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Solution Manual For 3 Pedrotti Introduction to Optics Ophthalmic Chapter 1-28

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Solution Manual For 3 Pedrotti Introduction to Optics Ophthalmic Chapter 1-28

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Solution Manual For
3 Pedrotti Introduction to Optics Ophthalmic
Chapter 1-28


ChaPTER 1 NaTuRE Of LIghT


34
1-1. a) h h 6.63 × 10− J · s 6.63 10−34 m
λ= = = = ×
p m v (0.05 kg) (20 m/s)
34
h h 6.63 × 10− J · s −10
b) λ = = √ = = 3.88 × 10 m
2 m E [(2 · 9.11 × 10−31 kg) (10 · 1.602 × 10−19 J)]
p 1/2




word problems. The problems may require students to apply formulas, solve equations, or prove mathematical theorems.Theoretical Questions: In higher-level mathematics exams, students may
be asked to demonstrate their understanding of theoretical concepts, such as the proof of a mathematical theorem or the explanation of a mathematical concept.Multiple Choice Questions (MCQs):
While less common, some mathematics exams use MCQs to test students' quick recall of formulas, definitions, or theorems.3.3. Skills Tested in Mathematics ExamsProblem-Solving Ability: The
core skill tested in mathematics exams is problem-solving. Students must approach complex problems systematically, using appropriate methods to reach the correct solution.Logical Reasoning:
Mathematics is rooted in logical structures, and students are expected to demonstrate their ability to apply logic to prove statements or solve problems. This is especially important in subjects like
proof-based mathematics.Calculation and Accuracy: Mathematical exams test a student’s ability to perform accurate calculations and apply mathematical principles in the correct sequence to reach
a solution.Understanding


Energy n h ν n h c 100 6.63 × 10−34 J · s 3 × 108m/s
1-2. P = = = = 17
= 3.62 × 10− W
time t tλ (1 s) (550 × 10−9m)



1-3. The energy of a photon is given by E = h ν = h c/λ
6.63 × 10−34 J · s 3 × 108m/s 1 eV
At 380 nm = 5.23 10 J 3.27 eV
−19
λ= : E= 380 × 10−9 m × 1.60 × 10−19J =
34 8
6.63 × 10− J · s 3 × 10 m/s 19 1 eV
770 10 2.58 × 10− J 1.60 10 −19



2 22 h hc hc h 12
1-4. p = E/c = m c /c = m c = 2.73 × 10 − kg · m/s, λ = = = = = 2.43 × 10 − m
p



2 31 8 2 14 1 MeV
1-5. Ey=0 = m c = 9.109 × 10− kg 2.998 × 10 m/s = 8.187 × 10− J 1.602 × 10−19 J = .511 MeV


√ 2 2 4 2
√ 2 2
1-6. c p = E − m c , where E = EK + m c = (1 + 0.511))MeV. So c p = 1.511 − 0.511 MeV
That is, c p = 1.422 MeV and p = 1.422 MeV/c.



34 8
hc 6.626 × 10− J · s 2.998 × 10 m/s 1 eV 1Å 12, 400
Å · eV
1.602 × 10−19 J 10−10 m


!
h i 1
1 −1/2 2
√ 1 − ( − 1/2) v /c − 1 = m v
2 2 2 2 2 2 2
1-8. EK = m c —1 =mc 1 − v /c −1 ' mc
1 − v /c
2 2 2
1

, 1-9. The total energy of the proton is,
19
2 9 1.60 × 10 − J 1.67 × 10 −27 kg 3.00 × 10 m/s
8 2
−10 J
1 eV
h
2 2 4 10 2 27 2 8 2 −1/2
p 4.71 × 10− J − 1.67 × 10− kg 3.00 × 10 m
a) p = =
c 3.00 × 108 m/s
18
p = 1.49 × 10− kg · m/s
34 18 16
b) λ = h/p = 6.63 × 10− J · s / 1.49 × 10− kg · m/s = 4.45 × 10− m
34 8 10 16
c) λphoton = h c/E = 6.63 × 10− J · s 3.00 × 10 m/s / 4.71 × 10− = 4.22 × 10 − m


word problems. The problems may require students to apply formulas, solve equations, or prove mathematical theorems.Theoretical Questions: In higher-level mathematics exams, students may
be asked to demonstrate their understanding of theoretical concepts, such as the proof of a mathematical theorem or the explanation of a mathematical concept.Multiple Choice Questions (MCQs):
While less common, some mathematics exams use MCQs to test students' quick recall of formulas, definitions, or theorems.3.3. Skills Tested in Mathematics ExamsProblem-Solving Ability: The
core skill tested in mathematics exams is problem-solving. Students must approach complex problems systematically, using appropriate methods to reach the correct solution.Logical Reasoning:
Mathematics is rooted in logical structures, and students are expected to demonstrate their ability to apply logic to prove statements or solve problems. This is especially important in subjects like
proof-based mathematics.Calculation and Accuracy: Mathematical exams test a student’s ability to perform accurate calculations and apply mathematical principles in the correct sequence to reach
a solution.Understanding
2 4 2
Energy Energy = 1000 W/m 10− m 17
1-10. nphotons = = = 2.77 × 10
hν h c/λ (6.63 × 10−34 J) (3.00 × 108 m/s)/(550 × 10−9 m)


n1 Ee/h ν1 Ee λ1/h c λ1
1-11. = = =
n2 Ee/h ν2 Ee λ2/h c λ2


1-12. The wavelength range is 380 nm to 770 nm. The corresponding frequencies are
8 8
3.00 × 10 m/s
c 14 c 3.00 × 10 m/s 14
ν770 = = −9 = 3.89 × 10 Hz ν380 = = −9 = 7.89 × 10 Hz
λ 770 × 10 m λ 380 × 10 m


8 6
1-13. The wavelength of the radio waves is λ = c/ν = 3.00 10 m/s / 100 10 Hz = 3 m. The length of the
half wave antenna is then λ/2 = 1.5 m.
8 6
1-14. The wavelength is λ = c/ν = 3.0 10 m/s / 90 10 Hz = 3.33 m. The length of each of the rods is then
λ/4 = 0.83 m.
3 8 4 4
1-15. a) t = Dl/c = (90 × 10 /3.0 × 10 s = 3.0 × 10− s. b) Ds = vs t = (340 ) 3.0 × 10− m = 0.10 m
Φe 500 W Φe 500 W 6 2
1-16. a) Ie = = = 39.8 W/sr b) Me = = = 10 W/m
∆ω 4 π sr A −4
5 × 10 m 2

Φe Φe 500 W 2 2 2
c) Ee = = = 2= 9.95 W/m e) Φe = Ee A = 9.95 W/m π (0.025 m) = .0195 W
4 π r2 4 π(2 m)

1-17. a) The half angle divergence θ1/2 can be found from the relation

tan(θ )≈θ = rspot = 0.0025 m = 1.67 × 10−4 rad = .0096◦
1/2 1/2
Lroom 15 m

2 2
Aspot πr π (0.0025 m)
b) The solid angle is ∆ω = 2 2 2
8
= 8.73 × 10− sr.
room room (15 m)
Φe Φe 0.0015 W 2
c) The irradiance on the wall is Ee = = = = 76.4 W/m .
Aspot 2 2
π spot π (0.0025 m)




2

, d) The radiance is (approximating differentials as increments)

Φe 0.0015 W W
Le ≈ = 8.75 × 10
10
(8.73 × 10−8 sr) π (0.00025 m) m2 · sr
2
laser




word problems. The problems may require students to apply formulas, solve equations, or prove mathematical theorems.Theoretical Questions: In higher-level
mathematics exams, students may be asked to demonstrate their understanding of theoretical concepts, such as the proof of a mathematical theorem or the explanation of a mathematical
concept.Multiple Choice Questions (MCQs): While less common, some mathematics exams use MCQs to test students' quick recall of formulas, definitions, or theorems.3.3. Skills Tested in
Mathematics ExamsProblem-Solving Ability: The core skill tested in mathematics exams is problem-solving. Students must approach complex problems systematically, using appropriate methods
to reach the correct solution.Logical Reasoning: Mathematics is rooted in logical structures, and students are expected to demonstrate their ability to apply logic to prove statements or solve
problems. This is especially important in subjects like proof-based mathematics.Calculation and Accuracy: Mathematical exams test a student’s ability to perform accurate calculations and apply
mathematical principles in the correct sequence to reach a solution.Understanding




ChaPTER 2 GeoMETRICAL OPTICS

Σ Σ
dop i ni xi
2-1. t = =
c c
2-2. Referring to Figure 2 12 and with lengths in cm,
2 2 1/2 2 2 1/2
n x +y +n y + so + s − x) = no so + ni si
2 2 1/2 2 2
1/2
(1) x + y + 1.5 y + (30 − x) = 20 + 1.5 (10) = 35
2
2 2 2 2 1/2
2.25 y + (30 − x) = 35 − x + y
2 2 2 2 1/2
1.25 x + y + 70 x + y − 135 x + 800 = 0

Using a calculator to guess and check or using a computer algebra system, (like the free program Maxima,
for example) one can numerically solve this equation for x for given y values. Doing so results in,
x (cm) 20 20.2 20.4 20.8 21.6 22.4 23.2 24.0 24.8 25.6 26.4 27.2
y (cm) 0 ± 1.0 ± 1.40 ± 1.96 ± 2.69 ± 3.20 ± 3.58 ± 3.85 ± 4.04 ± 4.14 ± 4.18 ± 4.13




3

, 2-4. See the figure below. Let the height of the person be h = h1 + h2.

h1 The person must be able to see the top of
2
his head and the bottom of his feet. From
h1
the figure it is evident that the mirror
height is:
mirror
h2 hmirror = h − h1/2 − h2/2 = h/2

h2 The mirror must be half the height of the
2 person. So for a person of height six ft
person, the mirror must be 3 ft high.


2-5. Refer to the figure.

45◦
p

At Top: (1) sin 45 = 2 sin θ′ θ′ = 30
30◦ √ √
At Side: 2 sin 60◦ = (1) sin θ′, sin θ′ = 1.5 > 1
60◦
Thus total internal reflection occurs.
At Bottom: reverse of Top: θ′ = 45◦

45◦




word problems. The problems may require students to apply formulas, solve equations, or prove mathematical theorems.Theoretical Questions: In higher-level mathematics
exams, students may be asked to demonstrate their understanding of theoretical concepts, such as the proof of a mathematical theorem or the explanation of a mathematical concept.Multiple
Choice Questions (MCQs): While less common, some mathematics exams use MCQs to test students' quick recall of formulas, definitions, or theorems.3.3. Skills Tested in Mathematics
ExamsProblem-Solving Ability: The core skill tested in mathematics exams is problem-solving. Students must approach complex problems systematically, using appropriate methods to reach the
correct solution.Logical Reasoning: Mathematics is rooted in logical structures, and students are expected to demonstrate their ability to apply logic to prove statements or solve problems. This is
especially important in subjects like proof-based mathematics.Calculation and Accuracy: Mathematical exams test a student’s ability to perform accurate calculations and apply mathematical
principles in the correct sequence to reach a solution.Understanding




2-6. The microscope first focuses on the scratch using direct rays. Then it focuses on the image I2 formed in a
two step process: (1) reflection from the bottom to produce an intermediate image I1 and (2) refraction
through the top surface to produce an image I2. Thus, I1 is at 2t from top surface, and I2 is at the
2t 2t 3
apparent depth for I , serving as the object: s′ = or n = = = 1.60
n s′ 1.87
7.60/4
2-7. Refer to Figure 2 33 in the text. By geometry, tan θc = so θc = 40.18◦
1 2.25
Snell’s law: n sin θc = (1) sin 90◦ n = = 1.55
sin 40.18◦

2-8. Referring to the figure one can see that,
t
s = AB sin (θ − θ ) and AB = . Therefore,
cos θ2
t sin (θ1 − θ2) . For t = 3 cm, n2 = 1.50, θ1 = 50◦,
s=
cos θ2 1 1
Snell’s law gives, sin θ2 = sin θ1 = sin 50◦.
n2 1.5
◦ − 30.71◦)
3 sin (50
Then, θ2 = 30.71◦ and s = = 1.153cm.
cos 30.71◦

1 1 1
2-9. Image of near end: s = 60cm, = = , s′ − 24cm
60 s′ − 40
4
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