Diffraction
1.
A laser beam (λ = 632.8 nm) is incident on two slits 0.200 mm apart. How
far apart are the bright interference fringes on a screen 5.00 m away from the
double slits?
2.
In a Young’s double-slit experiment, a set of parallel slits with a separation
of 0.100 mm is illuminated by light having a wavelength of 589 nm, and the
interference pattern is observed on a screen 4.00 m from the slits. (a) What is the
difference in path lengths from each of the slits to the location of a third-order
bright fringe on the screen? (b) What is the difference in path lengths from the
two slits to the location of the third dark fringe on the screen, away from the
center of the pattern?
4.
Light of wavelength 460 nm falls on two slits spaced 0.300 mm apart.
What is the required distance from the slit to a screen if the spacing between the
first and second dark fringes is to be 4.00 mm?
29.
Helium–neon laser light (λ = 632.8 nm) is sent through a 0.300-mm-wide
single slit. What is the width of the central maximum on a screen 1.00 m from the
slit?
30.
Light of wavelength 600 nm falls on a 0.40-mm-wide slit and forms a
diffraction pattern on a screen 1.5 m away. (a) Find the position of the first dark
band on each side of the central maximum. (b) Find the width of the central
maximum.
31.
Light of wavelength 587.5 nm illuminates a slit of width 0.75 mm. (a) At
what distance from the slit should a screen be placed if the first minimum in the
diffraction pattern is to be 0.85 mm from the central maximum? (b) Calculate the
width of the central maximum.
35.
Three discrete spectral lines occur at angles of 10.1°, 13.7°, and 14.8°,
respectively, in the first-order spectrum of a diffraction-grating spectrometer. (a)
If the grating has 3 660 slits/cm, what are the wavelengths of the light? (b) At
what angles are these lines found in the second-order spectra?
Solutions:
24.1
ybright  ym 1  ym 
 632.8 10

L
d
9
m
 m  1 
L
d
m
L
d
  5.00 m   1.58 10
2
3
0.200 10 m
24.2
m  1.58 cm
(a) For a bright fringe of order m, the path difference is   m  , where
m  0,1,2, At the location of the third order bright fringe, m  3 and
  3  3 589 nm   1.77  103 nm  1.77  m
1

(b) For a dark fringe, the path difference is    m    , where m  0,1,2,
2

At the third dark fringe, m  2 and
1
5
   2      589 nm   1.47  103 nm  1.47  m

24.4
2
2
 L
1
 m   , the spacing between the first and second dark fringes is
d 
2
 L  3 1  L
y 
. Thus, the required distance to the screen is
  
d  2 2 d
From ydark 
3
-3
y d  4.00 10 m  0.300 10 m 

L



24.29
460 109 m
2.61 m
The distance on the screen from the center to either edge of the central maximum
is

y  L tan  L sin  L  
 a
 632.8 109 m 
3
  1.00 m  
  2.11 10 m =2.11 m m
3
 0.300 10 m 
The full width of the central maximum on the screen is then
2y  4.22 m m
24.30
(a) Dark bands occur where sin  m   a . At the first dark band, m  1 , and
the distance from the center of the central maximum is

y1  L tan  L sin  L  
 a
 600 109 m 
3
  1.5 m  
  2.25 10 m  2.3 m m
3
 0.40 10 m 
(b) The width of the central maximum is 2y1  2 2.25 m m   4.5 m m
24.31
(a) Dark bands (minima) occur where sin  m   a . For the first minimum,
m  1 and the distance from the center of the central maximum is
y1  L tan  L sin  L   a . Thus, the needed distance to the screen is
 0.75 103 m 
 a
L  y1     0.85 103 m  
  1.1 m
-9

 587.5 10 m 
(b) The width of the central maximum is 2y1  2 0.85 m m   1.7 m m
24.35
The grating spacing is d 
1
1
cm 
m and dsin  m 
3660
3.66  105
(a) The wavelength observed in the first-order spectrum is   dsin , or
9
1m
  10 nm
5 
 3.66 10   1 m
  
This yields:
at 10.1°,   479 nm ;
and
at 14.8°,   698 nm

 104 nm
si
n





 3.66

 sin

at 13.7°,   647 nm ;
(b) In the second order, m  2 . The second order images for the above
wavelengths will be found at angles  2  sin1  2 d  sin 1  2sin1 
This yields:
for   479 nm ,  2  20.5 ;
and
for   698 nm ,  2  30.7
for   647 nm ,  2  28.3 ;