This preview shows pages 1–3. Sign up to view the full content.
3424
Chapter 34
34.84.
IDENTIFY
and
SET UP:
Treating each of the goblet surfaces as spherical surfaces, we have to pass, from left to
right, through four interfaces. Apply
abba
nnnn
s
sR
−
+=
′
to each surface.
The image formed by one surface serves as
the object for the next surface.
EXECUTE: (a)
For the empty goblet:
1
1
1
1.50
0.50
12 cm
4.00 cm
s
ss
R
s
−
′
⇒+ =
⇒=
′′
∞
.
2
2
2
1.50
1
0.50
0.60 cm
12 cm
11.4 cm
64.6 cm.
11.4cm
3.40cm
s
−
′
=
−
=−
⇒
+
=
⇒
′
−
3
3
3
1
1.50
0.50
64.6 cm
6.80 cm
71.4 cm
9.31cm.
71.4 cm
3.40 cm
s
′
=+=⇒+
=
⇒
=
−
′
−
4
4
4
1.50
1
0.50
9.31cm
0.60 cm
9.91cm
37.9 cm.
9.91cm
4.00 cm
s
−
′
=
⇒
=
−
′
−
The final image is
37.9 cm
2(4.0 cm)
29.9 cm
−=
to the left of the goblet.
(b)
For the winefilled goblet:
1
1
1
1.50
0.50
12 cm
4.00 cm
s
R
s
−
′
∞
.
2
2
2
1.50
1.37
0.13
0.60 cm
12 cm
11.4 cm
14.7 cm.
s
−
′
=
−
⇒
+
=⇒
=
′
−
3
3
3
1.37
1.50
0.13
6.80 cm
14.7 cm
7.9 cm
11.1cm.
7.9 cm
3.40 cm
s
′
=−=
−
⇒
+
=
⇒
=
′
−−
4
4
4
1.50
1
0.50
0.60 cm
11.1cm
10.5 cm
3.73 cm
10.5 cm
4.00 cm
s
−
′
−
⇒
+
=
⇒
=
′
. The final image is 3.73 cm to the
right of the goblet.
EVALUATE:
If the object for a surface is on the outgoing side of the light, then the object is virtual and the object
distance is negative.
34.85.
IDENTIFY:
The image formed by refraction at the surface of the eye is located by
s
−
′
.
SET UP:
1.00
a
n
=
,
1.35
b
n
=
.
0
R
>
.
For a distant object,
s
≈ ∞
and
1
0
s
≈
.
EXECUTE: (a)
s
≈∞
and
2.5 cm
s
′ =
:
1.35
1.35 1.00
2.5 cm
R
−
=
and
0.648 cm
6.48 mm
R
=
=
.
(b)
0.648 cm
R
=
and
25 cm
s
=
:
1.00
1.35
1.35 1.00
25 cm
0.648
s
−
′
.
1.35
0.500
s
=
′
and
2.70 cm
27.0 mm
s
′ =
=
. The
image is formed behind the retina.
(c)
Calculate
s
′
for
s
and
0.50 cm
R
=
:
1.35
1.35 1.00
0.50 cm
s
−
=
′
.
1.93 cm
19.3 mm
s
′ =
=
. The image is formed in
front of the retina.
EVALUATE:
The cornea alone cannot achieve focus of both close and distant objects.
34.86.
IDENTIFY:
Apply
s
−
′
and
a
b
ns
m
′
to each surface.
The overall magnification is
12
mm
m
=
. The
image formed by the first surface is the object for the second surface.
SET UP:
For the first surface,
1.00
a
n
=
,
1.60
b
n
=
and
15.0 cm
R
= +
.
For the second surface,
1.60
a
n
=
,
1.00
b
n
=
and
R
=→ ∞
.
EXECUTE: (a)
1
1.60
0.60
36.9 cm.
12.0 cm
15.0 cm
s
R
s
−
′
⇒
+ =
−
The object distance for the far end
of the rod is 50.0 cm
( 36.9 cm)
86.9 cm.
=
The final image is 4.3 cm to the left of the vertex of the
hemispherical surface.
1.60
1
0
54.3 cm.
86.9 cm
s
R
s
−
′
⇒
+=⇒=
−
(b)
The magnification is the product of the two magnifications:
1
2
36.9
1.92,
1.00
1.92.
(1.60)(12.0)
a
b
m
m
m
′
−
=
=
⇒
=
=
EVALUATE:
The final image is virtual, erect and larger than the object.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document Geometric Optics
3425
34.87.
IDENTIFY:
Apply Eq.(34.11) to the image formed by refraction at the front surface of the sphere.
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics

Click to edit the document details