Physics 303L HW 11 - delgado (dd28287) hw11 hegelich...

delgado (dd28287) – hw11 – hegelich – (55795) 1
This print-out should have 28 questions.
Multiple-choice questions may continue on
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before answering.
001 10.0points
What must be the minimum length of a plane
mirror in order for you to see a full view of
yourself?
1. One fourth of your height
What is the spatial length of this pulse as it
travels through glass with index of refraction
n = 1 . 3?
4. Twice of your height
= 1 . 497 × 10
− 6
m
and the distance between the leading edge and
the rear edge of the pulse is 1 . 497 × 10
− 6
m ,
the required spatial length.
003(part2of2)10.0points
What is the spatial length of this pulse as it
travels through glass with index of refraction
n = 1 . 3?
002(part1of2)10.0points
Laser scientists have succeeded in generating
pulses of light lasting as short as 10
− 15
s.
Imagine that you create a pulse which lasts
4 . 99 × 10
− 15
s.
What is the spatial length of this pulse
along its direction of travel in a vacuum? The
speed of light in a vacuum is 3 × 10
8
m / s .
004 10.0points
An object, slanted at an angle of 45
◦
, is
placed in front of a vertical plane mirror, as
shown.
Mirror
Explanation:
Which of the following shows the appar-
ent position and orientation of the object’s
image?

delgado (dd28287) – hw11 – hegelich – (55795) 2
2.
image
correct
3.
image
4.
image
5.
image
6.
image
7.
image
8.
image
Explanation:
For a plane mirror, the line connecting a
point of the object and corresponding point of
the image is perpendicular to the mirror, and
the corresponding distances from the points
to the mirror are the same.
Thus, the image should be
image
object
005 10.0points
A coin is at the bottom of a beaker. The
beaker is filled with 4 . 6 cm of water ( n
1
=
1 . 33) covered by 2 . 4 cm of liquid ( n
2
= 1 . 4)
floating on the water.
How deep does the coin appear to be from
the upper surface of the liquid (near the top
of the beaker)?
Correct answer: 5 . 17293 cm.
Explanation:
For small angles
sin θ ≈ tan θ =
x
ℓ
.
The appearance of the width of the coin x
remains the same. Applying Snell’s Law,
n
1
sin θ
1
= n
2
sin θ
2
n
i
x
ℓ
i
= n
f
x
ℓ
f
ℓ
f
=
ℓ
i
n
i
since n
f
≈ 1 for air and the apparent distance
d = ℓ
f
. Thus
d =
ℓ
2
n
2
+
ℓ
1
n
1
=
2 . 4 cm
1 . 4
+
4 . 6 cm
1 . 33
= (1 . 71429 cm) + (3 . 45865 cm)
= 5 . 17293 cm .
The coin appears to be closer to the surface
by
(4 . 6 cm) + (2 . 4 cm) − (5 . 17293 cm)
= 1 . 82707 cm .
Alternate Solution: