General Physics II
Spring 2006
General Physics II
PHY 112 Test 2
Name
Dr. Satinder S. Sidhu
SOLUTIONS
Attempt all questions. Show essential work to claim full credit.
1. One end of a string with a linear mass density of 3.40 ! 10 "4 kg/m is connected to a vibrator oscillating at 120 Hz. The other end is passed over a pulley located 3.0 m from the vibrator, and connected to a hanger. The mass on the hanger is adjusted so that a three-loop standing wave forms on the string. a) What is the wavelength of these waves? The four nodes of this standing wave are identified as N1 , N 2 , etc. Distance between two consecutive nodes is 1 meter. ! = 1.0 m " ! = 2.0 m 2 2.0 m
3
b) What is the wave speed? 240 m/s
5
Wave speed, v = f ! = (120 Hz ) ( 2.0 m ) = 240 m/s .
c) What is the tension in the string? 19.6 N Solve the equation v =
(
10
T , giving the unknown tension as µ
)
T = µv 2 = 3.40 ! 10 "4 kg/m ( 240 m/s ) = 19.6 N 2
d) What is the mass of the hanging block? 2.0 kg The tension is provided by the weight of the hanging block and equals the weight. T = Mg ! M =
T 19.6 N = = 2.0 kg g 9.8 m/s 2
2
2. A light beam passes from air to water through a layer of a mystery fluid. The angles it makes with the normal lines, and the indices of refraction of air and water, are shown in the figure below. a) Calculate the index of refraction nM for the mystery fluid. nM sin 30° = nW sin 26.75°
nM =
nW sin 26.75° (1.333) ( 0.45 ) = = 1.20 sin 30° 0.5
1.20 b) What is the speed of light in this fluid, given that light travels at 3.0 ! 10 8 m/s in vacuum? 2.50 ! 10 8 m/s Speed of light in the mystery fluid, v =
8
2
c 3.0 ! 10 8 m/s = = 2.50 ! 10 8 m/s . n 1.20
c) Calculate the angle ! the beam makes with the normal in air. 36.9°
6
nA sin ! = nM sin 30° n sin 30° (1.20 ) ( 0.500 ) sin ! = M = = 0.600 " ! = sin #1 ( 0.600 ) = 36.9° nA 1.000
3. The figure below shows a diamond (index of refraction 2.419) immersed in a liquid. A beam of light is incident on the interface from inside diamond. It is observed that the beam can emerge into the liquid only if the angle ! is less than 37.5°. Which of the following liquids is the diamond kept in? Liquid
Water
Ethyl alcohol
Glycerin
Refractive Index
1.333
1.361
1.473
Glycerin!
sin ! c =
nsmaller nlarger
" nsmaller = nlarger sin ! c = ( 2.419 ) ( sin 37.5° ) = 1.473 .
16
4. The lens shown in the figure has a focal length of 62.5 cm in air. What is the index of refraction of the material of which it is made? For light incident from the left, centers of curvature of both surfaces lie on their R-sides. Therefore, both radii of curvature are positive.
60 cm C1
R1 = 60 cm and R2 = 100 cm .
We use these in the lens maker’s equation:
C2
100 cm
" 1 1 1% = ( n ! 1) $ ! ' f # R1 R2 &
3.4
16
1 1 % 3 % n !1 " 1 " 5 = ( n ! 1) $ ! = ( n ! 1) $ ! = ' # 60 100 & # 300 300 '& 150 62.5 n !1=
150 = 2.4 " n = 2.4 + 1 = 3.4 62.5
Note: This high a refractive index is not physically impossible, just unusually large. Perhaps some exotic substance needs to be synthesized. 5. A bare light bulb is situated at a distance of 100 cm from a wall. You are given a converging lens of 16-cm focal length. Where should you position the lens (between the bulb and the wall) to obtain a wellfocused image of the bulb on the wall? Note that there are two possible positions; describe the image formed for each of them, indicating the type (real or virtual) and magnification of the image. p
q
100 cm
See below As can be seen from the figure, p + q = 100 ! q = 100 " p . Substituting this into
16
1 1 1 + = p q f
1 1 1 which simplifies to the quadratic p 2 ! 100 p + 1600 = 0 . The two solutions + = p 100 ! p 16 are p = 20 cm and p = 80 cm . The image details are in the following table: Object distance p Image distance q Magnification M = ! q p Image type 20 cm 80 cm Real –4.0 (Inverted; enlarged) 80 cm 20 cm Real –0.25 (Inverted; reduced)
gives
6. A small candle is located at a 50-cm distance from the center of a shiny globe of 20-cm radius. 20 cm
a) Calculate the position of the image of the candle formed in the globe. Radius is negative because the center of curvature lies on the V-side of the mirror. R = !20 cm "
f =
50 cm 7.5 cm
R = !10 cm 2 9
–7.5 cm
The object-to-mirror distance p is the distance from the candle to the nearest point on the globe, giving p = 50 ! 20 = 30 cm . We insert these numbers into the imaging equation and solve for the mirror-to-image distance q . 1 1 1 1 1 4 = ! = ! =! q f p !10 30 30
" q = !7.5 cm
This virtual image appears on the V-side, 7.5 cm beneath the surface of the globe, as shown above. b) Is the image real or virtual? Justify your answer. 2
Virtual Virtual, as indicated by the negative sign of q . c) If the candle is 6 cm tall, how high is its image?
3
1.5 cm Magnification, M = !
q !7.5 =! = +0.25 . Image height is 0.25 times object height; p 30
h! = Mh = ( 0.25 ) ( 6 ) = 1.5 cm d) Is the image erect or inverted? Justify your answer. 2
Erect Erect, as indicated by the positive sign of M . fn =
n T 2L µ
v=
T µ
n=
c v
c = f!
f =
R 2
1 1 1 + = p q f
Node-node distance = ! 2
n1 sin !1 = n2 sin ! 2
M=
h! "q = h p
v = f!
sin ! c =
nsmaller nlarger
" 1 1 1% = ( n ! 1) $ ! ' f # R1 R2 &
General Physics II
PHY 112 Test 2
Name
Dr. Satinder S. Sidhu
SOLUTIONS
Attempt all questions. Show essential work to claim full credit.
1. One end of a string with a linear mass density of 3.40 ! 10 "4 kg/m is connected to a vibrator oscillating at 120 Hz. The other end is passed over a pulley located 3.0 m from the vibrator, and connected to a hanger. The mass on the hanger is adjusted so that a three-loop standing wave forms on the string. a) What is the wavelength of these waves? The four nodes of this standing wave are identified as N1 , N 2 , etc. Distance between two consecutive nodes is 1 meter. ! = 1.0 m " ! = 2.0 m 2 2.0 m
3
b) What is the wave speed? 240 m/s
5
Wave speed, v = f ! = (120 Hz ) ( 2.0 m ) = 240 m/s .
c) What is the tension in the string? 19.6 N Solve the equation v =
(
10
T , giving the unknown tension as µ
)
T = µv 2 = 3.40 ! 10 "4 kg/m ( 240 m/s ) = 19.6 N 2
d) What is the mass of the hanging block? 2.0 kg The tension is provided by the weight of the hanging block and equals the weight. T = Mg ! M =
T 19.6 N = = 2.0 kg g 9.8 m/s 2
2
2. A light beam passes from air to water through a layer of a mystery fluid. The angles it makes with the normal lines, and the indices of refraction of air and water, are shown in the figure below. a) Calculate the index of refraction nM for the mystery fluid. nM sin 30° = nW sin 26.75°
nM =
nW sin 26.75° (1.333) ( 0.45 ) = = 1.20 sin 30° 0.5
1.20 b) What is the speed of light in this fluid, given that light travels at 3.0 ! 10 8 m/s in vacuum? 2.50 ! 10 8 m/s Speed of light in the mystery fluid, v =
8
2
c 3.0 ! 10 8 m/s = = 2.50 ! 10 8 m/s . n 1.20
c) Calculate the angle ! the beam makes with the normal in air. 36.9°
6
nA sin ! = nM sin 30° n sin 30° (1.20 ) ( 0.500 ) sin ! = M = = 0.600 " ! = sin #1 ( 0.600 ) = 36.9° nA 1.000
3. The figure below shows a diamond (index of refraction 2.419) immersed in a liquid. A beam of light is incident on the interface from inside diamond. It is observed that the beam can emerge into the liquid only if the angle ! is less than 37.5°. Which of the following liquids is the diamond kept in? Liquid
Water
Ethyl alcohol
Glycerin
Refractive Index
1.333
1.361
1.473
Glycerin!
sin ! c =
nsmaller nlarger
" nsmaller = nlarger sin ! c = ( 2.419 ) ( sin 37.5° ) = 1.473 .
16
4. The lens shown in the figure has a focal length of 62.5 cm in air. What is the index of refraction of the material of which it is made? For light incident from the left, centers of curvature of both surfaces lie on their R-sides. Therefore, both radii of curvature are positive.
60 cm C1
R1 = 60 cm and R2 = 100 cm .
We use these in the lens maker’s equation:
C2
100 cm
" 1 1 1% = ( n ! 1) $ ! ' f # R1 R2 &
3.4
16
1 1 % 3 % n !1 " 1 " 5 = ( n ! 1) $ ! = ( n ! 1) $ ! = ' # 60 100 & # 300 300 '& 150 62.5 n !1=
150 = 2.4 " n = 2.4 + 1 = 3.4 62.5
Note: This high a refractive index is not physically impossible, just unusually large. Perhaps some exotic substance needs to be synthesized. 5. A bare light bulb is situated at a distance of 100 cm from a wall. You are given a converging lens of 16-cm focal length. Where should you position the lens (between the bulb and the wall) to obtain a wellfocused image of the bulb on the wall? Note that there are two possible positions; describe the image formed for each of them, indicating the type (real or virtual) and magnification of the image. p
q
100 cm
See below As can be seen from the figure, p + q = 100 ! q = 100 " p . Substituting this into
16
1 1 1 + = p q f
1 1 1 which simplifies to the quadratic p 2 ! 100 p + 1600 = 0 . The two solutions + = p 100 ! p 16 are p = 20 cm and p = 80 cm . The image details are in the following table: Object distance p Image distance q Magnification M = ! q p Image type 20 cm 80 cm Real –4.0 (Inverted; enlarged) 80 cm 20 cm Real –0.25 (Inverted; reduced)
gives
6. A small candle is located at a 50-cm distance from the center of a shiny globe of 20-cm radius. 20 cm
a) Calculate the position of the image of the candle formed in the globe. Radius is negative because the center of curvature lies on the V-side of the mirror. R = !20 cm "
f =
50 cm 7.5 cm
R = !10 cm 2 9
–7.5 cm
The object-to-mirror distance p is the distance from the candle to the nearest point on the globe, giving p = 50 ! 20 = 30 cm . We insert these numbers into the imaging equation and solve for the mirror-to-image distance q . 1 1 1 1 1 4 = ! = ! =! q f p !10 30 30
" q = !7.5 cm
This virtual image appears on the V-side, 7.5 cm beneath the surface of the globe, as shown above. b) Is the image real or virtual? Justify your answer. 2
Virtual Virtual, as indicated by the negative sign of q . c) If the candle is 6 cm tall, how high is its image?
3
1.5 cm Magnification, M = !
q !7.5 =! = +0.25 . Image height is 0.25 times object height; p 30
h! = Mh = ( 0.25 ) ( 6 ) = 1.5 cm d) Is the image erect or inverted? Justify your answer. 2
Erect Erect, as indicated by the positive sign of M . fn =
n T 2L µ
v=
T µ
n=
c v
c = f!
f =
R 2
1 1 1 + = p q f
Node-node distance = ! 2
n1 sin !1 = n2 sin ! 2
M=
h! "q = h p
v = f!
sin ! c =
nsmaller nlarger
" 1 1 1% = ( n ! 1) $ ! ' f # R1 R2 &
