Solutions Manual Introduction to Optics 3rd Edition By Frank L Pedrotti, Leno M Pedrotti

Solutions Manual
Introduction to Optics 3rd Edition
By
Frank L Pedrotti,
Leno M Pedrotti,
Leno S Pedrotti

( All Chapters Included - 100% Verified Solutions )




1

, C hapter 1 Nature of Light



h h 6. 63 × 1 0 − 3 4 J · s
1 -1 . a) λ = = = = 6. 63 × 1 0 − 3 4 m
p m v ( 0. 05 kg) ( 20 m/ s)
h h 6. 63 × 1 0 − 3 4 J · s
b) λ = =√ = = 3. 88 × 1 0 − 1 0 m
p 2 m E [ ( 2 · 9. 1 1 × 1 0 − 3 1 kg) ( 1 0 · 1 . 602 × 1 0 − 1 9 J) ]






Energy n h ν n h c 1 00 6. 63 × 1 0 − 3 4 J · s 3 × 1 0 8 m/s
1 -2 . P = = = = = 3. 62 × 1 0 − 1 7 W
time t tλ ( 1 s) ( 550 × 1 0 − 9 m)



1 -3. The energy of a photon is given by E = h ν = h c/ λ



6. 63 × 1 0 − 3 4 J · s 3 × 1 0 8 m/s
1 eV
At λ = 380 nm: E = = 5. 23 × 1 0 − 1 9 J = 3. 27 eV
380 × 1 0
− 9 m
1 . 60 × 1 0− 1 9J
− 34
6. 63 × 1 0 J · s 3 × 1 0 m/s
8
1 eV
At λ = 770 nm: E = = 2. 58 × 1 0 − 1 9 J = 1 . 61 eV
770 × 1 0 m − 9 1 . 60 × 1 0 − 1 9 J



h hc hc h
1 -4. p = E/ c = m c 2 / c = m c = 2. 73 × 1 0 − 2 2 kg · m/s, λ= = = = = 2. 43 × 1 0 − 1 2 m
p E m c2 m c




2
1 MeV
1 -5. Ev = 0 = m c 2 = 9. 1 09 × 1 0 − 3 1 kg 2. 998 × 1 0 8 m/ s = 8. 1 87 × 1 0 − 1 4 J = . 51 1 MeV
1 . 602 × 1 0 − 1 9 J


√ √
1 -6. c p = E 2 − m 2 c 4 , where E = EK + m c 2 = ( 1 + 0. 51 1 ) ) MeV. So c p = 1 . 51 1 2 − 0. 51 1 2 MeV
That is, c p = 1 . 422 MeV and p = 1 . 422 MeV/ c.




  
hc 6. 626 × 1 0 − 3 4 J · s 2. 998 × 1 0 8 m/ s 1 eV 1 Å 1 2, 400

1 -7. λ = = = Å · eV
E E 1 . 602 × 1 0 − 1 9 J 1 0− 1 0 m E


!
1 h
− 1 /2 i 
 1
1 -8. EK = m c 2 p −1 = m c2 1 − v 2 / c2 − 1 ' m c 2 1 − ( − ) v 2 / c 2 − 1 = m v 2
1 − v 2 / c2 2




1
2

, 1 -9. The total energy of the proton is,
 −19

9 1 . 60 × 1 0 J

2
E = EK + m p c = 2 × 1 0
2
+ 1 . 67 × 1 0 − 2 7 kg 3. 00 × 1 0 8 m/s = 4. 71 × 1 0 − 1 0 J
1 eV
q h
2
2
2 i − 1 /2
E 2 − m 2p c 4 4. 71 × 1 0 − 1 0 J − 1 . 67 × 1 0 − 2 7 kg 3. 00 × 1 0 8 m
a) p = =
c 3. 00 × 1 0 8 m/ s
p = 1 . 49 × 1 0 − 1 8 kg · m/ s



b) λ = h/ p = 6. 63 × 1 0 − 3 4 J · s / 1 . 49 × 1 0 − 1 8 kg · m/ s = 4. 45 × 1 0 − 1 6 m




c) λ p h ot on = h c/ E = 6. 63 × 1 0 − 3 4 J · s 3. 00 × 1 0 8 m/s / 4. 71 × 1 0 − 1 0 = 4. 22 × 1 0 − 1 6 m





Energy Energy 1 000 W/m 2 1 0 − 4 m 2
1 -1 0. n ph ot on s = = = = 2. 77 × 1 0 1 7
hν h c/ λ ( 6. 63 × 1 0 − 3 4 J) ( 3. 00 × 1 0 8 m/ s) / ( 550 × 1 0 − 9 m)


n 1 E e / h ν 1 Ee λ 1 / h c λ 1
1 -1 1 . = = =
n 2 E e / h ν 2 Ee λ 2 / h c λ 2


1 -1 2 . The wavelength range is 380 nm to 770 nm. The corresponding frequencies are

c 3. 00 × 1 0 8 m/ s c 3. 00 × 1 0 8 m/s
ν7 7 0 = = = 3. 89 × 1 0 1 4 Hz ν3 8 0 = = = 7. 89 × 1 0 1 4 Hz
λ 770 × 1 0 m
− 9 λ 380 × 1 0 − 9 m




1 -1 3. The wavelength of the radio waves is λ = c/ ν = 3. 00 × 1 0 8 m/s / 1 00 × 1 0 6 Hz = 3 m. The length of the
half wave antenna is then λ/ 2 = 1 . 5 m.



1 -1 4. The wavelength is λ = c/ ν = 3. 0 × 1 0 8 m/ s / 90 × 1 0 6 Hz = 3. 33 m. The length of each of the rods is then
λ/ 4 = 0. 83 m.



1 -1 5. a) t = D l / c = ( 90 × 1 0 . 0 × 1 0 8 s = 3. 0 × 1 0 − 4 s. b) D s = v s t = ( 340 ) 3. 0 × 1 0 − 4 m = 0. 1 0 m

Φe 500 W Φ 500 W
1 -1 6. a) Ie = = = 39. 8 W/ sr b) Me = e = = 1 0 6 W/ m 2
∆ω 4 π sr A 5 × 1 0 − 4 m2
Φ Φe 500 W

c) Ee = e = 2
= 2 = 9. 95 W/m 2 e) Φ e = Ee A = 9. 95 W/ m 2 π ( 0. 025 m) 2 = . 01 95 W
A 4πr 4 π( 2 m)

1 -1 7. a) The half angle divergence θ can be found from the relation

r sp ot 0. 0025 m
tan( θ ) ≈ θ = = = 1 . 67 × 1 0 − 4 rad = . 0096 ◦
L ro om 15 m

2
A sp ot π r2 π ( 0. 0025 m)
b) The solid angle is ∆ ω = 2 = 2 s p ot = 2 = 8. 73 × 1 0 − 8 sr.
L ro om L ro om ( 1 5 m)
Φe Φe 0. 001 5 W
c) The irradiance on the wall is Ee = = 2 = = 76. 4 W/m 2 .
A s p ot π r sp ot π ( 0. 0025 m) 2
d) The radiance is ( approximating differentials as increments)

Φe 0. 001 5 W W
Le ≈ =
= 8. 75 × 1 0 1 0 2
∆ω ∆A laser cosθ ( 8. 73 × 1 0 sr) π ( 0. 00025 m) cos( 0)
− 8 2 m · sr



2
3

, C hapter 2 G eometrical O ptics

P P
d op ni xi
2 -1 . t = = i
c c
2 -2 . Referring to Figure 2 1 2 and with lengths in cm,

1 /2 
1/2
n0 x2 + y2 + n i y 2 + s o + s i − x) 2 = no s o + ni s i

1 /2  
2 1 /2
( 1 ) x2 + y2 + 1 . 5 y 2 + ( 30 − x) = 20 + 1 . 5 ( 1 0) = 35
  
1 /2  2
2. 25 y 2 + ( 30 − x) 2 = 35 − x 2 + y 2


1 /2
1 . 25 x 2 + y 2 + 70 x 2 + y 2 − 1 35 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 2 0. 4 20. 8 21 . 6 22. 4 2 3. 2 2 4. 0 24. 8 2 5. 6 2 6. 4 2 7. 2
y ( cm) 0 ± 1 . 0 ± 1 . 40 ± 1 . 96 ± 2. 69 ± 3. 2 0 ± 3. 58 ± 3. 85 ± 4. 04 ± 4. 1 4 ± 4. 1 8 ± 4. 1 3


2 -3. Refer to the figure for the relevant parameters.

√
d = d 0 = 30 2 + 2. 5 2 = 30. 1 04 cm d
t
d0
Fermat: d + d 0 = s + s 0 − t + m t
d + d0 = s + s 0 + t ( m − 1 )
2 ( 30. 1 0399) = 60 = t ( 1 . 52 − 1 )
t = 4 mm
s 0 = 3 0 cm s 0 = 30 cm
n = 1 . 52


2 -4. See the figure below. Let the height of the person be h = h 1 + h 2 .

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

mirror h m irror = h − h − h = h/ 2
h2

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 ◦
Top
√
At Top: ( 1 ) sin 45 = 2 sin θ 0 ⇒ θ 0 = 30
30 ◦ √ √
At Side: 2 sin 60 ◦ = ( 1 ) sin θ 0 , sin θ 0 = 1 . 5 > 1
Side Thus total internal reflection occurs.
60◦
At Bottom: reverse of Top: θ 0 = 45 ◦

45◦
Bottom




3
4