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PHYS 3101
Exam 2
Spring 2009
Dr. Aktas
Name :________________________
Student # : _________________________
You have Six questions, 20 points each.
This is a closed book exam. I understand I am not to use any notes or
information other than on this exam sheet. I may use a pocket calculator but
only for the purpose of numerical calculation. I accept the
responsibility to know and observe the requirements of the UNC-
Charlotte Code of Student Academic Integrity.
_________________
_
Signature
Good luck
Show all of your work. Do not skip steps. First write down the relevant
equations then substitute the numbers if necessary.
- Shown from above in figure below is one corner of a rectangular box filled
with water. A laser beam starts from side A of the container and enters the
water at position x. You can ignore the thin walls of the container.
a. If x = 15 cm, does the laser beam refract back into the air through side B
or reflect from the side B back into the water?
b. Find the minimum value of x for which the laser beam passes through the
side B and emerges into the air. Index of refraction of water is 1.330.
- Light of wavelength 600 nm passes through a double slit and is viewed on a
screen 2.0 m behind the slits. Each slit is 0.040 mm wide and they are separated by 0.
mm. Calculate the positions of first five bright interference fringes relative to the center.
How many bright fringes are seen in the central diffraction envelope?
- Calculate the minimum thickness of a soap bubble film ( n = 1.33 ) that results in
constructive interference in the reflected light if the film is illuminated with light whose
wavelength in air is 600 nm.
- Assume that the light leaving the slits in a double slit experiment has the electric
field components given as,
E
1
= E
0
sin ω t
E
2
= E
0
sin( ω t + φ)
By using the method of phasors show that the intensity variation on the screen is
I = 4 I
0
cos
2
π d
λ D
y
Where I 0
is the intensity of incident beam, d is the slit separation, D is the distance
between the screen and the slit. Use small angle approximation.
Some useful formulas
y ( x , t ) = A sin( kx − ω t + ϕ)
cos α + cos β = 2 cos(
α + β
)cos(
α − β
ω = 2 π f =
2 π
T
, k =
2 π
λ
, v =
T
μ
v =
ω
k
= λ f =
λ
T
f = f
v ± v
D
v ± v
S
sin( α + β) = sin α cos β + sin β cos α
Δ φ
2 π
Δ d
λ
, c =
E
m
B
m
ε
0
μ
0
S =
μ
0
E ×
B , I = S
ave
, Δ p =
Δ U
c
, F =
Δ p
Δ t
F =
IA
c
, p
r
F
A
, ε
0
= 8.85 × 10
− 12
F / m , μ
0
= 1.26 × 10
− 6
H / m
€
λ
n
=
λ
n
, ±
1
f
= ±
1
o
±
1
i
, m = −
i
o
=
′ h
h
, n
1
sin θ
1
= n
2
sin θ
2
c
2
= a
2
2
− 2 ab cos θ , θ : Angle between a and b.
Double slit path diffrence = d sin θ
Single slit path diffrence = a sin θ
Thin film path diffrence = 2 t
for small θ , sin θ ≈ tan θ
