Home Work 13
1. Figure 1 depicts a simplistic optical fiber: a plastic core (n1 = 1.58) is surrounded by a plastic sheath (n2 =1.53). A light ray is incident on one end of the fiber at angleθ. The ray is to undergo total internal reflection at point A, where it encounters the core–sheath boundary. (Thus there is no loss of light through that boundary.) What is the maximum value ofθthat allows total internal reflection at A?
Sol
When examining Fig. 33-62, it is important to note that the angle (measured from the central axis) for the light ray in air, q, is not the angle for the ray in the glass core, which we denote q ' . The law of refraction leads to
1
sin 1 sin
n
assumingnair 1.The angle of incidence for the light ray striking the coating is the complement of q ', which we denote as q'comp and recall that
sincomp cos 1sin2.
In the critical case, q'comp must equal qc specified by Eq. 33-47. Therefore,
n
n n
2 1
2
1
2
1 1 1
F
H G I K J
sincomp sin sin
which leads to the result: sin n12n22. With n1 = 1.58 and n2 = 1.53, we obtain
sin1
c
158. 2153. 2h
23 2. .2. In Fig. 2, a light ray in air is incident on a flat layer of material 2 that has an index of refraction n2 = 1.5. Beneath material 2 is material 3 with an index of refraction n3.The ray is incident on the air–material 2 interface at the Brewster angle for that interface. The ray of light refracted into material 3 happens to be incident on the material 2–material 3 interface at the Brewster angle for that interface. What is the value of n3?
Sol
Since the layers are parallel, the angle of refraction regarding the first surface is the same as the angle of incidence regarding the second surface (as is suggested by the notation in Fig. 33-66). We recall that as part of the derivation of Eq. 33-49 (Brewster’s angle), the refracted angle is the complement of the incident angle:
2 (1)c 90 1.
We apply Eq. 33-49 to both refractions, setting up a product:
3 3
2
B1 2 B 2 3 1 2
1 2 1
(tan ) (tan ) (tan )(tan ).
n n
n
n n n
Now, since θ2 is the complement of θ1 we have
tan tan ( )
tan .
2 1
1
c 1
Therefore, the product of tangents cancel and we obtain n3/n1 = 1. Consequently, the third medium is air: n3 = 1.0.
3. An isotropic point source emits light at wavelength 500 nm, at the rate of 200 W. A light detector is positioned 400 m from the source. What is the maximum rate ∂ B/ ∂ t at which the magnetic component of the light changes with time at the detector’s location?
Sol
From the equation immediately preceding Eq. 33-12, we see that the maximum value of B/t is wBm . We can relate Bm to the intensity:
2 0 m m
c I B E
c c
,
and relate the intensity to the power P (and distance r) using Eq. 33-27. Finally, we relate w to wavelength l using w = kc = 2pc/l. Putting all this together, we obtain
0 6 max
2 2
3.44 10 T/s 4
P
B c
t c r
.
4. A small laser emits light at power 5.00 mW and wavelength 633 nm. The laser beam is focused (narrowed) until its diameter matches the 1266 nm diameter of a sphere placed in its path. The sphere is perfectly absorbing and has density 5.00 x 103 kg/m3.What are (a) the beam intensity at the sphere’s location, (b) the radiation pressure on the sphere, (c) the magnitude of the
corresponding force, and (d) the magnitude of the acceleration that force alone would give the sphere?
Sol
(a) We note that the cross section area of the beam is d 2/4, where d is the diameter of the spot (d = 2.00). The beam intensity is
I P
d
2
3
9 2
9 2
4
5 00 10
2 00 633 10 4
3 97 10 /
.
. /
. .
W m
W / m
b gc h
(b) The radiation pressure is
p I
r c
3 97 10
2 998 10 13 2
9 2
8
.
. W / m .
m / s Pa.
(c) In computing the corresponding force, we can use the power and intensity to eliminate the area (mentioned in part (a)). We obtain
F d
p P
I p
r
F
r rH G IK J
F
H GIK J
2 3
2
11
4
5 00 10 13 2
167 10
. .
W Pa .
3.97 10 W / m9 N.
c hb g
(d) The acceleration of the sphere is
a F
m
F d
r r
( / )
( .
)[( . )(
. .
3
11
9 3
6
6 167 10
2 00 633 10 314 10
N)
kg / m m)]
m / s
3 3
2
5. A beam of partially polarized light can be considered to be a mixture of polarized and unpolarized light. Suppose we send such a beam through a polarizing filter and then rotate the filter through 360° while keeping it perpendicular to the beam. If the transmitted intensity varies by a factor of 5.0 during the rotation, what fraction of the intensity of the original beam is associated with the beam’s polarized light?
Sol
Let I0 be the intensity of the incident beam and f be the fraction that is polarized. Thus, the intensity of the polarized portion is f I0. After transmission, this portion contributes f I0 cos2 to the intensity of the transmitted beam. Here is the angle between the direction of polarization of the radiation and the polarizing direction of the filter. The intensity of the unpolarized portion of the incident beam is (1– f )I0 and after transmission, this portion contributes (1 – f )I0/2 to the transmitted intensity. Consequently, the transmitted intensity is
2
0 0
cos 1(1 ) . I f I 2 f I
As the filter is rotated, cos2 varies from a minimum of 0 to a maximum of 1, so the transmitted intensity varies from a minimum of
Imin 1( f I)
2 1 0
to a maximum of
max 0 0 0
1 1
(1 ) (1 ) .
2 2
I f I f I f I
The ratio of Imax to Imin is I
I
f f
max min
.
1 1
Setting the ratio equal to 5.0 and solving for f, we get f = 0.67.
6. When the rectangular metal tank in Fig. 33-51 is filled to the top with an unknown liquid, observer O, with eyes level with the top of the tank, can just see corner E. A ray that refracts toward O at the top surface of the liquid is shown. If D = 85.0 cm and L = 1.10 m, what is the index of refraction of the liquid?
Sol
Note that the normal to the refracting surface is vertical in the diagram. The angle of refraction is
2 = 90° and the angle of incidence is given by tan 1 = L/D, where D is the height of the tank and L is its width. Thus
1 1
1
1.10 m
tan tan 52.31 .
0.850 m L
D
The law of refraction yields
n1 n2 2
1
100 90
52 31 126
F H
G I K J
sin
sin ( . ) sin
sin . . ,
where the index of refraction of air was taken to be unity.
Fig. 1 Fig. 2 Fig. 3
