General Physics II
General Physics II
PHY 112 Test 3
Name
Spring 2006 Dr. Satinder S. Sidhu
Solutions
Attempt all questions. Show essential work to claim full credit.
1. A double-slit system is used to measure the wavelength of light. The system has slit spacing d = 15 µm and slit-to-screen distance L = 2.2 m . If the m = 1 maximum in the interference pattern occurs 7.1 cm from screen center, what is the wavelength of light illuminating the slits?
Calculate the angular direction of the m = 1 maximum (first bright fringe from the center) from the given distances: 7.1 " 10 #2 m tan ! = = 0.0323 . 2.2 m
For an angle this small, sin ! ! tan ! = 0.0323 . Now use the known direction and slit spacing to calculate the wavelength.
(
)
15 # 10 $6 m ( 0.0323) d sin " != = = 4.84 # 10 $7 m = 484 nm m 1
484 nm
20
2. A 0.44-µm thick film is illuminated by white light. Its index of refraction is 1.5. What wavelength within the visible spectrum (400 – 700 nm) will be intensified in the reflected beam? If we assume that the medium surrounding the film has a lower index of refraction, then there will be a phase reversal at the surface that light strikes first, but not at the second. On the other hand, if the film is immersed in a medium with a higher index of refraction, then the phase reversal occurs at the second surface, but not at the first. There is exactly one phase reversal in either situation. Strong reflection occurs when the condition for constructive interference is satisfied. From that condition we have
(
)
2 (1.5 ) 0.44 " 10 #6 m 1.32 " 10 #6 m 1320 nm 2nt != = = = m + 12 m + 12 m + 12 m + 12
The table below shows the intensified wavelengths for different values of the integer m , with the one falling in the visible range highlighted and bold-faced. 0 2640
m ! (nm)
1 880
2 528
3 377
528 nm
10
3. We wish to coat a flat slab of glass ( n = 1.50 ) with a transparent material ( n = 1.25 ) so that normallyincident light of wavelength 600 nm is not reflected. What is the thinnest coating that will accomplish this end? In this case the materials are so layered that there is a phase-reversal at both surfaces—air-to-coating and also coating-to-glass. Overall reflection is suppressed when waves reflected from these surfaces interfere destructively. The required film thickness is calculated from the condition for destructive interference as t=
( m + 12 ) ! = 2n
( m + ) 2 (1.25 ) = ( m + )( 240 nm ) . 1 2
600 nm
1 2
Minimum film thickness corresponds to m = 0 and turns out to be 120 nm.
120 nm
10
4. Two small particles each have a mass of 5.00 g. One carries a 50 µC charge and the other a !20 µC charge. They are released from rest at a 0.50-m separation. What initial acceleration will each experience? These oppositely-charged particles will attract each other with a force whose magnitude is given by Coulomb’s Law as Fe = ke
(
)(
)
2 50 ! 10 )6 C 20 ! 10 )6 C q1 q2 # 9 N"m & = 9 ! 10 = 36 N . %$ r2 C2 (' ( 0.50 m )2
Since both particles have the same mass, this force will produce in each the same acceleration, of initial magnitude a=
F 36 N = = 7.2 ! 10 3 m/s 2 . m 5.00 ! 10 "3 kg
Each charge will be accelerated towards the other one. Positive
15
7.2 ! 10 3 m/s 2 Negative
5
7.2 ! 10 3 m/s 2 5. An electron moving parallel to the x-axis has an initial speed of 3.0 ! 10 6 m/s at the origin O. It reaches point P, 2 cm away from the origin, with its speed reduced to 1.0 ! 10 5 m/s . Calculate the potential difference between the point P and the origin O. Electron’s charge is !e = !1.6 " 10 !19 C and mass me = 9.11 ! 10 "31 kg .
We apply conservation of energy to this situation, equating the total energy at O to that at P as K O + UO = K P + U P
! K O + qVO = K P + qVP
! q (VP " VO ) = K O " K P .
(1)
Kinetic energy of the electron at each point is calculated from its speed and mass as
K O = 12 mvO2 =
1 2
( 9.11 ! 10
K P = 12 mvP2 =
1 2
( 9.11 ! 10
)(
)
)(
)
"31
kg 3.0 ! 10 6 m/s
"31
kg 1.0 ! 10 5 m/s
2
= 4.0995 ! 10 "18 J , and
2
= 4.555 ! 10 "21 J .
Now use these results in Equation (1) to get q (VP ! VO ) = K O ! K P = 4.0995 " 10 !18 J ! 4.6 " 10 !21 J = 4.094945 " 10 !18 J and
K O ! K P 4.094945 " 10 !18 J VP ! VO = = = !25.6 J/C = !25.6 V . q !1.6 " 10 !19 C The negative sign indicates that the potential at P is lower than the potential at O.
–25.6 V
20
6. Four point charges, each of magnitude 2.0 nC, are placed at the corners of a square 10 cm on a side. Three of the charges are positive and one is negative. Find the magnitude and direction of the electric field experienced by the negative charge. The electric fields produced by the three charges are shown on the figure, the subscript on each matching the label on the corresponding source charge. The lengths of the field vectors are ! in correct mutual proportions. The magnitudes of fields E A and ! EC are equal, being 2 # )9 q # C& 9 N " m & 2.0 ! 10 = 9.0 ! 10 = 1800 N/C . r 2 %$ C2 (' %$ ( 0.10 m )2 (' ! The magnitude of field EB is
EA = EC = ke
2 # q # 2.0 ! 10 )9 C & 9 N"m & EB = ke 2 = % 9.0 ! 10 = 900 N/C . r C2 (' %$ ( 0.1414 m )2 (' $ ! ! ! ! Vector E AC , the sum of fields E A and EC , is in the same direction as field EB . The magnitude of this vector is
! EAC = E AC = 1800 2 + 1800 2 = 1800 2 = 2545.6 N/C . ! ! Because vectors EB and E AC are in the same ! direction, their sum—labeled simply as E —is also in that same direction. The magnitude of this vector ! ! is the sum of the magnitudes of EB and E AC .
E = E = EAC + EB = 2545.5 + 900 = 3445.6 N/C This is the electric field experienced by the negative charge.
Magnitude
3445.6 N/C Direction
45° clockwise of up.
16 4
PHY 112 Test 3
Name
Spring 2006 Dr. Satinder S. Sidhu
Solutions
Attempt all questions. Show essential work to claim full credit.
1. A double-slit system is used to measure the wavelength of light. The system has slit spacing d = 15 µm and slit-to-screen distance L = 2.2 m . If the m = 1 maximum in the interference pattern occurs 7.1 cm from screen center, what is the wavelength of light illuminating the slits?
Calculate the angular direction of the m = 1 maximum (first bright fringe from the center) from the given distances: 7.1 " 10 #2 m tan ! = = 0.0323 . 2.2 m
For an angle this small, sin ! ! tan ! = 0.0323 . Now use the known direction and slit spacing to calculate the wavelength.
(
)
15 # 10 $6 m ( 0.0323) d sin " != = = 4.84 # 10 $7 m = 484 nm m 1
484 nm
20
2. A 0.44-µm thick film is illuminated by white light. Its index of refraction is 1.5. What wavelength within the visible spectrum (400 – 700 nm) will be intensified in the reflected beam? If we assume that the medium surrounding the film has a lower index of refraction, then there will be a phase reversal at the surface that light strikes first, but not at the second. On the other hand, if the film is immersed in a medium with a higher index of refraction, then the phase reversal occurs at the second surface, but not at the first. There is exactly one phase reversal in either situation. Strong reflection occurs when the condition for constructive interference is satisfied. From that condition we have
(
)
2 (1.5 ) 0.44 " 10 #6 m 1.32 " 10 #6 m 1320 nm 2nt != = = = m + 12 m + 12 m + 12 m + 12
The table below shows the intensified wavelengths for different values of the integer m , with the one falling in the visible range highlighted and bold-faced. 0 2640
m ! (nm)
1 880
2 528
3 377
528 nm
10
3. We wish to coat a flat slab of glass ( n = 1.50 ) with a transparent material ( n = 1.25 ) so that normallyincident light of wavelength 600 nm is not reflected. What is the thinnest coating that will accomplish this end? In this case the materials are so layered that there is a phase-reversal at both surfaces—air-to-coating and also coating-to-glass. Overall reflection is suppressed when waves reflected from these surfaces interfere destructively. The required film thickness is calculated from the condition for destructive interference as t=
( m + 12 ) ! = 2n
( m + ) 2 (1.25 ) = ( m + )( 240 nm ) . 1 2
600 nm
1 2
Minimum film thickness corresponds to m = 0 and turns out to be 120 nm.
120 nm
10
4. Two small particles each have a mass of 5.00 g. One carries a 50 µC charge and the other a !20 µC charge. They are released from rest at a 0.50-m separation. What initial acceleration will each experience? These oppositely-charged particles will attract each other with a force whose magnitude is given by Coulomb’s Law as Fe = ke
(
)(
)
2 50 ! 10 )6 C 20 ! 10 )6 C q1 q2 # 9 N"m & = 9 ! 10 = 36 N . %$ r2 C2 (' ( 0.50 m )2
Since both particles have the same mass, this force will produce in each the same acceleration, of initial magnitude a=
F 36 N = = 7.2 ! 10 3 m/s 2 . m 5.00 ! 10 "3 kg
Each charge will be accelerated towards the other one. Positive
15
7.2 ! 10 3 m/s 2 Negative
5
7.2 ! 10 3 m/s 2 5. An electron moving parallel to the x-axis has an initial speed of 3.0 ! 10 6 m/s at the origin O. It reaches point P, 2 cm away from the origin, with its speed reduced to 1.0 ! 10 5 m/s . Calculate the potential difference between the point P and the origin O. Electron’s charge is !e = !1.6 " 10 !19 C and mass me = 9.11 ! 10 "31 kg .
We apply conservation of energy to this situation, equating the total energy at O to that at P as K O + UO = K P + U P
! K O + qVO = K P + qVP
! q (VP " VO ) = K O " K P .
(1)
Kinetic energy of the electron at each point is calculated from its speed and mass as
K O = 12 mvO2 =
1 2
( 9.11 ! 10
K P = 12 mvP2 =
1 2
( 9.11 ! 10
)(
)
)(
)
"31
kg 3.0 ! 10 6 m/s
"31
kg 1.0 ! 10 5 m/s
2
= 4.0995 ! 10 "18 J , and
2
= 4.555 ! 10 "21 J .
Now use these results in Equation (1) to get q (VP ! VO ) = K O ! K P = 4.0995 " 10 !18 J ! 4.6 " 10 !21 J = 4.094945 " 10 !18 J and
K O ! K P 4.094945 " 10 !18 J VP ! VO = = = !25.6 J/C = !25.6 V . q !1.6 " 10 !19 C The negative sign indicates that the potential at P is lower than the potential at O.
–25.6 V
20
6. Four point charges, each of magnitude 2.0 nC, are placed at the corners of a square 10 cm on a side. Three of the charges are positive and one is negative. Find the magnitude and direction of the electric field experienced by the negative charge. The electric fields produced by the three charges are shown on the figure, the subscript on each matching the label on the corresponding source charge. The lengths of the field vectors are ! in correct mutual proportions. The magnitudes of fields E A and ! EC are equal, being 2 # )9 q # C& 9 N " m & 2.0 ! 10 = 9.0 ! 10 = 1800 N/C . r 2 %$ C2 (' %$ ( 0.10 m )2 (' ! The magnitude of field EB is
EA = EC = ke
2 # q # 2.0 ! 10 )9 C & 9 N"m & EB = ke 2 = % 9.0 ! 10 = 900 N/C . r C2 (' %$ ( 0.1414 m )2 (' $ ! ! ! ! Vector E AC , the sum of fields E A and EC , is in the same direction as field EB . The magnitude of this vector is
! EAC = E AC = 1800 2 + 1800 2 = 1800 2 = 2545.6 N/C . ! ! Because vectors EB and E AC are in the same ! direction, their sum—labeled simply as E —is also in that same direction. The magnitude of this vector ! ! is the sum of the magnitudes of EB and E AC .
E = E = EAC + EB = 2545.5 + 900 = 3445.6 N/C This is the electric field experienced by the negative charge.
Magnitude
3445.6 N/C Direction
45° clockwise of up.
16 4
