Version 035/AACAD – midterm 04 – Turner – (59070)
1
This
printout
should
have
19
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
The reflecting surfaces of two intersecting flat
mirrors are at an angle of 62
◦
, as shown in
the figure. A light ray strikes the horizontal
mirror at an angle of 49
◦
with respect to the
mirror’s surface.
62
◦
49
◦
φ
Figure is not drawn to scale.
Calculate the angle
φ
.
1. 54.0
2. 68.0
3. 70.0
4. 74.0
5. 48.0
6. 76.0
7. 58.0
8. 46.0
9. 62.0
10. 56.0
Correct answer: 56
◦
.
Explanation:
Basic Concept:
θ
incident
=
θ
reflected
Solution:
θ
1
θ
2
φ
θ
Figure is to scale.
The sum of the angles in a triangle is 180
◦
.
In the triangle on the left we have angles
θ ,
180
◦
−
θ
1
2
,
and
180
◦
−
θ
2
2
,
so
180
◦
=
θ
+
180
◦
−
θ
1
2
+
180
◦
−
θ
2
2
,
or
θ
1
+
θ
2
= 2
θ .
(1)
In the triangle on the right we have angles
θ
1
,
θ
2
,
and
φ .
180
◦
=
θ
1
+
θ
2
+
φ ,
so
θ
1
+
θ
2
= 180
◦
−
φ .
(2)
Combining Eq. 1 and 2, we have
φ
= 180
◦
−
2
θ
= 180
◦
−
2 (62
◦
)
=
56
◦
.
As a matter of interest,
in the upperhalf
of the figure the angles (clockwise) in the
triangles from left to right are
41
◦
,
41
◦
,
and 98
◦
;
82
◦
,
28
◦
,
and 70
◦
;
110
◦
,
21
◦
,
and 49
◦
;
131
◦
,
21
◦
,
and 28
◦
;
and in the lowerhalf of the figure the angles
(counterclockwise) in the triangles from left
to right are
21
◦
,
21
◦
,
and 138
◦
;
42
◦
,
28
◦
,
and 110
◦
;
70
◦
,
41
◦
,
and 69
◦
;
111
◦
,
41
◦
,
and 28
◦
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Version 035/AACAD – midterm 04 – Turner – (59070)
2
002
10.0 points
Consider a prism with the shape shown in the
diagram. Its index of refraction is labeled by
n
p
, and it is submerged in a special liquid
which has an index of refraction
n
ℓ
, where the
difference
n
p
−
n
ℓ
= 0
.
58.
The light ray is
perpendicularly incident from the liquid into
the prism as shown in the diagram.
Notice
that the incident angle at both points A and
B is 45
◦
.
n
p
A
B
n
l
45
°
45
°
Find the index of refraction of the prism for
a critical angle of 45
◦
.
1. 1.46811
2. 1.67296
3. 1.87782
4. 1.39983
5. 1.77539
6. 1.57054
7. 1.98024
8. 1.70711
9. 1.63882
10. 1.60468
Correct answer: 1
.
98024.
Explanation:
Let :
∆
n
= 0
.
58
θ
c
= 45
◦
.
n
p
sin
θ
c
= (
n
p
−
∆
n
) sin 90
◦
n
p
1
√
2
=
n
p
−
∆
n
n
p
parenleftbigg
1
−
1
√
2
parenrightbigg
= ∆
n
n
p
=
√
2 ∆
n
√
2
−
1
=
√
2 (0
.
58)
√
2
−
1
=
1
.
98024
.
003
10.0 points
At what distance from a 16 W electromag
netic wave point source (like a light bulb) is
the amplitude of the electric field 46 V
/
m?
μ
o
c
= 376
.
991 Ω
.
1. 0.673562
2. 0.426956
3. 0.68853
4. 5.84808
5. 0.438178
6. 32.8634
7. 1.57762
8. 0.803902
9. 4.42627
10. 1.99489
Correct answer: 0
.
673562 m.
Explanation:
Let :
P
= 16 W
,
E
max
= 46 V
/
m
,
and
μ
0
c
= 376
.
991 Ω
.
The wave intensity (power per unit area) is
given by
I
=
S
av
=
E
2
max
2 (
μ
o
c
)
=
(46 V
/
m)
2
2 (376
.
991 Ω)
= 2
.
80643 W
/
m
2
.
The constant
μ
o
c
(impedance of free space)
turns out to be 120
π
Ω.
You should verify that (V
/
m)
2
Ω = W
/
m
2
.
With this we can find the intensity, which
multiplied by the area of a circle of radius
R
must gives us the source’s power
P
, therefore
4
π R
2
I
=
P
= 16 W
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Turner
 Physics, Light, Wavelength, Correct Answer

Click to edit the document details