Version PREVIEW – HW 3 – Savrasov – (39824)
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Wavelength Comparison
001
10.0 points
The speed oF light is 2
.
998
×
10
8
m
/
s.
The ratio oF the wavelength oF visible light
oF wavelength 600 nm to that oF an ±M radio
wave oF Frequency 100 MHz is about
1.
10
−
1
2.
10
−
7
correct
3.
10
−
3
4.
10
−
14
5.
10
−
10
Explanation:
The wavelength oF the ±M radio wave is
λ
FM
=
c
f
FM
=
2
.
998
×
10
8
m
/
s
100 MHz
= 2
.
998 m
.
Then the ratio oF the wavelength oF visible
light to that oF the ±M radio wave is about
1
×
10
−
7
.
keywords:
Perpendicular Mirrors
002
(part 1 oF 2) 10.0 points
Consider the case in which light ray A is in
cident on
mirror 1
, as shown in the fgure.
The re²ected ray is incident on
mirror 2
and
subsequently re²ected as ray B. Let the an
gle oF incidence (with respect to the normal)
on
mirror 1
be
θ
A
= 50
◦
and the point oF
incidence be located 20 cm From the edge oF
contact between the two mirrors.
90
◦
x
A
B
mirror 1
mirror 2
θ
A
θ
B
What is the angle oF the re²ection oF ray B
(with respect to the normal) on
mirror 2
?
Correct answer: 40
◦
.
Explanation:
IF the angle oF the incident ray A is
θ
A
, the
angle oF re²ection must also be
θ
A
. Since the
mirrors are perpendicular to each other, angle
θ
B
is equal to 90
◦

θ
A
θ
B
= 90
◦

θ
A
= 90
◦

50
◦
= 40
◦
.
003
(part 2 oF 2) 10.0 points
Determine the angle between the rays A and
B.
1.
50
◦
2.
0
◦
3.
80
◦
4.
40
◦
5.
180
◦
correct
6.
100
◦
7.
90
◦
Explanation:
ReFer to the fgure below.
Consider the
angle each ray makes with
mirror 1
. Ray A
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makes an angle of
φ
A
with
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 Fall '09
 RichardScalettar
 Light, Snell's Law, Total internal reflection

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