Home Work 17
13-1 Manufacturers of wire (and other objects of small dimension)
sometimes use a laser to continually monitor the thickness of the
product. The wire intercepts the laser beam, producing a diffraction
pattern like that of a single slit of the same width as the wire
diameter (Fig. 36-39). Suppose a helium–neon laser, of wavelength
632.8 nm, illuminates a wire, and the diffraction pattern appears on
a screen at distance L = 2.60 m. If the desired wire diameter is 1.37 mm, what is the observed
distance between the two tenth-order minima (one on each side of the central maximum)?
Sol:
From y = mL/a we get
(632.8nm)(2.60)
 mL  L
y   
m 
[10  (10)]  24.0 mm .

1.37 mm
 a  a
17-2 Floaters. The floaters you see when viewing
a bright, featureless background are diffraction
patterns of defects in the vitreous humor that fills
most of your eye. Sighting through a pinhole
sharpens the diffraction pattern. If you also view a
small circular dot, you can approximate the defect’s size. Assume that the defect diffracts light as
a circular aperture does. Adjust the dot’s distance L from your eye (or eye lens) until the dot and
the circle of the first minimum in the diffraction pattern appear to have the same size in your
view. That is, until they have the same diameter D’ on the retina at distance L’ = 2.0 cm from the
front of the eye, as suggested in Fig. 36-44a, where the angles on the two sides of the eye lens
are equal. Assume that the wavelength of visible light is λ= 550 nm. If the dot has diameter D
= 2.0 mm and is distance L = 45.0 cm from the eye and the defect is x = 6.0 mm in front of the
retina (Fig. 36-44b), what is the diameter of the defect?
Sol:
From Fig. 36-44(a), we find the diameter D on the retina to be
D  D
L
2.00 cm
 (2.00 mm)
 0.0889 mm .
L
45.0 cm
Next, using Fig. 36-44(b), the angle from the axis is
 D / 2 
1  0.0889 mm / 2 
  tan 1 
  tan 
  0.424 .
 x 
 6.00 mm 
Since the angle corresponds to the first minimum in the diffraction pattern, we have
sin   1.22 / d , where  is the wavelength and d is the diameter of the defect. With
  550 nm, we obtain
d
1.22 1.22(550 nm)

 9.06 105 m  91 m .
sin 
sin(0.424)
17-3 Figure 36-49 is a graph of intensity versus angular
position θ for the diffraction of an x-ray beam by a
crystal. The horizontal scale is set byθs = 2.00°. The
beam consists of two wavelengths, and the spacing
between the reflecting planes is 0. 94 nm. What are the
(a) shorter and (b) longer wavelengths in the beam?
Sol:
We use Eq. 36-34.
(a) From the peak on the left at angle 0.75° (estimated from Fig. 36-49), we have
b
gb g
 2d sin  1  2 0.94 nm sin 0.75  0.025 nm  25 pm.
This is the shorter wavelength of the beam. Notice that the estimation should be viewed
as reliable to within ±2 pm.
(b) We now consider the next peak:
b
g
  2d sin  2  2 0.94 nm sin 115
.   0.038 nm  38 pm.
This is the longer wavelength of the beam. One can check that the third peak from the
left is the second-order one for 1.
17-4 In Fig. 36-50, first-order reflection from the reflection planes shown
occurs when an x-ray beam of wavelength 0.260 nm makes an angleθ= 63.8°
with the top face of the crystal. What is the unit cell size a0?
Sol:
The angle of incidence on the reflection planes is  = 63.8° – 45.0° =
18.8°, and the plane-plane separation is d  a0
2 . Thus, using 2d sin  = ,
we get
a0  2d 
2
0.260 nm

 0.570 nm.
sin 
2 sin 18.8
17-5 A diffraction grating having 180 lines/mm is illuminated with a light signal containing only
two wavelengths, λ1 = 400 nm and λ2 = 500 nm. The signal is incident perpendicularly on the
grating. (a) What is the angular separation between the second-order maxima of these two
wavelengths? (b) What is the smallest angle at which two of the resulting maxima are
superimposed? (c) What is the highest order for which maxima for both wavelengths are present
in the diffraction pattern?
Sol
(a) Since d = (1.00 mm)/180 = 0.0056 mm, we write Eq. 36-25 as
 m 
1
  sin (180)(2)
 d 
  sin 1 
where   4  104 mm and    5  104 mm. Thus,    2   1  21
. .
(b) Use of Eq. 36-25 for each wavelength leads to the condition
m11  m2  2
for which the smallest possible choices are m1 = 5 and m2 = 4. Returning to Eq. 36-25, then, we
find
4
mm) 
 m11 
1  10
1

sin

  sin  0.36   21.

 d 
 0.0056 mm 
  sin 1 
(c) There are no refraction angles greater than 90°, so we can solve for “mmax” (realizing it might
not be an integer):
mmax 
d sin 90 d
0.0056 mm


 11
2
 2 104 mm
where we have rounded down. There are no values of m (for light of wavelength 2) greater than
m = 11.
17-6 The D line in the spectrum of sodium is a doublet with wavelengths 589.0 and 589.6 nm.
Calculate the minimum number of lines needed in a grating that will resolve this doublet in the
second-order spectrum.
Sol
Letting R = / = Nm, we solve for N:
N
 589.6 nm  589.0 nm  / 2  491.
m 2  589.6 nm  589.0 nm 


17-7 With a particular grating the sodium doublet (589.00 nm and 589.59 nm) is viewed in the
third order at 10° to the normal and is barely resolved. Find (a) the grating spacing and (b) the
total width of the rulings.
Sol
(a) From d sin  = m we find
d
m avg
sin 

b
g
3 589.3 nm
 10
.  104 nm = 10 m.
sin 10
(b) The total width of the ruling is
L  Nd 
 d
b589.3 nmgb10 mg  3.3  10 m = 3.3 mm.
RI
F
d

G
J
HmK m 3b589.59 nm  589.00 nmg
avg
3