sin sin sin : chain nair θair nglass θglass nwater water 33.11: As shown below, the angle between the beams and the prism is A/2 and the angle between the beams and the vertical is A
Trang 133.1: a) 2.04 10 m s.
47 1
s m 10 00
n
c v
47 1
) m 10 50 6 (
λ
n
Hz 10 5.80
s m 10 00 3
14
8 vacuum
f c
b) λ (5.803.00101410Hz)(1.52)m s 3.40 10 7 m.
8
fn
c
33.3: a) 1.943.00 10108 m/sm/s 1.54.
8
v
c n
b) λ0 λ (1.54 (3.55 10 7 m) 5.47 10 7 m.
n
7 Benzene
water water Benzene
Benzene water
n
n n
5 47 equal
always are
angles reflected and
Incident
:
33.5
b) 2 2 arcsin sin 2 arcsin 11..6600sin42.5 66.0.
b
a b
n
33.6: 11.25.5010m9s 2.17108 m s
t
d
v
38 1 m/s 10 2.17
m/s 10 00 3
8
8
v
c
n
33.7: n asina n bsinb
s m 10 51 2 194 1 / ) m/s 10 00 3 ( so
194 1 1 48 sin
7 62 sin 00 1 sin
sin
8
n c v v c
n
n
n
b
a a
Trang 233.8 (a)
Apply Snell’s law at both interfaces
33.9: a) Let the light initially be in the material with refractive index n a and let the third and final slab have refractive index nb Let the middle slab have refractive index n1
1
1sin sin
: interface
b b
n
: interface
sin sin
gives equations two
the
b) For N slabs, where the first slab has refractive index na and the final slab has
sin sin
, , sin sin
, sin sin
, index
refractive n b n a a n1 1 n1 1 n2 2 n N2 N2 n b b
of angle
on the depends travel
of direction final
The sin sin
gives
This n a a n b b
incidence in the first slab and the indicies of the first and last slabs
33.10: a) arcsin sin air arcsin 11..3300sin35.0 25.5 .
water
air
n n
b) This calculation has no dependence on the glass because we can omit that step in the
sin sin
sin
:
chain nair θair nglass θglass nwater water
33.11: As shown below, the angle between the beams and the prism is A/2 and the angle
between the beams and the vertical is A, so the total angle between the two beams is 2A
Trang 333.12: Rotating a mirror by an angle while keeping the incoming beam constant leads
rotation
mirror
the from arose 2
of deflection additional
an where 2
2 becomes beams
outgoing
and
incoming between
angle the Therefore
by angle incident
in the increase
an
to
θ
θ
33.13: arcsin sin arcsin 1.581.70sin62.0 71.8.
b
a b
n
33.14: arcsin sin arcsin 1.521.33sin45.0 38.2.
b
a b
n
53.2 is angle the Therefore surface
the
of
tilt
the
of because 15
additional
an is vertical the
from angle the so surface, the
to normal
the
from
33.15: a) Going from the liquid into air:
1.48
42.5 sin
1.00
a
n
58.1 35.0
sin 1.00
1.48 arcsin sin
arcsin :
b
a
n
n
b) Going from air into the liquid:
22.8 35.0
sin 1.48
1.00 arcsin sin
b
a b
n
33.16:
c
θ
θ
circle, largest
for the
so
escapes, light
no angle, critical
If
2 2
2
c
1 w
1
c
air
c
w
m 401 m)
(13.3
m 11.3 6
48 tan ) m 0 10 ( m
0 10
/
tan
48.6 1.333
1 sin ) /
1
(
sin
00 1 ) 00 1 ( 00 1 ( 90 sin sin
π
πR
A
R
R
n
θ
n
θ
n
Trang 433.17: For glasswater,θcrit 48.7
1.77 sin48.7
1.333 sin
so , 90 sin sin
crit
a b
a
n n
n n
33.18: (a)
00 1 90 sin ) 00 1 ( sin : AC
at occurs reflection
internal
41.1
1.00 sin
(1.52)
θ
the is answer this
so angle, critical than the
less thus and smaller is
larger, is
largest that can be
(b) Same approach as in (a), except AC is now a glass-water boundary
61.3
1.333 sin
1.52
1.333 90
sin
θ θ
n θ n
90 61.3 28.7
33.19: a) The slower the speed of the wave, the larger the index of refraction—so air has a larger index
of refraction than water.
s m 1320
s m 344 arcsin arcsin
arcsin
b)
water
air
v
v n
n
a
b
c) Air For total internal reflection, the wave must go from higher to lower index of refraction—in this case, from air to water.
24.4 2.42
1.00 arcsin arcsin
a
b
n
n
:
33.20
40 1 40
1 54.5 tan tan
a
b
n
n
:
33.21
35.6 54.5
sin 1.40
1.00 arcsin sin
arcsin
b
a b
n
Trang 5: so and , 0 37 page,
next
on the picture the
:
33.22
1.77
37 sin
53 sin 1.33 sin
sin
b
a a
n
1.65
tan31.2
1.00 tan
tan
p
b a
a
b p
n n
n
n
:
33.23
58.7 31.2
sin 1.00
1.65 arcsin sin
arcsin
b
a b
n
58.9 1.00
1.66 arctan arctan
air
In
)
a
b p
n
n
:
33.24
51.3 1.33
1.66 arctan arctan
In water
a
b
n
2
1 : filter first e Through th
a)
:
33.25
285 0 ) 0 41 ( cos 2
1 : filter second
0
b) The light is linearly polarized.
max
2 maxcos cos (22.5 ) 0.854
:
33.26
max
2 maxcos cos (45.0 ) 0.500
max
2 maxcos cos (67.5 ) 0.146
polarized is
light the and m W 10.0 is
intensity filter the
first
0 2
1 I
:
where , cos is
filter second after the
intensity The
filter
first the of axis
the
I
I
37.0 25.0
62.0 and
m
W
Trang 633.28: Let the intensity of the light that exits the first polarizer be I1 , then, according to repeated
application of Malus’ law, the intensity of light that exits the third polarizer is
)
0 23 0 62 ( cos ) 0 23 ( cos cm
W
0
1
incident intensity
the also is which ,
) 0 23 0 62 ( cos ) 0 23 ( cos
cm W 0 75 that
see
we
I
on the third polarizer after the second polarizer is removed Thus, the intensity that exits the third polarizer after the second polarizer is removed is
cm W 32.3 )
23.0 (62.0
cos
)
(23.0
cos
) (62.0 cos cm
W
2 2
2 2
125 0 ) (45.0 cos ,
250 0 ) 0 45 ( cos 2
1 , 2
1
2 3 0
2 0 2 0
:
33.29
0
) (90.0 cos 2
1 , 2
1
0 2 0
I
33.30: a) All the electric field is in the plane perpendicular to the propagation direction,
and maximum intensity through the filters is at to the filter orientation for the case of minimum intensity Therefore rotating the second filter by 90 when the situation
originally showed the maximum intensity means one ends with a dark cell
b) If filter P1 is rotated by 90, then the electric field oscillates in the direction pointing
toward the P2 filter, and hence no intensity passes through the second filter: see a dark cell
c) Even if P2 is rotated back to its original position, the new plane of oscillation of the electric field, determined by the first filter, allows zero intensity to pass through the second filter
33.31: Consider three mirrors, M1 in the (x,y)-plane, M2 in the (y,z)-plane, and M3 in the
(x,z)-plane A light ray bouncing from M1 changes the sign of the z-component of the velocity, bouncing from M2 changes the x-component, and from M3 changes the
y-component Thus the velocity, and hence also the path, of the light beam flips by 180
46.6 9.73
sin 344
1480 arcsin sin
arcsin sin
arcsin
a)
a
b a
b
a b
v
v n
:
33.32
1480
344 arcsin arcsin
b
a
v
v
33.33: a) n1sin1 n2sin2 andn2sin2 n3sin3,son1sin1 n3sin3
with material
in the noral the respect to with
angle same the makes light the
and
sin
sin so , sin sin
and sin sin
b) / ) sin
(
sin
3
3
1 1 1 1 2 2 2
2 3 3 3
1 1
3
n
n n
n n
n n
1
n as it did in part (a).
c) For reflection, r a These angles are still equal if becomes the incident r angle; reflected rays are also reversible
Trang 733.34: It takes the light an additional 4.2 ns to travel 0.840 m after the glass slab is
inserted into the beam Thus,
ns
4.2 m 0.840 ) 1 ( m 0.840 m
c
n c
n c
We can now solve for the index of refraction:
2.50
1 m
0.840
) s m 10 (3.00 s) 10
n
The wavelength inside of the glass is
nm
200 nm 196 2.50
nm 490
38 1
00 1 arcsin 90
arcsin
b
a
n
00 1
) 6 43 sin(
38 1 arcsin
sin arcsin
sin
a
b b a
b b a
n n
2 n
n
b b b a a
sin
51.7 2
1.80 arccos 2 )
80 1 ( 2 cos
2
2 sin (1.80) 2
cos 2 sin 2 2 2 sin sin
(1.00)
a a
a a
a a
a
33.37: The velocity vector “maps out” the path of the light beam, so the geometry as
shown below leads to:
, arccos
arccos
r a r
r a
a r
a r
v
v v
v v
sign chosen by inspection Similarly, arcsin arcsin
x x x
x
r a r
r a
a
v v v
v v
v
Trang 833.38:
m 10 5.40
m 0.00250 m
10 5.40
m) 0.00250 m
(0.0180 λ
λ λ
# ) λ (#
λ
glass
10 3.52
(1.40) 4
33.39:
7 40 sin(
1 0
1 arcsin arcsin
7 40 m
00310 0
2 / ) m 00534 0 ( arctan
n n
n
a
b
Note: The radius is reduced by a factor of two since the beam must be incident at crit, then reflect
on the glass-air interface to create the ring
m 2 1
m 5 1 arctan
a
36 51
sin 1.33
1.00 arcsin sin
b
a b
n
So the distance along the bottom of the pool from directly below where the light
enters to where it hits the bottom is:
x(4.0m)tanb (4.0m)tan362.9m
xtotal 1.5m x1.5m2.9m4.4m
cm 16.0
cm 4.0 arctan and
27 cm 16.0
cm 8.0
14 sin
27 sin 00 1 sin
sin sin
b
a a b b b a a
n n n
n
33.42: The beam of light will emerge at the same angle as it entered the fluid as seen by
following what happens via Snell’s Law at each of the interfaces That is, the emergent
beam is at 42.5 from the normal
333 1
000 1 arcsin 90
sin
w
a
n
The ice does not come into the calculation sincenairsin90nicesin c n wsini
b) Same as part (a)
45 sin
90 sin 33 1 sin
sin sin
a
b b a b b a a
n n n
n
Trang 933.45:
b
a a b
b b a a
n
n n
arcsin sin
sin
6 44 00
1
) 0 25 ( sin 66
1
So the angle below the horizontal is b 25.044.625.019.6,and thus the angle between the two emerging beams is 39.2
90 sin
60 sin 62 1 sin
sin sin
a
b b a b b a a
n n n
n
90 sin
2 57 sin 52 1 sin
sin sin
a
b b a b b a a
n n n
n
33.48: a) For light in air incident on a parallel-faced plate, Snell’s Law yields:
sin
sin sin
sin sin
b) Adding more plates just adds extra steps in the middle of the above equation that always cancel out The requirement of parallel faces ensures that the angle n n and the chain of equations can continue
c) The lateral displacement of the beam can be calculated using geometry:
cos
) sin(
cos and
) sin(
b
b a b
b a
t d
t L L
d
80 1
0 66 sin arcsin
sin arcsin
n
b
5 30 cos
) 5 30 0 66 sin(
) cm 40
2
d
33.49: a) For sunlight entering the earth’s atmosphere from the sun BELOW the
horizon, we can calculate the angle as follows:
n asina n bsinb (1.00)sina nsinb, where n b n is the atmosphere’s index of refraction But the geometry of the situation tells us:
sinb R Rh sina R nRh a b arcsinR nRharcsinR Rh
m 10 2.0 m 10 64
m 10 6.4 arcsin
m) 10 2.0 m 10 6.4
m) 10 (6.4 (1.0003)
6 4
6
6
0.22 This is about the same as the angular radius of the sun, 0.25
Trang 1033.50: A quarter-wave plate shifts the phase of the light by 90 Circularly polarized light is out of phase by 90, so the use of a quarter-wave plate will bring it back into phase, resulting in linearly polarized light
8
1 ) sin (cos 2
1 ) 90 ( cos cos 2
0
2 0
2 2
I
b) For maximum transmission, we need 2 90,so 45
33.52: a) The distance traveled by the light ray is the sum of the two diagonal segments:
2 2 2
2
1
x
Then the time taken to travel that distance is just:
c
y x l y
x c
d
t
2 2 2 2 2
2 1
b) Taking the derivative with respect to x of the time and setting it to zero yields:
sin
sin )
(
) (
0 )
( ) ( ) (
1
) ( ) (
1
2 1 2 1
2 2 2 2
1 2
2 2 2 2 2
2 1 2
2 2 2 2 2
2 1 2
y x l
x l y
x x
y x l x l y
x x c dx dt
y x l y
x dt
d c dx dt
33.53: a) The time taken to travel from point A to point B is just:
2
2 2
2 1
2 2 1 2
2 1
1
v
x l h v
x h v
d v
d
Taking the derivative with respect to x of the time and setting it to zero yields:
sin sin
) (
) ( and
But
) (
) ( )
( 0
2 2 1 1 2 2
2
2 2
2 1
1 2
2 1
1
2 2
2 2 2 2 1 1 2
2 2
2 1
2 2 1
n x
l h
x l n x
h
x n n
c v n
c v
x l h v
x l x
h v
x v
x l h v
x h dt
d dx
dt
Trang 1133.54: a) n decreases with increasing λ, so n is smaller for red than for blue So beam a
is the red one
b) The separation of the emerging beams is given by some elementary geometry
, tan tan
tan tan
v r
v r
v
r
x d
d d
x
x
x
separation as they emerge from the glass 2.92mm
20 sin
mm 00
geometry, we also have
cm 9 5 34 tan 7 35 tan
mm 92 2 tan
tan
: so , 5 34 66
1
70 sin arcsin and
7 35 61
1
70 sin arcsin
v r
v r
x d
2 sin sin
sin
2
sin 2
2 sin 2
sin 2
A n A
A A
2
sin 2
sin 2
so , A n A
b) From part (a), n AA
2 sin arcsin 2
9 38 0 60 2
0 60 sin ) 52 1 ( arcsin
c) If two colors have different indices of refraction for the glass, then the deflection angles for them will differ:
0 5 2 47 2 52 2
52 0 60 2
0 60 sin ) 66 1 ( arcsin 2
2 47 0 60 2
0 60 sin ) 61 1 ( arcsin
2
violet
red
Trang 12Direction of ray A: by law of reflection
Direction of ray B:
At upper surface: n1sin n2sin
The lower surface reflects at Ray B returns to upper surface at angle of
incidence :n2 sin n1sin
Thus
Therefore rays A and B are parallel
33.57: Both l-leucine and d-glutamic acid exhibit linear relationships between
concentration and rotation angle The dependence for l-leucine is:
Rotation angle (0.11100ml g)C(g/100ml),and for d-glutamic acid is: Rotation angle (0.124100ml g)C(g/100 ml)
33.58: a) A birefringent material has different speeds (or equivalently, wavelengths) in
two different directions, so:
) (
4
λ 4
1 λ λ
4
1 λ λ
λ λ and
λ
λ
2 1
0 0
2 0
1 2
1 2
0 2 1
0
1
n n D D
n D n D
D n
) 635 1 875 1 ( 4
m 10 89 5 ) (
4
2 1
n n D