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‫كلية العلوم‬
‫نموذج إجابة مادة ضوء فيزيائى‬
)‫فيزياء (تخلفات‬-‫الفرقة الثانية رياضة‬
‫ صالح عيد حمزة‬/.‫د‬
2011/6/9 ‫تاريخ االمتحان‬
1. a) Young's double-slit experiment
Interference in light waves from two sources was first
demonstrated by Thomas Young in 1801. We can describe Young's
experiment with the help of Figure (4). The screen is located a distance L
from the double slit S1 and S2, which are separated by a distance d and the
source is monochromatic. under these conditions, the waves emerging
from S1 and S2 have the same frequency and amplitude and are in phase.
Note that, in order to reach P, a wave from S2 travels farther than a wave
from S1 by a distance d sin  . This distance is called the path difference,
 , where
  r2  r1  d sin  ,
(1)
This equation assumes that r1 and r2 are parallel, which is approximately
true because L is much greater than d. The value of this path difference
determines whether or not the two waves are in phase when they arrive at
P. If the path difference is either zero or some integral multiple of the
wavelength, the two waves are in phase at P and constructive interference
results. Therefore, the condition for bright fringes, or constructive
interference, at P is
  d sin   m
(m  0,  1,  2, ...).
(2)
The number m is called the order number. The central bright fringe at
  0 (m  0) is called the zero-order maximum. The first maximum on
1
either side, when m  1, is called the first- order maximum, and so
forth.
P
S1
y
'

d
S
O
L
S2

Po
a
Fig. (4): Geometric construction for describing
Young's double-slit experiment.
When the path difference is an odd multiple of

, the two waves
2
o
at P are 180 out of phase and will give rise to destructive interference.
Therefore, the condition for dark fringes, or destructive interference, at P
is
  d sin   (m  12 )
(m  0,  1,  2, ...).
(3)
It is useful to obtain expressions for the positions of the bright and dark
fringes measured vertically from O to P. Since the angle  is small, we
can use the approximation sin   tan  , or
sin   tan  
y
,
L
(4)
Using this result together with Eq. (2), we see that the position of the
bright fringes measured from O are given by
2
y bright  m
L
,
d
(5)
Similarly, using Eq. (3) and (4), we find that
y dark  (m  12 )
L
,
d
(6)
The spacing or distance between the centers of two adjacent bright (or
dark) fringes is given by
y m 1  y m 
L
,
d
(7)
1. (b) Huygens' principle
Huygens' principle is a construction for using knowledge of an
earlier wave front to determine the position of a new wave front at some
instant. In Huygens' construction, all points on a given wave front are
taken as point sources for the production of spherical secondary waves,
called wavelets, which propagate outward with speeds characteristic of
waves in that medium, see Fig. (1). After some time has elapsed, the new
position of the wave front is the surface tangent to the wavelets.
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A
C
S
D
B
E
Incident
wave
Fig. (1): A plane wave is incident on a slit. The slit
behave as a point source emitting spherical waves
≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈ ≈
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2. (a) The Michelson interferometer
The interferometer, invented by the American physicist A. A.
Michelson (1852-1931),splits a light beam into two parts and then
recombines them to form an interference pattern. The device can be used
to measure wavelengths or other lengths accurately.
M1
t
M2
S
G1
G2
T
Fig. (14): A diagram of the Michelson interferometer
A schematic diagram of the interferometer is shown in Fig. (14). A
beam light provided by a monochromatic source is split into two rays by
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a partially silvered mirror G1 inclined at 45o to the incident light beam.
One ray is reflected vertically upward toward mirror M1 , while the
second ray is transmitted horizontally through G1 toward mirror M 2 .
After reflecting from M1 and M 2 , the two rays recombine to produce an
interference pattern, which can be viewed through a telescope T. The
glass plate G 2 , equal in thickness to mirror G1 , is placed in the path of
the horizontal ray in order to ensure that the two rays travel the same
distance through glass.
The interference condition for the two rays is determined by the
difference in their optical path length. when the two rays are viewed as
shown, the image of M 2 is at M'2 parallel to M1 , see Fig. (15). Hence,
M'2 and M1 form the equivalent of an air film. The effective thickness
of the air film is varied by moving mirror M1 along the direction of the
light beam with a finely threaded screw. Under these conditions, the
interference pattern is a series of bright and dark circular rings. If the
mirror M1 and the image of M 2 ( M'2 ) is not parallel the gap between
M1 and M'2 is a wedge shaped film and the fringes is parallel lines as in
Fig. (16which known as localized fringes.
The condition for constructive interference is
2nt  (m  12 )
(m  0, 1, 2, ...) .
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(1)
and the condition for destructive interference is
2nt  m
(m  0, 1, 2, ...).
(2)
in the case of air film, n  1.
t
T
M 2
M1
Fig. (15a): Circular rings in Michelson interferometer
t1
t2
T
t3
M1
M 2
Fig. (15b): Localized rings in Michelson interferometer
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3. a) The diffraction grating
The diffraction grating consists of a large number of equally
spaced parallel slits. a grating can be made by cutting parallel lines on a
glass plate with a precision ruling machine. In a transmission grating, the
space between any two lines is transparent to the light and hence acts as a
separate slit. Gratings with many lines very close to each other can have
very small slit spacings. For example, a grating ruled with
5000 lines / cm has a slit spacing
d
1
cm  2  10 4 cm ,
5000
(1)
A section of a diffraction grating is illustrated in Fig. (11). A plane
wave is incident from the left, normal to the plane of grating. The pattern
observed on the screen is the result of the combined effects of
interference and diffraction. Each slit produce diffraction, and the
diffracted beams interfere with each other to produce the final pattern.
For some arbitrary direction  the waves must travel different path
lengths before reaching the point P. From Fig. (11), note that the path
difference between waves from any two adjacent slits is equal to d sin  .
Therefore, the constructive interference at angle  is
d sin   m,
(m  0, 1, 2, ...) .
(2)
This expression can be used to calculate the wavelength from a
knowledge of the grating spacing and the angle of deviation  .
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P
d
)a(
2
1 m0
1
2
2
1
m0
1
2
I
)b(

)c(
Fig. (11): (a)Side view of a diffraction grating.
(b) Intensity distribution
(c) The corresponding photography
The intensity distribution for diffraction grating used a
monochromatic source is shown in Fig. (11b).
3. b) Monochromatic light from a helium neon laser (   632.8 nm ) is
incident normally on a diffraction grating containing 6000 lines / cm .
Find the angles at which the first order, second order, and third order
maximum can be observed.
-------------------------------- Solution ---------------------------------
  632.8 109 m ,
d
d sin   m
1
cm  1.667  10 4 cm  1.667  10 6 m ,
6000
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 For the first order m  1, we get
 632.8  109
sin 1  
 0.3797
d 1667  10 6
1  22.31o
 For the second order m  2 , we get
 2  632.8  109
sin 2  
 0.7592
d
1667  10 6
2  49.41o
 For the third order m  3 , we get
 3  632.8  109
sin 3  
 1.139
d
1667  10 6
Since sin  cannot exceed unity, this does not represent a realistic
solution. Hence, only zeroth, first, and second order maxima are observed
for this situation.
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