Tài liệu Physics exercises_solution: Chapter 34 pptx

34.1: If up is the
y

-direction and right is the
x

-direction, then the object is at
),,(atis),,(
00200
yxPyx 


and mirror 1 flips the
y
-values, so the image is at
),(
00
yx
which is
.
3
P

34.2:
Using similar triangles,
m.3.24
m0.350
m0.350m28.0
m040.0
mirror
tree

mirrortree
mirror
tree
mirror
tree



d
d
hh
d
d
h
h
34.3:
A plane mirror does not change the height of the object in the image, nor does the
distance from the mirror change. So, the image is
cm2.39
to the right of the mirror, and
its height is
cm.85.4
34.4: a)
cm.0.17
2
cm0.34
2

R
f

b) If the spherical mirror is immersed in water, its focal length is unchanged—it
just depends upon the physical geometry of the mirror.
34.5:
a)
b)
cm,0.33
cm5.16
1
cm0.22
21111







s
sfss
to the left of the
mirror.
cm,20.1
cm16.5
cm33.0
cm)600.0(




s

s
yy
and the image is inverted and real.
34.6:
a)
b)
cm,60.6
cm5.16
1
cm0.22
21111







s
sfss
to the right of the
mirror.
cm,240.0
cm16.5
cm).606(
cm)600.0(







s
s
yy
and the image is upright and
virtual.
34.7:
m.75.1
m1058.5
1
m75.1
11111
10









s
sfss
m.1013.2)m106794)(1014.3(
1014.3
1058.5
75.1
4311

11
10








myym
34.8:
cm40.1
cm0.21
1
cm00.3
21111
,cm00.3







s
sfss
R
(in the
ball). The magnification is

.0667.0
cm0.21
cm40.1





s
s
m
34.9: a)
.Also.
111111
sf
f
s
s
m
fs
sf
s
fs
fs
sfsfss















b) For
,0,0





sfsf
so the image is always on the outgoing side and is
real. The magnification is
,0


sf
f
m
since
.sf

c) For
,12 



f
f
mfs
which means the image is always smaller and
inverted since the magnification is negative.
For
.102 
f
f
mffsfsf
d) Concave mirror:
,00





sfs
and we have a virtual image to the right
of the mirror.
,1
f
f
m
so the image is upright and larger than the object.
34.10: For a convex mirror,
.00 







fs
fs
fs
sf
sf
Therefore the image is
always virtual. Also
,0







sf
f
sf
f
sf
f
m
so the image is erect, and
,since1 fsfm 
so the image is smaller.

34.11:
a)
b)
.0,for0




sfss
c)
.0for0 fss




d) If the object is just outside the focal point, then the image position approaches
positive infinity.
e) If the object is just inside the focal point, the image is at negative infinity, “behind”
the mirror.
f) If the object is at infinity, then the image is at the focal point.
g) If the object is next to the mirror, then the image is also at the mirror.
h)
i) The image is erect if
.fs

j) The image is inverted if
.fs

k) The image is larger if
.20 fs



l) The image is smaller if
.0or2


sfs
m) As the object is moved closer and closer to the focal point, the magnification
INCREASES to infinite values.
34.12:
a)
a)
0.for0 

sfs
b)
.0andsfor0 

sfs
c) If the object is at infinity, the image is at the outward going focal point.
d) If the object is next to the mirror, then the image is also at the mirror. For the
answers to (e), (f), (g), and (h), refer to the graph on the next page.
e) The image is erect (magnification greater than zero) for
.fs 
f) The image is inverted (magnification less than zero) for
.fs 
g) The image is larger than the object (magnification greater than one) for
.02  sf
h) The image is smaller than the object (magnification less than one) for
.2and0 fss 

34.13:
a)
b)
cm,45.5
cm0.12
1
cm0.20
21111







s
sfss
to the right of the
mirror.
cm,409.0
cm12.0
cm5.45
cm)9.0(






s

s
yy
and the image is upright and virtual.
34.14: a)
,00.4
0
.
12
0.48





s
s
m
where
s

comes from part (b).
b)
cm.0.48
cm0.12
1
cm0.32
21111








s
sfss
Since
s

is negative,
the image is virtual.
c)
34.15:
cm.67.20
00.1
cm50.3
309.1
0







s
ss
n
s
n

ba
34.16: a)
cm,26.50
00.1
cm00.7
33.1
0







s
ss
n
s
n
ba
so the fish appears
cm26.5
below the surface.
b)
cm,8.240
00.1
cm0.33
33.1
0








s
ss
n
s
n
ba
so the image of the fish
appears
cm8.24
below the surface.
34.17: a) For
,ndwith,and0













baba
annR
we have:
.)()(

abba
b
a
bbabb
nnnn
n
n
nn

But
,and,,
R
h
s
h
s
h





so subbing them in one finds:
.
)(

R
nn
s
n
s
n
abba




Also, the magnification calculation yields:
.tanandtan
sn
sn
y
y
m
s
yn
s
yn
s
y
s
y
b
aba
ba















b) For

















abbaba
nnnnR :havewe,andwith,and0
,
)(
and,,But.)(
R
nn
s
n
s
n
R
h
s
h
s
h
nn
abba
ab









so subbing them

in one finds:
.
)(
R
nn
s
n
s
n
abba




Also, the magnification calculation yields:
.tantan
sn
sn
y
y
m
s
yn
s
yn
nn
a
aba
bbaa









34.18: a)
cm.00.8
cm00.3
60.060.11











s
sR
nn
s
n
s
n
abba

b)
cm.7.13
cm00.3
60.060.1
cm0.12
1









s
sR
nn
s
n
s
n
abba
c)
cm.33.5
cm00.3
60.060.1
cm00.2
1










s
sR
nn
s
n
s
n
abba
34.19:
.
)(
)()(
)(
b
n
Rs
Rs
s
s
n
Rs
Rs

Rs
sR
n
R
nn
s
n
s
n
ba
abba














.52.1)60.1(
cm)3.00cm0.90(
cm)3.00cm601(
cm160
cm.090





a
n
34.20:
cm.8.14
cm00.4
60.060.1
cm0.24
1









s
sR
nn
s
n
s
n
abba
mm,0.578mm50.1

)cm0.24)(60.1(
cm8.14























y
sn
sn
y
b

a
so the image height
is
mm,578.0
and is inverted.
34.21:
cm.35.8
cm00.4
60.060.1
cm0.24
1











s
sR
nn
s
n
s
n
abba

mm,0.326mm50.1
)cm0.24)(60.1(
cm)35.8(


















 



y
sn
sn
y
b

a
so the image height is
mm,326.0
and is erect.
34.22: a)
cm,0.14
cm0.14
33.000.1
cm0.14
33.1












s
sR
nn
s
n
s
n
abba

so the fish
appears to be at the center of the bowl.
.33.1
cm)0.17)(00.1(
cm)0.17)(33.1(






















sn
sn
m

b
a
b)
cm,4.56
cm0.14
33.033.100.1













s
sR
nn
s
n
s
n
abba
which is outside
the bowl.
34.23: For

:cm18

s
a)
cm.0.63
cm0.18
1
cm0.14
11111







s
sfss
b)
.50.3
0
.
18
0.63



s
s
m

c) and d) From the magnification, we see that the image is real and inverted.
For
:cm00.7

s
a)
cm.0.14
cm00.7
1
cm0.14
11111







s
sfss
b)
.00.2
00
.
7
0.14






s
s
m
c) and d) From the magnification, we see that the image is virtual and erect.
34.24: a)
cm,0.48
cm0.12
1
cm0.16
11111





f
ffss
and the lens is
diverging.
b)
cm,638.0
16.0
12.0)(
cm)850.0(



















s
s
yy
and is erect.
c)
34.25:
.25.3
400.0
30.1
s
s
y
y
m





Also:
cm75.1525.31
cm00.7
1
cm00.7
11111











s
s
s
s
ssfss
(to the left).
cm,85.4
cm75.15
1
cm0.7
11




s
s
and the image is virtual
).0since(


s
34.26:
:Also.711.0
20.3
50.4
ss
s
s
y
y
m






cm217
0.90
11
711.0
1111










s
ssfss
(to the right).
cm,154cm)217(711.0



s
and the image is real (since
).0


s
34.27:























cm50.3
1
cm00.5
1
)48.0(
1
cm0.18
111
)1(
11
21
sRR
n
ss
cm3.10





s
(to the left of the lens).
34.28: a) Given
m0.0741m6.0081.00m00.6and,0.80









ssssss
m.5.93and


s
b) The image is inverted since both the image and object are real
).0,0(



ss
c)
m,0732.0

m93.5
1
m0741.0
1111



f
ssf
and the lens is converging.
34.29:






















cm00.8
1
cm00.4
1
)60.0(
11
)1(
1
21
RR
n
f
cm.13.3cm,44.4




f
cm.44.4cm;44.4
cm
3.13;cm3.13;cm3.13;cm44.4;cm44.4;cm3.13
87
654321













ff
ffffff
34.30:
We have a converging lens if the focal length is positive, which requires:
.0
11
0
11
)1(
1
2121




















RRRR
n
f
This can occur in one of three ways:
(i)
0,0)ii(}0,{}{
212121
 RRRRRR
(iii)
}.0,{|}||{|
2121
 RRRR
Hence the three lenses in Fig. (35.29a).
We have a diverging lens if the focal length is negative, which requires:
.0
11
0
11
)1(
1
2121




















RRRR
n
f
This can occur in one of three
ways:
.0,0)iii(0)ii(}0,{}{)i(
21212121
 RRRRRRRR
Hence the three lenses in Fig. (34.29b).
34.31:
a) The lens equation is the same for both thin lenses and spherical mirrors, so the
derivation of the equations in Ex. (34.9) is identical and one gets:
.alsoand,

111111
sf
f
s
s
m
fs
sf
s
fs
fs
sfsfss














b) Again, one gets exactly the same equations for a converging lens rather than a
concave mirror because the equations are identical. The difference lies in the
interpretation of the results. For a lens, the outgoing side is not that on which the object
lies, unlike for a mirror. So for an object on the left side of the lens, a positive image

distance means that the image is on the right of the lens, and a negative image distance
means that the image is on the left side of the lens.
c) Again, for Ex. (34.10) and (34.12), the change from a convex mirror to a
diverging lens changes nothing in the exercises, except for the interpretation of the
location of the images, as explained in part (b) above.
34.32:
cm.0.7
cm0.17
1
cm0.12
11111





s
sfss
,cm34.0
4
.
2
cm800.0
4.2
2
.
7
)0.17(










m
y
y
s
s
m
so the object is
cm34.0
tall, erect, same side.
34.33:
cm.3.26
cm0.17
1
cm0.48
11111





s
sfss
cm24.1

646
.
0
cm800.0
646.0
3
.
26
0.17








m
y
y
s
s
m
tall, erect, same side.
34.34: a)
cm,1.11
cm0.36
1
cm0.16
11111




f
ffss
converging.
b)
,cm8.1
16
36
cm)80.0(


















s

s
yy
so the image is inverted.
c)
34.35: a)
mm.60m0600.0
m240
m024.0
m600


















y
y
ss

s
s
y
y
m
.mm60
mm60
1
mm100.6
1111
5





f
ssf
So one should use the
85-mm lens.
b)
mm.150m15.0
m9.6
m0.036
m0.40














y
y
ss
mm.149
mm150
1
mm1040
1111
3





f
ssf
So one should use the
135-mm lens.
34.36:
.m0869.0
m085.0
11

m90.3
1111







s
sfss
mm,39.0mm1750
90
.
3
0869.0




y
s
s
y
so it will not fit on the 24-mm

36-mm film.
34.37:
.cm1020
cm0.20

1
cm4.20
11111



s
sfss
34.38:
mm.37.2m0.0372m)7.70(
m109.50
m00.5
3






y
s
f
y
s
s
y
34.39: a)
.104.1
mm200,000
mm28

4


 m
s
f
s
s
m
b)
.103.5
mm200,000
mm105
4


 m
s
f
s
s
m
c)
.105.1
mm200,000
mm300
3


 m

s
f
s
s
m
34.40: a)
cm.12
11
1






fss
b)
cm.8cm12cm0.4
2
s
c)
cm,24
cm12
11
cm8
1111
2
2












s
sfss
to the right.
d)
cm.12
11
1


 fss
cm.4cm12cm0.8
2
s
cm.6
cm12
11
cm4
1111
2
2












s
sfss
34.41: a)
mm.75
4
mm300
4
44 
f
D
D
f
f
b)
,
8
8
f
Df


so the diameter is 0.5 times smaller, and the area is 0.25 times
smaller. Therefore only a quarter of the light entered the aperture, and the film must be
exposed four times as long for the correct exposure.
34.42:
The square of the aperture diameter (~ the area) is proportional to the length of the
exposure time required.
.
250
1
mm23.1
mm8
30
1
2






















ss
34.43:
a) A real image is formed at the film, so the lens must be convex.
b)
mm0.0.50with,and
1
so
111










f
fs
sf
s
sf
fs
sfss

For
mm.56smm,450cm45




s
For
mm.50,





fss
The range of distances between the lens and film is
mm.56tomm50
34.44: a)
cm.15.3m153.0
m150.0
1
m00.9
11111



s
sfss
b)



 8.58
153
.
0
00.9
s
s
m
dimensions are
m).2.12m41.1(mm)36mm24(



m
34.45: a)
.cm4.36m364.0
m75.2
1
power
1
1


f
The near-point is normally at
,cm80
cm4.36
11
cm25

1111
:cm25







s
sfss
in front of the eye.
b)
.cm9.76m769.0
m30.1
1
power
1
1




f
The far point is ideally at
infinity, so:
.cm9.76
cm9.76
111111












s
sfss

34.46:
.cm710.0
40.0
cm60.2
40.1
cm0.40
1





R
RR
nn
s
n

s
n
abba
34.47: a)
33.2
1
power
m600.0
1
m25.0
1111





fssf
diopters.
b)
67.1
1
power
m600.0
11111








fssf
diopters.
34.48: a) Angular magnification
.17.4
cm00.6
cm0.25cm0.25

f
M
b)
.cm84.4
cm00.6
1
cm0.25
11111





s
sfss
34.49: a)
.cm06.6
cm00.8
1
m0.25
11111






s
sfss
b)
mm13.4)13.4(mm)00.1(13.4
cm06.6
cm0.25




 ymy
s
s
m
34.50:
.cm00.8mm0.80
rad025.0
mm00.2



y
f
f
y
34.51:

cm00.4
1
50.6
1111
50.650.6









ssss
ss
s
s
m
.cm0.2250.6,cm38.3
00.4
1
50.6
1
1
1











sss
s
34.52: a)
.317
mm)0.26mm)(00.5(
mm)0.5mm160(mm)250(
mm)250(
21
1





ff
s
M
b)
.mm1015.3
317
mm10.0
4






m
y
y
y
y
m
34.53:
a) The image from the objective is at the focal point of the eyepiece, so
cm9.17cm80.1cm7.19
21


fds
oe
.cm837.0
cm800.0
1
cm9.17
11111



s
sfss
b)
.4.21
cm837.0

cm9.17
1



s
s
m
c) The overall magnification is
.297
cm80.1
cm0.25
)4.21(
cm0.25
2
1

f
mM
34.54: Using the approximation
,
1
fs

and then
,
1
1
1
f

s
m


we have:
.64
mm9.1
mm122
mm9.1;mm122mm9.1mm120:mm9.1
.31
mm4
mm124
mm4;mm124mm4mm120:mm4
.5.8
mm16
mm136
mm16;mm136mm16mm120:mm16
1
1
1





















s
s
m
ssf
s
s
mssf
s
s
m
ssf
The eyepiece magnifies by either 5 or 10, so:
a) The maximum magnification occurs for the 1.9-mm objective and 10x
eyepiece:
.640)10)(64(
1

e
mmM
b) The minimum magnification occurs for the 16-mm objective and 5x eyepiece:

.43)5)(5.8(
1

e
mmM
34.55: a)
.33.6
cm0.15
cm0.95
2
1

f
f
M
b)
,m950.0
m950.0
11
m3000
1111







s
sfss

so the height of an
image of a building is
.m019.0m)0.60(
3000
950.0




y
s
s
y
c)
.rad127.0)3000()0.60(33.6)30000.60arctan(33.6 


M
34.56:
m71.1m0900.0m80.1
2121










fdfdff
ssss
.0.19
00.9
171
2
1

f
f
M
34.57:
.m4.19m)02.1)(0.19()0.19(0.19  Df
D
f
34.58:
.m1040.4m)18(
180
)014.0(
3














θf
s
f
y
s
s
yy
34.59: a)
.m661.0m650.0
2
211
 ffd
R
f
b)
.1.59
m011.0
m650.0
2
1

f
f
M
34.60:








Rf
ffss
m50.1
1
m12.0m75.0
1
m3.1m75.0
1111
m.0.32


f
So the smaller mirror must be convex (negative focal length) and have a radius of
curvature equal to 3.0 m.
34.61: If you move away from the mirror at
s,m40.2
then your image moves away
from the mirror at the same speed, but in the opposite direction. Therefore you see the
image receding at
s,m80.4
the sum of your speed and that of the image in the mirror.
34.62:
a) There are three images formed.
34.63: The minimum length mirror for a woman to see her full height h, is
2h

, as
shown in the figure below.
34.64:
.m2.3m00.425.1
m00.4
25.2




 ss
s
s
s
s
m
So the mirror is
7.20 m from the wall. Also:
.m43.4
m20.7
1
m2.3
12211



R
RRss
34.65: a)
133.0

0.60
m00.8
0.60
00.6
360




 s
s
s
y
y
m
m is where the filament
should be placed.
b)
.m261.0
m00.8
1
m133.0
12211



R
RRss
34.66:
.m0894.0

m180.0
21
m0.13
1211









s
sRss
 
m.0103.0
0.13
0894.0
m50.1



















s
s
yy
b)
The height of the image is less then 1% of the true height of the car, and is less
than the image would appear in a plane mirror at the same location. This gives the
illusion that the car is further away then “expected.”
34.67: a)
and0

R )0(imagerealaso,0



ss
is produced for virtual object
positions between the focal point and vertex of the mirror. So for a 24.0 cm radius mirror,
the virtual object positions must be between the vertex and 12.0 cm to the right of the
mirror. b) The image orientation is erect, since
.0






s
s
s
s
m
c)
34.68:
The derivations of Eqs. (34.6) and (34.7) are identical for convex mirrors, as long
as one recalls that R and
s

are negative. Consider the diagram below:
We have:
,and
1211
2
211
s
s
y
y
m
fRss
R
fs
Rs












since
s

is not on the outgoing side of the mirror.
34.69: a)
,cm46
cm4.19
21
cm0.8
1211







s
sRss
so the image is virtual.

b)
,8.5
0
.
8
46





s
s
m
so the image is erect, and its height is:
.mm29mm)0.5)(8.5()8.5(




yy
c) When the filament is 8 cm from the mirror, there is no place where a real
image can be formed.
34.70:
sR
Rss
smss
s
s
m















4
321
2
5
,0,0since
5
2
2
5
.
4
3
and
10
3
RsRs 



34.71: a)
fss
111



and taking its derivative with respect to s we have


























mm
ds
sd
m
s
s
ds
sd
ds
sd
ss
fsds
d
But.
1111
s
1
0
2
2
2
22
.
2
m
Images are always inverted longitudinally.

b) (i) Front face:
.cm000.120
cm000.150
21
cm000.200
1211







s
sRss
Rear face:
.cm964.119
cm000.150
21
cm100.200
1211







s
sRss

(ii) Front face:
 





2
2
600000.0,600000.0
000
.
200
000.120
mm
s
s
m
.360.0

Rear face:
 





2
2
599520.0,599520.0

100
.
200
964.119
mm
s
s
m
.359425.0

(iii) So the front legs are magnified by 0.600000, the back legs by 0.599520, and
the side legs by 0.359425, the average of the front and back longitudinal magnifications.
34.72:
R
nn
s
n
s
n
abba




and taking its derivative with respect to s we have:
.
0
2
2
2

2
2
2
2
22
a
b
a
b
b
a
b
a
b
aabba
n
n
m
n
n
n
n
s
s
n
n
s
s
ds
sd

ds
sd
s
n
s
n
R
nn
s
n
s
n
ds
d





























But
.
2
a
b
n
n
mmm
ds
sd






34.73: a) R < 0 for convex so

















dt
ds
ds
sd
dt
sd
v
R
s
sR
s
R
s
s
2

211
 












2
2
2
2
Rs
sR
Rs
R
v
 
 


  
s.m1065.8
m25.1m0.102

m25.1
sm50.2
2
3
2
2
2
2








Rs
R
vv
b)
 
 


  
s.m142.0
m25.1m0.22
m25.1
sm50.2
2

2
2
2
2






Rs
R
vv
Note: The signs are somewhat confusing. If a real object is moving with v > 0,
this implies it is moving away from the mirror. However, if a virtual image is moving
with v > 0, this implies it is moving from “behind” the mirror toward the vertex.
34.74:
In this context, the microscope just takes an image and makes it visible. The real
optics are at the glass surfaces.
.31.1
mm50.2
mm780.0mm50.2
0
1
0











s
s
n
ss
n
s
n
s
n
ba

Note that the object and image are measured from the front surface of the second
plate, making the image virtual.
34.75:
a) Reflection from the front face of the glass means that the image is just h
below the glass surface, like a normal mirror.
b) The reflection from the mirrored surface behind the glass will not be affected
because of the intervening glass. The light travels through a distance 2d of glass, so the
path through the glass appears to be
,
2
n
d
and the image appears to be
n

d
h
2

behind the
front surface of the glass.
c) The distance between the two images is just
.
2
n
d
34.76:
a) The image from the left end acts as the object for the right end of the rod.
b)
.cm3.28
cm0.6
60.060.1
cm0.23
1









s
sR

nn
s
n
s
n
abba
So the second object distance is
.cm7.11cm3.28cm0.40
2
s
Also:
  
.769.0
0.2360.1
3.28
1



sn
sn
m
b
a
c) The object is real and inverted.
d)
.cm5.11
cm0.12
60.01
cm7.11

60.1
222












s
sR
nn
s
n
s
n
abba
Also:




  
.21.157.1769.057.1
7.11

5.1160.1
212




 mmm
sn
sn
m
b
a
e) So the final image is virtual, and inverted.
f)




mm.82.121.1mm50.1





y
34.77: a)
.cm3.28
cm0.6
60.060.1
cm0.23

1









s
sR
nn
s
n
s
n
abba
So the second object distance is
.cm3.3cm3.28cm0.25
2
s
Also:
  
.769.0
0.2360.1
3.28
1




sn
sn
m
b
a
b) The object is virtual.
c)
.cm87.1
cm0.12
60.01
cm3.3
60.1
2
222














s

sR
nn
s
n
s
n
abba
Also:




  
.693.0901.0769.0901.0
3.3
87.160.1
21
2
2
2




 mmm
sn
sn
m
b
a

d) So the final image is real and inverted.
e)




mm.04.1693.0mm50.1






ymy
34.78:
For the water-benzene interface to get the apparent water depth:
.cm33.70
50.1
cm50.6
33.1
0







s
ss

n
s
n
ba
For the benzene-air interface, to get the total apparent distance to the bottom:
.cm62.60
1
)cm60.2cm33.7(
50.1
0









s
ss
n
s
n
ba
34.79:
.00.21
2
1
2

1
2,but,











n
n
R
n
R
n
Rss
R
nn
s
n
s
n
abba
34.80: a)
.cm9.36
cm0.15

60.060.1
cm0.12
1









s
sR
nn
s
n
s
n
abba
So the
object distance for the far end of the rod is
cm.9.86)cm9.36(cm0.50



cm.3.540
1
cm9.86
60.1










s
sR
nn
s
n
s
n
abba
b) The magnification is the product of the two magnifications:
.92.100.1,92.1
)0.12)(60.1(
9.36
2121




 mmmm
sn
sn
m

b
a
34.81:
cm.00.9
cm00.4
80.080.11











s
sR
nn
s
n
s
n
abba
So the object distance
for the far side of the ball is
.cm00.1cm00.9cm00.8




,cm50.0
cm00.4
80.01
cm00.1
80.1














s
sR
nn
s
n
s
n
abba
which is
4.50 cm from the center of the sphere.

34.82:
.58.1
50.9
0.15
0
cm50.9
1
cm0.15
0





n
n
s
n
s
n
ba
When viewed from the curved end of the rod:
,cm1.21
cm0.10
58.01
cm0.15
58.1
R
11

















s
s
n
ss
n
R
nn
s
n
s
n
abba
so the image is 21.1 cm within the rod from the curved end.
34.83:
a) From the diagram:

)50.1(
190.0
sinBut.sin50.1
190.0
sin
RR
r
R
r
R






cm.127.0
50
.
1
cm190.0
 r
So the diameter of the light hitting the surface is
cm.254.02

r
b) There is no dependence on the radius of the glass sphere in the calculation
above.