Ch 27 HW Solutions
1.
SSM REASONING AND SOLUTION To decide whether constructive or destructive interference
occurs, we need to determine the wavelength of the wave. For electromagnetic waves, Equation 16.1 can
be written f   c , so that
c 3.00  10 8 m
 
 5.60  10 2 m
f 536  10 3 Hz
Since the two wave sources are in phase, constructive interference occurs when the path difference is an
integer number of wavelengths, and destructive interference occurs when the path difference is an odd
number of half wavelengths. We find that the path difference is 8.12 km – 7.00 km = 1.12 km. The number
of
wavelengths
in
this
path
difference
is
1.12  103 m / 5.60  102 m  2.00 .
Therefore,
constructive interference occurs .
3.
REASONING Let 1 and 2 be the distances from source 1 and source 2, respectively. For constructive
interference the condition is 2 – 1 = m, where  is the wavelength and m = 0, 1, 2, 3, …. For destructive
interference the condition is 2 – 1 = (m + 21 ), where m = 0, 1, 2, 3, …. The fact that the wavelength is
greater than the separation between the sources will allow us to decide which type of interference we have
and to locate the two places in question.
SOLUTION a. Since the places we seek lie on the line between the two sources and since the separation
between the sources is 4.00 m, we know that 2 – 1 cannot have a value greater than 4.00 m. Therefore,
with a wavelength of 5.00 m, the only way constructive interference can occur is at a place where m = 0.
But this would mean that 2 = 1, and there is only one place where this is true, namely, at the midpoint
between the sources. But we know that there are two places. As a result, we conclude that
destructive interference is occurring .
b. For destructive interference, the possible values for the integer m are m = 0, 1, 2, 3, …, and we must
decide
which
to
use.
The
condition
for
destructive
interference
is
1
2 – 1 = (m + 2 ). Since  = 5.00 m, all values of m greater than or equal to one result in 2 – 1 being
greater than 4.00 m, which we already know is not possible. Therefore, we conclude that m = 0, and the
condition for destructive interference becomes 2 – 1 = 21 . We also know that 2 + 1 = 4.00 m, since the
separation between the sources is 4.00 m. In other words, we have
 2   1  21  
1
2
b5.00 mg  2.50 m
and
 2   1  4.00 m
Adding these two equations gives
22 = 6.50 m
or
2 = 3.25 m
Since 2 + 1 = 4.00m, it follows that
1 = 4.00 m – 2 = 4.00 m – 3.25 m = 0.75 m
These results indicate that the two places are 3.25 m and 0.75 m from one of the sources.
5.
REASONING According to Equation 27.2, the
Double
wavelength  of the light is related to the angle  between
slit
the central bright fringe and the seventh dark fringe
according to
d sin 

m  12
Seventh
dark fringe
y

Central
bright
fringe
L
where d is the separation between the slits, and m = 0, 1, 2, 3, … The first dark fringe occurs when m = 0,
so the seventh dark fringe occurs when m = 6. The distance d is given, and we can determine the angle  by
using the inverse tangent function,   tan 1  y / L  , since both y and L are known (see the drawing).
SOLUTION We will first compute the angle between the central bright fringe and the seventh dark fringe
using the geometry shown in the drawing:
 y
 0.025 m 
  tan 1    tan 1 
  1.3
L
 1.1 m 
The wavelength of the light is

9.
d sin  1.4 104 m  sin1.3

 4.9  107 m
m  12
6  12
REASONING The drawing shows a top view of the double slit and the screen, as well as the position of
the central bright fringe (m = 0) and the second-order bright fringe (m = 2). The vertical distance y in the
drawing
can
be
obtained
from
the
tangent
function
as
y = L tan . According to Equation 27.1, the angle  is related to the wavelength  of the light and the
separation d between the slits by sin   m / d , where m = 0, 1, 2, 3, … If the angle  is small, then
tan   sin  , so that
 m 
y  L tan   L sin   L 
 
 d 




Second-order bright
fringe (m = 2)
Double
slit
We will use this relation to find the value of y when  =
585 nm.
y

L
Central bright
fringe (m = 0)
SOLUTION When  = 425 nm, we know that y = 0.0180 m, so
 m  425 nm  
0.0180 m  L 
 
d

Dividing Equation (1) by Equation (2) and algebraically eliminating the common factors of L, m, and d, we
find that
 m 
L

y

d





0.0180 m
 m  425 nm   425 nm
L

d

When  = 585 nm, the distance to the second-order bright fringe is
 585 nm 
y   0.0180 m  
  0.0248 m 
 425 nm 