57 through 68 GO 64, 65 SSM 59 Transmission through thin layers In Fig. 35-43, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray r 3 (the light does not reflect inside material 2) and r 4 (the light reflects twice inside material 2). The waves of r 3 and r 4 interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refraction, n 1, n 2, and n 3, the type of interference, the thin-layer thickness L in nanometers, and the wavelength λ in nanometers of the light as measured in air. Where λ is missing, give the wavelength that is in the visible range. Where L is missing, give the second least thickness or the third least thickness as indicated. Figure 35-43 Problem 57 through 68. Table 35-3 Problems 57 through 68: Transmission Through Thin Layers. See the setup for these problems. n 1 n 2 n 3 Type L λ 57 1.55 1.60 1.33 min 285 58 1.32 1.75 1.39 min 3rd 382 59 1.68 1.59 1.50 max 415 60 1.50 1.34 1.42 max 380 61 1.32 1.75 1.39 min 325 62 1.68 1.59 1.50 max 2nd 342 63 1.40 1.46 1.75 max 2nd 482 64 1.40 1.46 1.75 max 210 65 1.60 1.40 1.80 min 2nd 632 66 1.60 1.40 1.80 max 200 67 1.50 1.34 1.42 min 2nd 587 68 1.55 1.60 1.33 min 3rd 612
57 through 68 GO 64, 65 SSM 59 Transmission through thin layers In Fig. 35-43, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray r 3 (the light does not reflect inside material 2) and r 4 (the light reflects twice inside material 2). The waves of r 3 and r 4 interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refraction, n 1, n 2, and n 3, the type of interference, the thin-layer thickness L in nanometers, and the wavelength λ in nanometers of the light as measured in air. Where λ is missing, give the wavelength that is in the visible range. Where L is missing, give the second least thickness or the third least thickness as indicated. Figure 35-43 Problem 57 through 68. Table 35-3 Problems 57 through 68: Transmission Through Thin Layers. See the setup for these problems. n 1 n 2 n 3 Type L λ 57 1.55 1.60 1.33 min 285 58 1.32 1.75 1.39 min 3rd 382 59 1.68 1.59 1.50 max 415 60 1.50 1.34 1.42 max 380 61 1.32 1.75 1.39 min 325 62 1.68 1.59 1.50 max 2nd 342 63 1.40 1.46 1.75 max 2nd 482 64 1.40 1.46 1.75 max 210 65 1.60 1.40 1.80 min 2nd 632 66 1.60 1.40 1.80 max 200 67 1.50 1.34 1.42 min 2nd 587 68 1.55 1.60 1.33 min 3rd 612
57 through 68 GO 64, 65 SSM 59 Transmission through thin layers In Fig. 35-43, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray r3 (the light does not reflect inside material 2) and r4 (the light reflects twice inside material 2). The waves of r3 and r4 interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refraction, n1,n2, and n3, the type of interference, the thin-layer thickness L in nanometers, and the wavelength λ in nanometers of the light as measured in air. Where λ is missing, give the wavelength that is in the visible range. Where L is missing, give the second least thickness or the third least thickness as indicated.
Figure 35-43 Problem 57 through 68.
Table 35-3 Problems 57 through 68: Transmission Through Thin Layers. See the setup for these problems.
1. A charge of -25 μC is distributed uniformly throughout a spherical volume of radius 11.5 cm.
Determine the electric field due to this charge at a distance of (a) 2 cm, (b) 4.6 cm, and (c) 25 cm from
the center of the sphere.
(a) =
=
(b) E =
(c)Ẻ =
=
NC NC NC
1.
A long silver rod of radius 3.5 cm has a charge of -3.9
ис
on its surface. Here ŕ is a unit vector
ст
directed perpendicularly away from the axis of the rod as shown in the figure.
(a) Find the electric field at a point 5 cm from the center of the rod (an outside point).
E =
N
C
(b) Find the electric field at a point 1.8 cm from the center of the rod (an inside point)
E=0
Think & Prepare
N
C
1. Is there a symmetry in the charge distribution? What kind of symmetry?
2. The problem gives the charge per unit length 1. How do you figure out the surface charge density σ
from a?
1. Determine the electric flux through each surface whose cross-section is shown below.
55
S₂
-29
S5
SA
S3
+ 9
Enter your answer in terms of q and ε
Φ
(a) s₁
(b) s₂
=
-29
(C) Φ
զ
Ερ
(d) SA
=
(e) $5
(f) Sa
$6
=
II
✓
-29
S6
+39
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