Chapter 36, example problems:
(36.10) Light wave with electric field Ey (x, t) = Emax sin [(1.20 ×107 m−1) x − ω t]
passes through a slit. First dark band at ±28.6° from the center of the diffraction
pattern.
(a) Frequency of light f = ? We use 1.20 ×107 m−1 = k = 2 π / λ , and obtain λ =
523.6 nm . Thus f = 3 ×108 m/s / 523.6 nm = 5. 730 ×1014 Hz .
(b) We use a sin θ = m λ, and put in m = ±1, to get a = 1 × 523.6 nm / sin 28.6° =
1. 094 μm .
(c) Other dark bands also obey a sin θ = m λ, but with other interger values for m.
For m = ±2, we get θ = sin −1 (±2 × 523.6 nm / 1.094 μm) = ±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
slit. Screen 2.50 m in front of slit. Width of center maximum = 6.00 mm.
(a) λ = 500 nm.
a = m λ / sin θ ≅ m λ / θ (in rad.) = 1× 500 nm / (3.00 mm / 2.50 m)
= 4.17 ×10−4 m or 417 μm.
(b) λ = 50.0 μm.
a = m λ / sin θ ≅ m λ / θ (in rad.) = 1× 50.0 μm / (3.00 mm / 2.50 m)
= 4.17 ×10−2 m or 4.17 cm.
(c) λ = 0.500 nm.
a = m λ / sin θ ≅ m λ / θ (in rad.) = 1× 0.500 nm / (3.00 mm / 2.50 m)
= 4.17 ×10−7 m or 417 nm.
(36.18) Single-slit diffraction. Amplitude at O in Fig. 36.5a is E0. Draw phasor diagram
and determine amplitude graphically for:
(a) sin θ = λ / 2a (or a sin θ = λ / 2):
β = (2π / 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 / πr = 2 / π. Hence the amplitude for
this θ is (2 / π) E0 . (Note that in this case
E0 = πr.)
Thus for this θ, I = (2 / π)2 I0 , because the intensity is proportional to the
amplitude squared.
(b) sin θ = λ / a (or a sin θ = λ ) :
β = (2π / λ) a sin θ = 2π .
That is, the last little vector is rotated by
2π from the first vector, which means in
the same direction. Since the cord length
is now zero, the amplitude for this θ is 0.
(Note that in this case E0 = 2πr.)
Thus for this θ, I = 0 . Note that a sin θ = λ
is precisely the condition for the first minimum.
(c) sin θ = 3λ / 2a (or a sin θ = 3λ / 2):
β = (2π / λ) a sin θ = 3π .
That is, the last little vector is rotated by
3π 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 / 3πr = 2 / 3π. Hence the amplitude
for this θ is (2 / 3π) E0 . (Note that in this case E0 = 3πr.)
Thus for this θ, I = (2 / 3π)2 I0 = 0.0450 I0.
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 (=E0) 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 E0. 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,
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.
6
5
4
7
3
8
2
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 θ = 1λ, or, for very small θ, dθ = λ .
Then it is clear that d Δθ = Δλ, where Δλ denoted the visible range of λ, and Δθ
denotes to the corresponding range of θ, which is just the angular width of the
visible spectrum. Thus Δθ = Δλ / d = 300 nm / (1mm / 350) = 0.105 rad.
(b) The third order. The equation is now dθ = 3λ ⇒ Δθ = 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. Δλ = 0.18 nm. Second order,
dθ = 2λ ⇒ the angular separation of the two second-oder lines from the two
isotopes is Δθ = 2 Δλ / 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 = 1λ, or, for very small θ, Ndθ = 1λ. It gives a halfwidth 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 × 107 m).
Most distant star to see this sunspot? Denote this distance d. Then
1.609 × 107 m / d = 1.22 × 550 nm / 5.08 m, or d = 1.218 × 1014 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 Δθ = Δλ here [See Prob.
(36.32)], since the angle involved is not small. Instead, we must use d sinθ = λ
for λ1 = 400 nm and λ2 = 700 nm, and get sin θ1 = λ1 / d and sin θ2 = λ2 / d .
But θ2 = θ1 + 15.0°, and sin (θ1 + 15.0°) = sin θ1 cos 15.0° + cos θ1 sin 15.0° .
Hence (λ1 / d ) cos 15.0° + √[1−(λ1 / d ) 2 ] sin 15.0° = λ2 / d , or,
[1−(λ1 / d )2] sin2 15.0° = [(λ2 / d ) − (λ1 / d ) cos 15.0°]2 , or, after multiplying
both sides by d 2, and then solving for d 2 , we obtain
d 2 = [λ2 − λ1 cos 15.0°]2 / sin2 15.0° + λ12 = 1628391 nm2 , implying
d = 1276 nm = 1.276 × 10 −4 cm. Thus the number per centimeter is:
1 / 1.276 × 10 −4 cm = 7836 slits/cm .
(b) beginning and ending angles of this range:
θ1 = sin−1(λ1 / d ) = 18.27°, and θ2 = 33.27°.
You can also get θ2 from θ2 = sin−1(λ2 / d ) as a double check.
(36.61) Phasor diagram for eight equally-spaced, narrow (identical) slits. Consider
φ = 3π/4, 5π/4, 3π/2, 7π/4.
(a) If φ = 3π/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
5
2
6
3
4
⇒
7
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 = 5π/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° (counterclockwise), and then connect them end to end. These eight vectors are:
2
1
3
4
1
⇒
6
7
8
5
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 = 3π/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° (counterclockwise), and then connect them end to end. These eight vectors are:
1
3
2
4
1
⇒
6
8
7
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 = 7π/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:
1
5
2
6
3
1
4
⇒
7
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.
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.