption of light is due to polarizer effects. In fact most
polarizing films are not clear and thus there is also some absorption of light due to the
coloring of the Polaroid filters.
The experiment is set up according to Fig. 1. It must be made sure that the photocell is totally
illuminated when the polarization filter is set up.
Figure 1 :Experimental setup for investigating the type of polarization of the emergent light
(schematically).(a) halogen lamp (d) polarizer (f) analyzer (g) Si photo cell (h) translucent
screen
Questions
1) The polarization filter (analyzer) is then rotated in steps of 10° between the filter positions
+/- 90° and the corresponding photo cell current (most sensitive direct current range of the
digital multimeter) is determined.
Complete the table?
∅
90
80
70
60
50
40
30
20
10
0
10
20
30
40
50
60
70
𝑐𝑜𝑠 2 Ø
𝑰
𝑰/𝑰𝟎
2. Write the relationship between the intensity 𝐼 of light transmitted through the analyzer, and
the relative angle ∅ between the transmission axis of the analyzer and the plane of
polarization of the incident light, intensity 𝐼0 :
3 ‫الصفحة‬
Polarisation
80
90
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………
3. Describe in your own words what is mean by and draw a picture of unpolarized light and
polarized light.
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………
4. Plot the graphs of the intensity (I) of transmitted light as a function of the angle (∅) , in
radians, between the axis of analyzer and the plane of polarization (second polarizers) of light.
What is the shape of the plot of light intensity versus angle?
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
………………………………………………………….………
5. Plot the graphs of and intensity(𝑰/𝑰𝟎 ) as function of 𝑐𝑜𝑠 2 ∅.
What is the shape of the plot? Does the curve verify Malus’s law?
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………………………………………
…………………………………………………………………
4 ‫الصفحة‬
Polarisation
We see in Figure 2 that the intensity reaches a minimum at approximately 1.6 radians (_ 90◦),
when
the axis of the polariser is perpendicular to the polarisation of the light, as is expected. The
maxima
occur near where the relative angle is either 0◦ or 180◦.
5 ‫الصفحة‬
Polarisation
Figure 3 approximately verifies Malus’ law (equation 2), that the intensity of the transmitted
light is
proportional to the cosine of the relative angle. Towards the maxima there are deviations such
that
the graph appears to split into two lines. During the experiment it was found that the value of
the
intensities at the maxima were not equal, as should have occurred. This is believed to be due
to experimental
factors such as additional reflections of the laser light on parts of the apparatus, and perhaps
some inhomogeneity of the polariser itself. The line of greater slope corresponds to the
maximum around
the 0◦ point, while the lesser slope corresponds to the 180◦ region.
6 ‫الصفحة‬
Polarisation
Download