Chapter 36
Questions: 1, 4, 6, 10, 13, 15
Problems: 3, 9, 19, 25, 29, 51, 55,
p A : (a) Show that for a horizontal parabolic mirror of the form y =
4xf x; any horizontal light ray will re‡ect o¤ of the mirror through a focal
point at x = xf ; y = 0 independent of the original height of the light ray.
(b) Show that a circle with a radius of curvature R = xf =2 approximates
this parabolic mirror near x = y = 0: Hint consider the condition x << 1 for
a circle tangent to the parabola at x = y = 0:
Questions
1. Your eye is detecting light that is diverging in such a way that it
represents a true image.
4. If the object is located at the center of curvature (a distance equal to
the radius of curvature) of the mirror then it is the same size as the image.
6. Focus an image from a distant object on a screen. The distance from
th e lens to thescreen is equal to the focal length.
10. (a) Yes, if the object is outside the radius of curvature, then the
image is real and reduced.
(b) Yes, if the object is outside the focal length (f = R=2), then the
image is real and enlarged.
(c) Yes, if the object is inside the focal length, then the image is virtual
and enlarged.
13. If you place the lightbulb at the focal point, then the light will be
re‡ected approximately into a parallel beam.
15. Since the …sh will appear to be closer, it will also appear to be smaller.
Problems
3.(a) For object distance, ` = 50cm, and image distance, `0 = 75cm, the
focal length is given by
1 1
` + `0
1
=
+ 0 =
;
f
` `
``0
``0
50: 75
f =
=
= 30cm.
` + `0
125
(b) The lens equation for this mirror with ` = 20cm is
1
1 1
= + 0;
f
` `
1
so that
1
1
=
0
`
f
20
`0 =
20
1
` f
=
;
`
`f
30
= 60cm
30
This means that the image is virtual and located 60cm behind the mirror.
9. The lens equation for this mirror with ` = 10cm and a focal length of
f = 17cm is
1
1 1
= + 0:
f
` `
(a) So that
1
1
=
0
`
f
`0 =
1
` f
=
;
`
f`
170
= 24:3cm
7
(b) The magni…cation is
M=
`0
= 2:43:
`
Thus the image height is
h0 = M h = 2:43
12 = 29:2mm.
(c) Since `0 < 0; the image is virtual
19. For these conditions M = 1; we have
M=
`0
=
`
1 ! `0 = `:
From the lens equation
1 1
2
1
1
=
+ 0 = = ;
f
` `
`
50
0
` = ` = 100cm.
25. Again using the lens equation we have
1
1 1
= + 0;
f
` `
2
so that
1
1
=
0
`
f
`0 =
1
` f
=
;
`
`f
`f
`
f
:
For the plane ‡ying at an altitude of ` = 103 m we …nd
`0 =
103 12
= 12:15m.
988
For the stars in the background ` ! 1 and
`0 = f = 12m.
So the image of the plane is 15cm farther from the lens than the image of
the stars.
29. For a total distance between the candle and screen of 70cm, we
assume that ` = x and `0 = 70 x: Then the lens equation is
(70
1 1
` + `0
1
=
+ 0 =
;
f
` `
``0
``0 = f (` + `0 ) ;
x) x = 17 70 = 1190:
The resulting quadratic equation is
x2
70x + 1190 = 0;
with solutions
p
4900
`1 = 70
0
`1 = 40:9cm
4760 =2 = 29:1cm,
p
`2 = 70 + 4900
`02 = 29:1cm.
4760 =2 = 40:9cm,
and
Basically the two solutions correspond to the object and image exchanging
locations.
3
51. Again using the lens equation we have
1
1
1
1
1
1
= + 0 ; and
= + 0;
f1
` `1
f2
` `2
where ` >> `0 so that
`01 ' f1 and `02 ' f2 :
Thus
f1
`0
`0 `
= 10 = 1 0 =
f2
`2
` `2
h01
h
h
h02
=
h01
:
h02
Thus the relative sizes are
110
h01
=
= 2:9:
0
h2
38
55. For a microscope the magni…cation is given by
25L
;
fo fe
M=
where the dimensions of all lengths are the same, often centimeters. Therefore
fe =
25
25L
=
fo M
:45
10
:833
=
= 1:85cm.
300
:45
A : (a) A horizontal light ray that approaches the mirror with a value of
y = yo will strike the mirror at xo = yo2 =4xf : The tangent to the mirror at
this location has a slope given by
q
dy
= xf =xo :
tan o =
dx
Now a simple drawing shows that the re‡ected ray has a slope of tan 2 o :
The algebraic expression for this re‡ected light ray is
p
2 xf =xo
2 tan o
y = x tan 2 o + b = x
+b=x
+ b;
1 tan2 o
1 xf =xo
p
4xf xo
y =
x + b:
xo xf
4
p
Now this light ray passes through y = 4xf xo at x = xo : This enables us to
determine b;
p
p
p
4xf xo
4xf xo
xo =
xf :
b = 4xf xo
xo xf
xo xf
This re‡ected light ray can now be expressed algebraically as
p
4xf xo
(x xf ) :
y=
xo xf
This light ray crosses the y axis at x = xf independent of xo and/or yo :
(b) A circle that is tangent to this parabola at x = y = 0 with a radius
of curvature R satis…es
(x R)2 + y 2 = R2 ;
or
y=
For x << R we have
q
R2
(x
y'
R)2 =
p
p
2xR
2xR;
which for R = 2xf reduces to
y=
p
4xf x;
which is the equation for the original parabola.
5
x2 :