Quiz 4-1 Solution
Multiple-Choice:
Select the correct answer (1 per question)
1. A substance has an index of
refraction of 1.46. Light is
passing through it at 53.0°. At
what
angle will it leave into the air?
The index of refraction for air is 1.0003.
A) It will not leave.
Answer:
A
B) 59.1°
C) 43.2°
D) 33.2°
qcritical
= sin-1 (1.008/1.46) = 43 degrees
53,90 degrees us geater than the critical angle so it will not leave (total
internal reflection)
2. What is the critical angle
for light traveling from crown glass (n = 1.52) into water (n = 1.33)?
A) 42°
B) 48°
C) 57°
D) 61°
Answer:
D
qcritical
= sin-1 (1.33/1.52) = 61.04 = 61 degrees
3. Lucite has an index of
refraction of 1.50. What is its
critical angle of incidence?
Assume other medium is air.
A) 1.16°
B) 15°
C) 41.8°
Answer:
C
D) 87.4°
qcritical
= sin-1 (1.0003/1.5) = 41.8 degrees
4. The critical angle for a
substance is measured at 53.7°.
Light enters from air at 45.0°.
At
what angle will it continue?
A) 34.7°
Answer:
A
B) 45.0°
C) 53.7°
D) It will not continue, but be totally reflected.
From Snell's law:
N1sinq1
= N2 sinq2
It is given that N1 is air (N1 = 1.0003) and
q1
= 45 degrees
We need to find
q2,
but first we need to find N2 (in order to use the above equation)
We are also given that
qC
for the substance is 53.7 degrees
sin
qC
= (N2/N1)
but in this case N2 is air and N1 is the
substance
sin 53.7 = 1.0003/N1
gives N1 = 1.0003/Sin 53.7 = 1.0003/0.806 = 1.2411
so this is the N2 that we use in Snell's equation
sinq2
= N1/N2sinq1
= 1.0003/1.2411 sin 45 = 0.806 x 0.71 = 0.572
Gives angle = 34.7
5. Light enters a substance
from air at 30.0° to the normal.
It continues through the substance
at 23.0° to the normal. What
would be the critical angle for this substance?
A) 53°
B) 51.4°
Answer:
B
C) 36.7°
D) 12.6°
Know
q1,
q2,
and N1. Need to find
qC
Since Sin
qC
= Nair/Nsubstance we need to find Nsubstance
We use Snell's law
N1sinq1
= N2 sinq2
and
solve for N2 (which is Nsubstance)
N2
=
N1sinq1
/sin
q2
= 1.0003xsin30/sin23 =
1.0003x0.5/0.39 = 1.28
Sin
qC
= Nair/Nsubstance
= 1.0003/1.28 =
0.78
qC
= 51.4
6. A beam of light, traveling
in air, strikes a plate of transparent material at an angle of
incidence of 56.0°. It is
observed that the reflected and refracted beams form an angle of
90.0°. What is the index of
refraction of this material?
A) 1.40
B) 1.43
C) 1.44
D) 1.48
Answer:
D
Angle of incidence = angle of reflection, so angle of reflection = 56
degrees also.
1.0003 sin56 = N2 sin34 gives N2 = 1.48
7. A beam of light traveling
in air is incident on a slab of transparent material.
The incident
beam and the refracted beam make angles of 40° and 26° to the normal.
What is the speed
of light in the transparent material?
A) 1.0 × 108 m/s
B) 2.0 × 108 m/s
Answer:
B
C) 2.3 × 108 m/s
D) 3.0 × 108 m/s
Use Snell's law to solve for the only unknown, N2
1.0003sin 40 = N2sin26
N2
= 1.0003x0.64/0.44 = 1.46667
N2
= c/v gives v = c/N2 = 3x108/1.46667 = 2.0 x108
m/s
N2
= c/v gives v = c/N2
8.
A light ray hits a plane surface at 20 degrees. What is the angle
between the incident and
reflected
rays.
(A)
40 degrees
Answer = A
(B)
Cannot be determined from the information given
(C)
70 degrees
(D)
30 degrees
Angle of
incidence = angle of reflection. Double this and get 40 degrees
Problems: Show your work or you may get no credit
|
1 Two mirrors meet at a 135 degree angle
as shown
on the right.
Light rays strike one mirror at 40 degrees
as illustrated.
At what
angle
f
do they leave the second
mirror?
Show your work. |
|
Look at
the triangle formed by the mirrors and the first reflected ray
Angle of
incidence = angle of reflection = 40 degrees
40
degrees + 135 degrees = 175 degrees
That
leaves 180 - 175 - 5 degrees for the angle of incidence of the second ray.
Once again, angle of incidence = angle =
reflection gives
f
= 5 degrees
2. A
diver shines a flashlight upward from beneath the water at a 42.5 degree
angle to the
vertical. At what angle does the light leave the water?
We use Snell's law to find the angle of refraction
in the water
1.33 sin 42.5 = 1.0008 sin
q2
sin
q2
= 1.33/1.0008 sin 42.5 = 1.33 x 0.68 = 0.90
gives
q2
=
64 degrees
3. A light beam coming from an
underwater spotlight exits the water at an angle of 66.0
degrees to the vertical. At what angle of incidence does it hit the
air-water interface from
below the surface?
We use Snell's law: N1sin
q1
= N2 sin
q2
1.33
sin
q1
= 1.0003 sin 66 = 1.0003 x 0.91 = 0.91
sin
q1
= 0.91/1.33 = 0.69
gives
q1
=
43.6
Fill in the Blanks
1. Three models of light have been discussed. List these 3 models below.
Ø
___________________________________
Ray
Ø
___________________________________
Particle
Ø
___________________________________
Wave
2. When light strikes a
surface, 3 options are possible. List these below.
Ø
___________________________________
Reflected
Ø
___________________________________
Absorbed
Ø
___________________________________
Pass through
3. Identify the following
terminology associated with analysis of a concave mirror.
Ø
The normal to the center of the mirror is called the
__________________________
Principle Axis
Ø
The point at which the above normal touches the surface of the mirror
is called the
_____________________ vertex
Ø
The point on the normal to the center of the mirror, which is
equidistant from all points on
the surface of the mirror is called the
_________________________________
center of curvature
Ø
The point where parallel rays come to a focus after reflection is
called the
__________________ of the mirror.
focal point
4. The type of mirror typically
used for shaving and makeup is a _______________________
mirror.
These mirrors make objects appear larger.
concave
5. The type of mirror typically
used on some vehicles to give a wider field of view is a
___________________ mirror. convex
6.
In a ____________________ mirror, the image is virtual, upright, the same
size as the
object, and
is as far behind the mirror as the object is in front.
plane
7. If the rays that converge to form an image actually pass through the
image, so the image
would appear on
film or a screen placed there, the image is said to be a ________________
image. Otherwise,
if the light rays do not actually pass through the image, the image is said
to
be a __________________ image.
real virtual