Chapter 36, example problems:
(36.10)
Light wave with electric field
E y
( x
, t
) =
E max sin [(1.20 × 10 7 m
− 1 ) x
− ω t
]
passes through a slit. First dark band at ± 28.6
° from the center of the diffraction
(a)
pattern.
Frequency of light f
= ? We use 1.20
523.6 nm . Thus f
= 3 × 10 8
× 10 7 m
− 1 = k
= 2 π /
m/s / 523.6 nm = 5. 730 × 10 14
λ , and obtain
Hz .
λ =
(b) We use a
sin
1. 094 μ m .
θ = m
λ , and put in m
= ± 1, to get a
= 1 × 523.6 nm / sin 28.6
° =
(c) Other dark bands also obey a
sin
For m
= ± 2, we get θ = sin
θ = m
λ , but with other interger values for m
− 1 ( ± 2 × 523.6 nm / 1.094 μ m) = ± 73.2
° . These are
. the second-order dark bands.
For m
= ± 3, we find no solution, because the magnitude of the argument of arcsine cannot be bigger than unity. Thus third order-dark bands and beyond can not be observed in this diffraction pattern. Thus the total number of dark bands observable in this diffraction pattern is four (two on each side of the center).
(36.11)
Monochromatic radiation, wavelength λ . From distant source. Passes through a
(a) slit. Screen 2.50 m in front of slit. Width of center maximum = 6.00 mm.
λ = 500 nm. a
= m
λ
/ sin
= 4.17 × 10
θ ≅ m
λ
/
θ (in rad.) = 1
− 4 m or 417 μ m.
× 500 nm / (3.00 mm / 2.50 m)
(b) λ = 50.0 μ m. a
= m
λ
/ sin
= 4.17 × 10
θ ≅ m
λ
/
θ (in rad.) = 1
− 2 m or 4.17 cm.
× 50.0 μ m / (3.00 mm / 2.50 m)
(c) λ =
0.
500 nm. a
= m
λ
/ sin
= 4.17 × 10
θ ≅ m
λ
/
θ (in rad.) = 1
− 7 m or 417 nm.
× 0.500 nm / (3.00 mm / 2.50 m)
(36.18)
Single-slit diffraction. Amplitude at O in Fig. 36.5a is
E and determine amplitude graphically for:
0
. Draw phasor diagram
(a) sin θ = λ / 2 a
(or a
sin θ = λ / 2):
β = (2 π / λ ) a
sin θ = π .
That is, the last little vector is rotated by
π from the first vector. Hence the phasor
diagram:
The ratio of cord length to arc length is
2r / π r = 2 / π . Hence the amplitude for
this
E
0
θ is (2
= π r.)
/ π )
E
0
. (Note that in this case
Thus for this θ ,
I
= (2 / π ) 2
I
0
, because the intensity is proportional to the amplitude squared.

(b) sin θ = λ / a
(or a
sin θ = λ ) :
β = (2 π / λ ) a
sin θ = 2 π .
That is, the last little vector is rotated by
2 π from the first vector, which means in
the same direction. Since the cord length
is now zero, the amplitude for this
(Note that in this case E
0
= 2 π r.)
θ is 0.
Thus for this θ ,
I
= 0 . Note that a
sin θ = λ is precisely the condition for the first minimum.
(c) sin θ = 3 λ / 2 a
(or a
sin θ = 3 λ / 2):
β = (2 π / λ ) a
sin θ = 3 π .
That is, the last little vector is rotated by
3 π from the first little vector, which means
in the opposite direction to the first little
vector. Thus the cord is just like the case
(a), yet the arc has gone around a circle
one and a half times. Their length ratio is
therefore 2r / 3 π r = 2 / 3 π . Hence the amplitude
for this θ is (2 / 3 π ) E
0
. (Note that in this case E
0
= 3 π r.)
Thus for this θ ,
I
= (2 / 3 π ) 2
I
0
= 0.0450
I
0
.
One aspect of the above drawings is misleading. The total number of little arrows
should be the same, and the length of each little arrow should be the same, only
their turning angles are different in the three cases. Thus the three circles do not
have the same radius, and should become smaller as sin θ increases. It is like
wrapping a string of a fixed length (=
E
0
) around a cylinder of smaller and smaller
radius. Depending on the radius of the cylinder, the string can cover half of the
cylinder (the first case), the full cylinder (the second case), or one-and-a half
times the cylinder (the third case). The length of the string is always
E
0
. The
radius of the cylinder in each case is decided by the value of β , which decides
how many times the cylinder should be covered by the string.

(36.22)
Interference pattern due to eight parallel, equally spaced, narrow slits. Phase
difference φ between light from adjacent slits is π /4. ⇒ Interference minimum.
Phasor diagram given by Fig. 36.14(b):
From the phasor diagram, it is clear that
destructive interference occurs between
light from the first slit and the fifth slit,
between the second slit and the sixth slit,
between the third slit and the seventh slit,
7
6
5
4
3
8 2 and between the fourth slit and the eighth
slit. Thus for this phase difference φ between
light from adjacent slits, light from the eight
slits are pair-wise cancelled, giving
I
= 0 for this φ .
1
(36.32)
Visible range. 400 – 700 nm. White light falls on a diffraction grating at normal
incidence. 350 slits/mm. Find angular width of visible spectrum in
(a) The first order:
First order bright line decided by d sin
Then it is clear that d
Δ
denotes to the corresponding range of
visible spectrum. Thus Δ
θ = 1 λ , or, for very small θ , d
θ = λ .
θ = Δ λ , where Δ λ denoted the visible range of λ , and Δ θ
θ , which is just the angular width of the
θ = Δ λ / d
= 300 nm / (1mm / 350) = 0.105 rad.
(b) The third order. The equation is now d
θ = 3 λ ⇒ Δ θ = 3 × 0.105 rad = 0.315 rad.
Thus higher order does give a larger angular width of the visible spectrum. But if it is too large, it could overlap with the neighboring orders.
(36.36)
Hydrogen, 656.45 nm. Deuterium, 656.27 nm. Δ d
θ = 2 λ ⇒ isotopes is Δ
λ = 0.18 nm. Second order, the angular separation of the two second-oder lines from the two
θ = 2 Δ λ / d
= 2 × 0.18 nm / d
. On the other hand, the half-width of each bright line is the same as the half-width of the center bright line, which is given by the θ in
Nd
sin θ = 1 λ, or, for very small θ ,
Nd
θ = 1 λ. It gives a half- width of 656.31 nm /
Nd
.
(We have used the average λ since the tiny difference is not important here.) Thus to resolve the two lines, we need to require
:
656.31 nm / Nd < 2 × 0.18 nm / d . We see that the factor d is cancelled out from the two sides, and we get:
N
> 656.31 nm / 0.36 nm = 1823.08. (We have used the average wavelength to get this answer. If we used 656.27 nm, we would get N > 1822.97, and if we used 656.45 nm, we would get N > 1823.47.
The average of the two answers is
N
> 1823.22. Actually, the two bright lines are of slightly different widths, so using the average of the two widths makes the best sense. This is equivalent to using the average wavelength.
(36.46)
Photography. Telephoto lens. f
= 135 mm. Maximum aperture f
/4.00.
s
= 11.5 m. λ = 550 nm. Bear 11.5 m away. ( f
/4.00 means that f
/
D
= 4.00
,
where
D
is the diameter of the aperture. Hence
D
= f
/ 4.00
. )

(a) Width of the smallest resolved feature = 11.5 m × angular width
= 11.5 m × (1.22 λ /
D
) = 11.5 m [1.22 × 550 nm / (135 mm/4.00)]
= 0.000229 m = 0.229 mm.
(b) f-stop changed to f
/22.0. Width of the smallest resolved feature
= 11.5 m × [1.22 × 550 nm / (135 mm / 22.0)]
= 0.00126 m = 1.26 mm.
(36.50)
Searching for starspots. Hale telescope. Mirror diameter = 200 in (= 5.08 m
). Focuses visible light. Large sunspot is about 10,000 mi (= 1.609 × 10 7 m).
Most distant star to see this sunspot? Denote this distance d
. Then
1.609 × 10 7 m / d
= 1.22 × 550 nm / 5.08 m, or d
= 1.218 × 10 14 m =
0.0129 light-years. Any star this close? No. The closest star is red dwarf
Proxima Centauri, and is 4.22 light-years away.
(36.56)
Diffraction grating design. First-order visible spectrum dispersed to an angular
range of 15.0
° .
(a) Find number of slits per centimeter. We can not use d
Δ
(36.32) ], since the angle involved is not small. Instead, we must use d sin for
But
λ
θ
1
= 400 nm and
2
=
Hence (
[1 both sides by d
2 d
2
− ( λ
= [
1
λ
λ
θ
1
/ d )
2
−
1
2
λ
] sin
1 / 1.276 × 10
− 4
2
,
λ
2
= 700 nm, and get sin
+ 15.0
° , and sin (
15.0
° = [(
1
cos 15.0
° ] 2 / sin
d
= 1276 nm
=
1.276 × 10
λ
θ
2
1
+ 15.0
° ) = sin
/ d
) cos 15.0
° + √ [1 − ( λ
1
cm = 7836 slits/cm
/ d
)
2
/ d ) − ( λ
15.0
° +
1
θ = Δ λ here
/ d
and sin cos 15.0
° + cos
/ d ) cos 15.0
° ] and then solving for d
2
λ
2
θ
θ
1
1
= λ
1
] sin 15.0
° = λ
1
2
, we obtain
2
2
θ
θ
/ d
, or,
[See Prob.
λ
θ = λ
2
=
2
/ d
.
1 sin 15.0
° .
, or, after multiplying
= 1628391 nm 2 , implying
− 4 cm. Thus the number per centimeter is:
.
(b) beginning and ending angles of this range:
θ
1
= sin
− 1 ( λ
1
/ d
) = 18.27
You can also get θ
2
from
°
θ
, and
2
= sin
θ
2
− 1
= 33.27
° .
( λ
2
/ d
) as a double check.
(36.61)
Phasor diagram for eight equally-spaced, narrow (identical) slits. Consider
φ = 3 π /4, 5 π /4, 3 π /2, 7 π /4.
(a) If φ = 3 π /4 = π /2 + π /4 = 90 ° + 45 ° , then we should draw eight vectors of the same length, each rotated from the previous one by 90 ° + 45 ° (counterclockwise), and then connect them end to end. These eight vectors are:
1 2
3
5
6 7
4
8
⇒
1
It is clear that destructive interference occurs between the first and the fifth slits, between the second and the sixth slits, between the third and the seventh slits, and between the fourth and the eighth slits. Therefore the net amplitude is zero, and the net intensity is zero. That is, it does correspond to an intensity minimum.

If φ = 5 π /4 = π + π /4 = 180 ° + 45 ° , then we should draw eight vectors of the same length, each rotated from the previous one by 180 ° + 45 ° (counterclock- wise), and then connect them end to end. These eight vectors are:
1
2
6
3
7
4
⇒
1
5 8
It is clear that destructive interference occurs between the first and the fifth slits, between the second and the sixth slits, between the third and the seventh slits, and between the fourth and the eighth slits. Therefore the net amplitude is zero, and the net intensity is zero. That is, it does correspond to an intensity minimum.
If φ = 3 π /2 = π + π /2 = 180 ° + 90 ° , then we should draw eight vectors of the same length, each rotated from the previous one by 180 ° + 90 ° (counterclock- wise), and then connect them end to end. These eight vectors are:
1
2 3
4
⇒
1
6
7
8
5
It is clear that destructive interference occurs between the first and the third slits, between the second and the fourth slits, between the fifth and the seventh slits, and between the sixth and the eighth slits. Therefore the net amplitude is zero, and the net intensity is zero. That is, it does correspond to an intensity minimum.
If φ = 7 π /4 = π + π /2 + π /4 = 180 ° + 90 ° + 45 ° , then we should draw eight vectors of the same length, each rotated from the previous one by 180 ° + 90 ° +
45 ° (counterclockwise), and then connect them end to end. These eight vectors are:
5
6 7
4
8
⇒
1
It is clear that destructive interference occurs between the first and the fifth slits, between the second and the sixth slits, between the third and the seventh slits, and between the fourth and the eighth slits. Therefore the net amplitude is zero, and the net intensity is zero. That is, it does correspond to an intensity minimum.
Each case is different, so it is impossible for you to attempt to remember the answers to all cases. However, once you learned the general concepts, it should not be difficult for you to do any such case, for any number of slits.