PHYSICS 9B
MIDTERM 2
November 25, 2008
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Problem 1. (2 points) When light travels from medium “X” to medium “Y” as shown in the figure:
A. both the speed and the wavelength increase
B. both the speed and the frequency increase
C. both the wavelength and the frequency are unchanged
D. both the speed and the wavelength decrease
X
Y
E. both the speed and the frequency decrease
Explanation:
According to the figure, the index of refraction n of “X” is larger than that of “Y”. Therefore, both the speed and
wavelength are increased by a factor of n . The frequency remains constant. Answer A.
Problem 2. (2 points) Consider a horizontal plane of thin film with a thickness t. This film is located in between
air and water (see sketch). Light is directed from the air downward through the film and into the water,
perpendicular to the surfaces. The index of refraction of the film n1. The index of refraction of water is n2.
The wavelength of the incident wave in the air is λ.
The smallest non-zero thickness tmin leading to no reflection is
A. tmin =
B. tmin =
C. tmin =
λ
2
λ
2n1
λ
2n2
D. tmin = λ
E. tmin =
F. tmin =
λ
n1
λ
n2
Explanation.
Solution.
Problem 3. (2 points) If the path difference between two waves coming to a particular point is 3/4
wavelengths, the corresponding phase difference is
3 π /8
3 π /4
3 π/2
2π
3π
Α.
B.
C.
D.
E.
Explanation. One wavelength corresponds to the phase difference of 2π. Consequently ¾ of the wavelength
corresponds to 2π*¾ = 3π/2. Answer C.
Problem 4. (2 points) X-ray diffraction is used to determine
A.
B.
C.
D.
E.
Velocity of light in the medium
Distances between atoms in solids
Spectrum of light wavelengths emitted by a source
Refractive index of the medium
All of the above
Solution. X-ray diffraction is used to determine distances between atoms in solids. Answer B.
Problem 5. (3 points) A cylindrical opaque drinking glass has a diameter 5.1 cm and height h, as shown in the
figure. An observer’s eye is placed as shown (the observer is just barely looking over the rim of the glass). When
empty, the observer can just barely see the edge of the bottom of the glass. When filled to the brim with a
transparent liquid, the observer can just barely see the center of the bottom of the glass. The liquid in the drinking
glass has an index of refraction of 1.29. Calculate the angle θr.
Problem 6. (3 points) Two thin parallel slits that are 0.0116 mm apart are illuminated by a laser beam of
wavelength 585 nm. The interference pattern is seen on a very large distant screen.
a) What is the total number of bright fringes (those indicating complete constructive interference) including the
central fringe and those on both sides of it?
b) At what angle relative to original direction of the beam will the fringe that is most distant from the central bright
fringe occur?
c) If the maximum intensity is 2 W/m2, what is the intensity at points of the screen corresponding to the path
difference of 2/3 wavelengths?
Explanation.
a) dsinθ=mλ therefore mλ/d=sinθ<1
λ/d==585x10-9/0.0116x10-3= 0.05431
mmax<d/λ=1/0.05431=19.8
mmax=19
Total number of fringes including the central one is 2 mmax +1=39.
b) Max. angle sinθmax =mmaxλ/d=19/19.8=0.959
θmax=73 degrees
c) The path difference of 2/3 wavelengths corresponds to the phase difference φ=2π*2/3=4π/3.
The intensity is given by
I = I 0 cos2 φ / 2 = I 0 cos2 2π / 3 = I 0 / 4 = 0.5 W / m 2
Problem 7. (3 points) You are looking at your own reflection in a shiny silvered Christmas tree ornament 0.75 m
away. The diameter of the ornament is 7.2 cm. Your height is approximately 1.7 m. Although not drawn to scale
the figure below illustrates the setting. In the following use paraxial approximation.
0.75 m
a) Draw a principal ray diagram showing the position of your image. State whether the image is virtual or real
and whether it is direct or inverted.
b) At what distance from the surface of the ornament the image appears?
c) What is the height of your image in the ornament?
Explanation
a) Here we obtain a virtual direct image
b) Note that the focal distance for convex mirrors is negative. Here f=-d/4=-7.2/4=-1.8 cm.
The distance from the surface (in cm) is s’=(1/f-1/s)-1=(-1/1.8-1/75)=-1.76 cm.
c) The lateral magnification is s’/s=1.76/75=0.0234. Therefore, the image height is 0.0234*170=3.978 cm.
Problem 8 (3 points) In a single slit diffraction experiment the wavelength of the incident light is 520 nm, the slit
width is 8 μm and the distance between the slit and the screen is 8.4 m. In the following use small angle
approximation tanθ= sinθ=θ.
a) Find the position y = y1 of the first intensity minimum.
b) Find the position y = y2 of the second intensity maximum
(approximate it by point between first and second intensity minima).
c) If the intensity of the first (central) maximum is 0.5 W/m2,
estimate the intensity at the second intensity maximum.
Solution
a) For single slit diffraction, destructive interference occurs when, a sinθ =mλ. The small angle approximation
gives us y1/L = tanθ = sinθ = mλ/a or for m=1 y1 = Lλ/a = 8.4 * 520 * 10-9/ (8 * 10-6) = 546 * 10-3 = 0.546 m.
b) The second minimum can be found at the position 2y1=1.092 m, therefore the second intensity maximum is
approximately located at y2=0.819 m.
c) The intensity is approximately given by
I
⎡ sin[π a sin θ / λ ⎤
⎡ sin[3π / 2) ⎤
I = I0 ⎢
= I0 ⎢
= 20 = 0.022 W / m 2
⎥
⎥
9π / 4
⎣ π a sin θ / λ ⎦
⎣ 3π / 2 ⎦
2
2
y2