Chapter 35
Questions: 2, 5, 6, 7, 8, 11, 14, 19, 21
Problems: 1, 2, 5, 9, 13, 15, 25, 27, 33, 35, 37, 41, 43, 55
Questions
2. Low frequency sound waves have large wavelenghts. For example a
10Hz sound wave has a wavelength of ~34m or approximately 100f t: This
is often large compared to objects that it encounters which nulli…es the ray
approximation. At high frequencies the wavelengths are smaller. For visible
light the wavelengths are MUCH smaller.
5. Smoothness means that the roughness is on a scale smaller than the
wavelength of the light.
6. Applying Snell’s law to a glass slab with parallel surfaces results in
n1 sin 1 = n2 sin 2 = n1 sin 3 : This expression makes it clear that 1 = 3
independent of n2 . However the dispersion in n2 can spread the beam before
it exits so dispersion can still exist.
7. In …gure 35-15a, the glass prism has interior
p angles of 45 . Since the
index of refraction for glass satis…es n > 1:5 > 1= 2 ' 1:414; there will be
total internal re‡ection. If however the index of refraction is n = 1:25; this
will no longer be the case and the light will emerge through the diagonal
face.
8. The beam will emerge from the second prism parallel to its original
direction. However it will be displaced and if the index of refraction is frequency dependent, there will be dispersion. That is a rainbow of colors will
emerge.
11. Since the velocity of the radio waves is dependent of the index of
refraction, v = c=n; the di¤erent arrival times indicates that the Earth’s
atmosphere has an index that is frequency dependent.
14. In the day time there is so much light transmitted through the window
that you don’t see the re‡ections.
19. Light that re‡ects o¤ of surfaces is often polarized. A large fraction
of this polarized light does not penetrate through the polarizing sunglasses.
21. The polarizing angle is found from tan = n2 =n1 : For
< 45 ;
tan < 1 or n2 < n1 .
Problems
1. Since the angle of re‡ection equals the angle of incidence, you should
rotate the mirror 75 o¤ of normal incidence, 15 grazing incidence, for a light
ray to rotate 30 .
1
2. (a) After the …rst re‡ection the beam makes a 30 angle with the upper
mirror. Hence it arrives perpendicular to the bottom mirror. It re‡ects and
retraces its original directions. This means it makes 3 re‡ections.
(b) It exits the mirror system in the same direction that it arrived.
5. Assume an arbitrary half angle (as opposed to 30 in problem 2).
Then after the …rst re‡ection the beam makes an angle with respect to the
upper mirror. This means that as it arrives at the lower mirror it will have
an angle
3 with this mirror. This means that it will have an angle of
3 after re‡ection as well. (Note that this is consistent with the answer
to problem 2.) A little geometry shows that this is an angle =
4 with
a horizontal axis. Since 2 = 75 , = 30 : This is less than ( = 37:5 ).
Hence this re‡ected beam never reaches the upper mirror.
(a) There are two re‡ections.
(b) The exit angle for the incident beam is 4 = 150 : This means that
the beam has been rotated by 2
4 = 210 :
9. A quarter wavelength in glass is = (4n) where n is the index of refraction. For n = 1:55 and = 780nm this depth is
d = 780=6:2nm ' 125:8nm.
13. From Snell’s law n1 sin
1
= n2 sin
2:
For this problem
1:52 sin 27:9 = n sin 31:5 ;
n = 1:52 sin 27:9 = sin 31:5 = 1:3613:
From table 35-1 the ‡uid is “ethyl alcohol”.
15. If the side of the cube is a; then the exit angle, , is determined from
p
a=2
= 1= 5:
sin = q
a2 + (a=2)2
The minimum index of refraction (larger indices bend the light toward the
center of the cube) is found from
p
p
sin 55 = n= 5 ! n = 5 sin 55 = 1:83
25. Total internal re‡ection at the n1 -n2 interface occurs when
where
n1 sin 1c = n2 ! 1c = sin 1 n2 =n1
1:
2
1
1c
Similarly the total internal re‡ection at the n2 -n3 interface occurs when
2c where
n2 sin 2c = n3 ! 2c = sin 1 n3 =n2
2:
2
To avoid internal re‡ection at both interfaces we must satisfy
sin
n2 =n1 and sin
1
2
n3 =n2
This results in
sin
1
sin
2
n3 =n1
To avoid internal re‡ection at an n1 -n3 interface where n3 < n1 we must
satisfy
sin 13 n3 =n1
If the initial incident angle satis…es sin 1 < n3 =n1 ; then the product sin 1 sin 2
will satisfy sin 1 sin 2 n3 =n1 as the maximum value for sin 2 is one and
there will be no internal re‡ection.
27. The incident angle is 45 : The condition for the critical angle becomes
p
n sin 45 = 1 ! n = 2:
p
Clearly if n is larger than 2 then Snell’s law will not be satis…ed and there
will be internal re‡ection.
33. The critical angle of incidence for the cone is determined by n sin c =
1: For a depth of h, the radius of subtended cone is found from
p
r = h tan c = h sin c = cos c = h sin c = 1 sin2 c ;
p
p
r = hn sin c = n2 n2 sin2 c = h= n2 1:
This results in a diameter of
p
d = 2h= n2
1:
35. The critical angle of incidence for blue light is
cb
= sin
1
(1=1:680) = :6376rad = 36:5 :
The critical angle of incidence for red light is
cr
= sin
1
(1=1:621) = :6648rad = 38:1 :
3
Thus over a range of 1:59 ; from 36:5 to 38:1 , the blue light is totally
re‡ected and the red light is not.
37. Taking the indices of refraction to be
n656 = 1:517 and n486 = 1:526;
the refracted angles are found from
sin
sin
With a
656
486
= 2:8
= 1:517 sin 40 = :975109 !
= 1:526 sin 40 = :980894 !
= 1:3472rad
= 1:3750rad
10 2 rad the separation at 1m is
d=1
= 2:8cm
41. The polarizing angle is found from
tan
= n2 =n1 = 1= (4=3) = 3=4 !
= 36:9
43. From Snell’s law n1 sin 1 = n2 sin 2 : In air n = 1:0003: For an
incident angle of = 35 and an angle of refraction of 22 we …nd
1:0003 sin 35 = n sin 22 ;
n = 1:5316 = c=v:
This means
v=3
108 =1:5316 = 1:96
108 m=s.
55. First draw an isosceles triangle with an apex angle of : Then de…ne
an incident angle 1 and a refracted angle 2 : On the opposite wall of the
triangle de…ne an incident angle of 3 with a refracted angle of 4 . Summing
the interior angles of the quadrangle formed by the base, two sides of the
triangle, and the interior light beam show that
2
+
3
= :
Projecting the original light beam through the triangle show that it forms
an angle = =2 +
1 with the opposite surface. Then
=2
4
+ + =2 +
4
1
= ;
or
=
1
+
4
:
Now via Snell’s law we have two conditions,
sin
n sin
1
3
= n sin 2 ;
= sin 4 :
Using these relations we substitute into the expression for
=
=
=
1
1
+ sin
+ sin
1
+ sin
1
1
1
(n sin 3 ) ;
(n sin [
n sin
5
and …nd
2 ]) ;
sin
1
sin
n
1
: