Physics 2220 – Module 13 Homework
01.
The distance from your eyes to your toes is 165 cm. You're standing 200 cm in front of a tall mirror.
How far is it from your eyes to the image of your toes?
Use the properties of a flat mirror. The height of the person, the distance from the person's eyes to the
image's toes, and the distance from the real toes to the image toes form a right triangle.
2
2
2
(2d) = (165 cm) + (400 cm)
(2d)2 = √ (165 cm)2 + (400 cm)2 = 433 cm
02.
A 150-cm-tall diver is standing completely submerged on the bottom of a swimming pool full of water.
You are sitting on the end of the diving board, almost directly over her. How tall does the diver appear to
be?
Treat the diver's head and toes as point sources. The
distance between the images of the point sources will
give the apparent height.
n air
s
n water head
n
s ' toes = air stoes
n water
s ' head =
Take the difference between these two equations.
(s ' head − s ' toes ) =
(
nair
nair
shead −
p
n water
n water toes
) (
)
nair
(s − stoes )
n water head
1.00
(s ' head − s ' toes ) =
(150 cm) = 113 cm
1.33
( s ' head − s ' toes ) =
03.
A goldfish lives in a 50-cm-diameter spherical fish bowl. The fish sees a cat watching it. If the cat's face
is 20 cm from the edge of the bowl, how far from the edge does the fish see it as being? (You can ignore
the thin glass wall of the bowl.)
Use the equation for the spherical lens assuming paraxial rays, using the cat's face as the object
distance.
1
1
=
s'
n2
04.
(
n1
n
n − n1
+ 2 = 2
s
s'
R
n2 − n 1 n 1
1 1.33 − 1.0
1.0
−
=
−
R
s
1.33
25 cm
20 cm
s ' = −36 cm
)
(
)
The fish sees the cat 36 cm from the edge of the bowl.
An object is 12 cm in front of a concave mirror with a focal length of 20 cm. Use ray tracing to locate the
image. Is the image upright or inverted?
The ray tracing should look something like this:
1
1 1
+ =
s' q
f
1
1 1
1
1
= − =
−
s'
f
s 20 cm 12 cm
s ' = −30 cm
The image is virtual and upright
05.
An object is 6 cm in front of a converging lens with a focal length of 10 cm. Use ray tracing to determine
the location of the image. Is the image upright or inverted?
The ray tracing should look something like this:
The image is virtual and upright
1
1
1
+
=
s s'
f
1
1 1
1
1
= − =
−
s'
f
s 10 cm 6 cm
s ' = −15 cm
06.
A 2.0-cm-tall object is 20 cm to the left of a lens with a focal length of 10 cm. A second lens with a focal
length of 5.0 cm is 30 cm to the right of the first lens.
(a)
Use ray tracing to find the position and height of the image. Do this accurately with a ruler or
paper with a grid. Estimate the image distance and image height by making measurements on
you diagram
The ray tracing should look something like this:
(b)
Calculate the image position and height. Compare with your ray-tracing answers in part (a).
Through the first lens:
1
1
1
+
=
s1 s ' 1 f 1
1
1
1
1
1
= − =
−
s ' 1 f 1 s1 10 cm 20 cm
s ' 1 = 20 cm
(Real Image)
M1 = −
s '1
20 cm
=−
= −1
s1
20 cm
The image from the first lens is located 20 cm to the right of the first lens. The two lenses are 30
cm apart. Therefore the object distance for the second lens is 10 cm in front of the second lens.
1
1
1
+
=
s2
s '2 f 2
1
1
1
1
1
= − =
−
s ' 2 f 2 s2 5.0 cm 10 cm
s '2 = 10 cm
(Real Image)
s '2
10 cm
=−
= −1
s2
10 cm
M = M 1 M 2 = (−1) (−1) = 1
M2 =−
The final image is real, upright, and has no magnification.
07.
A 2.0-cm-tall object is 20 cm to the left of a lens with a focal length of 10 cm. A second lens with a focal
length of – 5 cm is 30 cm to the right of the first lens.
(a)
Use ray tracing to find the position and height of the image. Do this accurately with a ruler or
paper with a grid. Estimate the image distance and image height by making measurements on
you diagram
The ray tracing should look something like this:
(b)
Calculate the image position and height. Compare with your ray tracing answers in part (a).
Through the first lens:
1
1
1
+
=
s1 s ' 1 f 1
1
1
1
1
1
= − =
−
s ' 1 f 1 s1 10 cm 20 cm
s ' 1 = 20 cm
(Real Image)
M1 = −
s '1
20 cm
=−
= −1
s1
20 cm
The image from the first lens is located 20 cm to the right of the first lens. The two lenses are 30
cm apart. Therefore the object distance for the second lens is 10 cm in front of the second lens.
1
1
1
+
=
s2 s '2
f2
1
1
1
1
1
= − =
−
s ' 2 f 2 s2 −5.0 cm 10 cm
s ' 2 = −3.3 cm
(Real Image)
s '2
−3.33 cm
=−
= 0.33
s2
10 cm
M = M 1 M 2 = (−1) (0.33) = −0.33
M2 = −
The final image is virtual, inverted, and 33% of the size of the object.
08.
A 15-cm-focal-length converging lens is 20 cm to the right of a 7.0-cm-focal-length converging lens. A
1.0-cm-tall object is distance L to the left of the first lens.
(a)
For what value of L is the final image of this two-lens system halfway between the two lenses?
Halfway between the two lenses is 10 cm. So this puts the final image at -10 cm (virtual with
respect to lens 2). So from here, work our way backwards. First, solve for the initial position of
the object for the second lens.
1
1
1
+
=
s2 s ' 2 f 2
1
1
1
1
1
= − =
−
s ' 2 f 2 s2 15.0 cm −10 cm
s ' 2 = 6.0 cm
(Real Image)
The object for the second lens is 6.0 cm in front of the second lens. The object for the second
lens and the image for the first lens is the same. This puts the image distance from the first lens
at 14 cm. Now solve for L, the object distance of the first lens.
1
1
1
1
1
= −
=
−
L f 1 s '1 7.0 cm 14 cm
L = 14 cm
(b)
What are the height and orientation of the final image?
Solve for the final magnification:
s '1
s1
s '2
s '1 s '2
14 cm
=
=
s2
s1 s 2
14 cm
( )( )
M = M1 M2 = −
−
(
So the final height is -1.7 cm. The image is inverted.
cm
= −1.7
) ( −10
6.0 cm )