This preview shows page 1. Sign up to view the full content.
Unformatted text preview: ses is
2 = S1 cos 2 θ
=
S2
A
= S1 cos 2 θ
=
S2
B
= S1 cos 2 θ
S2
=
C cos 0°
( 24 W/m ) = 24 W/m
=
( 24 W/m ) cos 90° 0 W/m
30.0° )
( 24 W/m ) cos ( 60.0° −=
2 2 2 2 2 2 2 2 18 W/m 2 ______________________________________________________________________________
45. REASONING In experiment 1, the light falling on the polarizer is unpolarized, so the
average intensity S transmitted by it is S = 1 S ′ , where S ′ is the intensity of the incident
2
light. The intensity of the light passing through the analyzer and reaching the photocell can
be found by using Malus’ law.
In the second experiment, all of the polarized light passes through the polarizer, so the
average intensity reaching the analyzer is S ′ . The intensity of the light passing through the
analyzer and reaching the photocell can again be found by using Malus’ law. By setting the
intensities of the light reaching the photocells in experiments 1 and 2 equal, we can
determine the number of additional degrees that the analyzer in experiment 2 must be
rotated relative to that in experiment 1. 1304 ELECTROMAGNETIC WAVES SOLUTION In experiment 1 the light intensity incident on the analyzer is S0 = 1 S ′ .
2
Malus' law (Equation 24.7) gives the average intensity S1 of the light reaching the photocell
as
= S0 cos 2 60.0° 1 S ′ cos 2 60.0°
S1
=2
In experiment 2 the incident light is polarized along the axis of the polarizer, so all the light
is transmitted by the polarizer. Thus, the light intensity incident on the analyzer is S0 = S ′ .
Malus' law again gives the average intensity S2 of the light reaching the photocell as
= S0 cos 2 θ S ′ cos 2 θ
S2 = Setting S1 = S2 , we have
1
2 S ′ cos 2 60.0° = S ′ cos 2 θ Algebraically eliminating S ′ from this equation, we have that
1 cos 2 60.0°
=
2 cos 2 θ or = cos −1
θ ) ( 1 cos 2 60.0=
°
2 69.3° The number of additional degrees that the analyzer must be rotated is 69.3° − 60.0° = 9.3° .
The angle θ is increased by the additional rotation.
_____________________________________________________________________________________________ 46. REASONING The rms value Erms of the electric field is related to the average intensity S
2
of the light by S = cε o Erms (Equation 24.5b). The average intensity S of the light transmitted by the polarizer is related to the incident intensity S0 by Malus’ law,
S = S0 cos 2 θ , where θ is the angle between the transmission axis and the direction of
polarization. These two relations will allow us to determine the rms value of the electric
field. SOLUTION Combining the two equations given above and solving for the rms value of the
electric field, we have =
Erms S0 cos 2 θ
=
cε 0 (15 W/m ) cos
m/s ) 8.85 × 10 2 2 25° = 68 N/C
2
2
C / N⋅m ______________________________________________________________________________ (3.0 × 10 8 −12 ( ) Chapter 24 Problems 47. 1305 SSM WWW REASONING The average intensity of light leaving each analyzer is
given by Malus' Law (Equation 24.7). Thus, intensity of the light transmitted through the
first analyzer is
S1 = S0 cos 2 27° Similarly, the intensity of the light transmitted through the second analyzer is
S 2 = S1 cos 2 27° = S0 cos 4 27° And the intensity of the light transmitted through the third analyzer is
S 3 = S2 cos 2 27° = S 0 cos 6 27° If we generalize for the Nth analyzer, we deduce that
S N = S N –1 cos 2 27° = S 0 cos 2 N 27° Since we want the light reaching the photocell to have an intensity that is reduced by a least
a factor of one hundred relative to the first analyzer, we want S N / S 0 = 0.010 . Therefore,
we need to find N such that cos 2 N 27° = 0.010 . This expression can be solved for N.
SOLUTION Taking the common logarithm of both sides of the last expression gives
log 0.010
= 20
2 log (cos 27 °)
______________________________________________________________________________
2 N log(cos 27 °) = log 0.010 or N= 48. REASONING No light intensity will pass through two adjacent polarizers that are in a
crossed configuration, that is, whose transmission axes are oriented perpendicular to one
another. With this in mind, let’s remove the polarizers one by one (see the following
drawing). When A is removed, no two of the remaining adjacent polarizers are crossed.
When B is removed, A and C are left in a crossed configuration. When C is removed, B and
D are left in a crossed configuration. When D is removed, no two of the remaining adjacent
polarizers are crossed. Thus, when sheet B or C is removed, the intensity transmitted on the
right is zero, and when sheet A or D is removed, the intensity transmitted on the right is
greater than zero. 1306 ELECTROMAGNETIC WAVES horizontal vertical
30.0º 60.0º Light D
C
B
A
We can anticipate that the greater intensity is transmitted on the right when sheet D is
removed. To begin with, we note that the polarization directions of the light striking B and C
are the same, no matter whether A or D is removed, with the result that B and C absorb the
same fraction of t...
View Full
Document
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details