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Midterm II
Physics 9B
Summer 2002
Session I
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1) Two converging lenses, L1 and L2 , are placed on an optical bench, 26 cm apart.
L1 has a 10 cm focal length and is placed to the left of L2 , which has a 17 cm focal
length. A 2.5 cm tall object is placed 7.5 cm “to the left” of L1 . (THE OBJECT IS
NOT BETWEEN THE TWO LENSES).
A)
B)
C)
D)
E)
F)
G)
What is the position of the image from the first lens? (4 pts.)
Is the image from part A) erect or inverted? (2 pts.)
How tall is the image from part A)? (3 pts.)
What is the object distance for lens L2 , of the image from L1 ? (2 pts)
Where is the final image? (4 pts.)
Is the final image erect or inverted, relative to the first object? (2 pts.)
How tall is the final image? (3 pts.)
1 1 1
10cm⋅ 7.5cm
+ = ⇒ s'=
= −30cm
s s' f
7.5cm− 10cm
s'
−30
b) m = − = −
= +4 ⇒ erect
s
7.5
c) 4 ⋅ 2.5cm ⇒10cm
d) s2 = 30cm+ 26cm= 56cm
17cm⋅ 56cm
e) s2 '=
= 24.4cm
56cm− 17cm
24.4cm
f) m = −
= −0.436 ⇒ inverted
56cm
g) 4.36 cm tall
a)
2) White light reflects at normal incidence from the top and bottom surface of a
glass plate (n=1.52). There is air above and below the plate. Constructive
interference is observed for light whose wavelength in air is 477.0 nm. What is the
thickness of the plate if the next longer wavelength for which there is constructive
interference is 540.6 nm? (12 pts).
There is just one half-cycle phase change upon reflection, so for constructive
1
1
interference: 2t = (m1 + )λ 1 = (m2 + m + )λ 2 . But the two different
2
2
wavelengths differ by just one m-value, m2 = m1 – 1.
1
1
λ + λ2
λ + λ2


⇒ m1 = 1
⇒  m1 + λ1 =  m1 + λ2 ⇒ m1 (λ2 − λ1 ) = 1
2
2
2
2(λ2 − λ1 )


⇒ m1 =
477.0 nm + 540.6 nm
= 8.
2(540.6nm − 477.0 nm)
1λ
17 (477.0nm)

⇒ 2t = 8 +  1 ⇒ t =
= 1334 nm.
2 n
4(1.52)

3) A ray of light is traveling in glass and strikes a glass/liquid interface. The angle
of incidence is 58O and the index of refraction of the glass is n=1.50.
A) What must the index of refraction of the liquid be in order for the angle of
refraction to equal the angle of incidence? (4 pts.)
B) What is the largest index of refraction that the liquid could have so that none
of the light is transmitted into the liquid? (4 pts.)
C) Suppose the angle of refraction is 79O, what is the index of refraction of the
liquid? (4 pts.)
D) What is the Brewster angle in question C above? (4 pts.)
a) 1.50
b) ng sinθ i = n l sin 90o ⇒ nl = 1.5sin 58o = 1.27
sin 79o
c) ng sinθ i = n l sin 79o ⇒ nl = ng
= 1.296
sin 58o
n
1.296 
d) tanθ B = g ⇒θ B = arctan
= 40.9o
 1.50 
nl
4) Natural light falls on a color filter, which passes only red light at a
wavelength, λ = 630 nm. This red light then impinges three consecutive
polarizing filters. The first filter’s axis is vertical, the second filter’s axis is
rotated 30O, counter-clockwise from vertical, and the third filter’s axis is
rotated 120O clockwise, relative to the second filter’s axis.
A) If the intensity of the red light coming from the red filter is IO, what is the
intensity of light remaining after the first polarizer? (4 pts.)
B) What fraction of IO remains after the 2nd polarizer? (4 pts.)
C) What fraction of IO remains after the 3rd polarizer? (4 pts.)
a) I1 =
Io
2
I
I  3
3
b) I 2 = I1 cos φ = o cos2 30o = o   = I o
2
2 2 
8
2
2
 1
3
c) I 3 = I 2 cos φ = I 2 cos 120 = I 2 −  =
I
 2
32 o
2
2
2
o
5) After the red light in problem 4 passes through the polarizing filters, the light
passes through a narrow slit, followed shortly thereafter by a pair of slits, which are
separated by 1.5 microns. Following the double slit, the light hits a screen 3m away.
A) At what height above the center of the screen will the 1st and 4th bright
fringes be found? (6 pts.)
B) What is the height of the 2nd dark fringe? (4 pts.)
C) If the intensity of the light at the center of the screen is I1 , what is the
intensity of the light at a height of 75 cm? (6 pts.)
D) If the second polarizing filter is removed, what will happen to the
interference pattern? (4 pts.)
There was a typo on the writing of this question. The slit spacing was supposed to be
15 microns not 1.5 microns, thereby allowing the use of the small angle formulas. The
answers below are written for 15 microns, but I accepted the use of 1.5 microns in the
grading.
a)
Rmλ
3m ⋅ 4 ⋅ 630nm
⇒ y4 =
= 50.4cm
d
15x10−6 m
3m ⋅ 630nm
⇒ y1 =
= 12.6cm
15x10−6 m
ym =
 1
3m ⋅ 1+  ⋅ 630nm
 2
b) y1d =
= 18.9cm
15x10−6 m
φ 
2πd
0.75m
where φ =
sinθ and further tanθ =
⇒θ = 0.245rad , thus
 2
λ
3m
2π ⋅15x10−6 m
φ=
sin(0.245) = 36.3rad and therefore I=I1 cos2 (18.1)=0.54*I1
630nm
c) I = I1 cos2
d) Pattern will disappear.
6) (4 pts. for each part.)
A) Light is a longitudinal electromagnetic wave, true or false?
False
B) The speed of light in vacuum is denoted by the letter c. To 1 or 2 significant
digits, what is the speed of light in vaccum?
2.99 x 108 m/s, or 3.0 x 108 m/s.
C) Light passes from air to water, i) does the frequency change? ii) does the
wavelength change? iii) as approrpriate does either the frequency or
wavelength get bigger or smaller?
i)
ii)
iii)
No
Yes
Wavelength gets shorter.
D) In fiber optic communications networks, light is sent through fibers over
great distances, which includes many changes in direction. What
phenomenon allows for most of the light to reach the end of the fiber?
Total Internal Reflection
E) In a Young’s double slit experiment, coherent, monochromatic light is
incident on the slits, and a fringe pattern is observed on a screen some
distance away. If the slits are moved closer together, what happens to the
spacing between adjacent bright fringes on the screen?
The fringes move apart.
F) If the screen in part E) is moved away from the slits, i) what happens to the
spacing between adjecent bright fringes? ii) what happens to the angular
distance between adjacent bright fringes?
i)
ii)
the spacing grows
the angular distance remains separation remains constant.