UNIVERSITY OF KWAZULU-NATAL,
PIETERMARITZBURG
SCHOOL OF CHEMISTRY AND PHYSICS
PHYS120: LAB TUTORIAL 3
OCTOBER 2014
SURNAME............................................................................ INITIALS .........................
STUDENT NUMBER .....................................................................................
PLACE A CROSS AGAINST YOUR PRAC GROUP IN THE TABLE BELOW
Monday Friday
A
B
A B
Geometrical And Physical Optics
Question 1
The wavelength of red light from a neon-helium laser in air is 633 nm. The wavelength of this
red light in the aqueous humour inside a human eyeball is 474 nm. Calculate the refractive
index of the aqueous humour, the speed of the light in it and the light’s frequency.
The refractive index of a medium is the ratio of the wavelength λ0 of light in a vacuum
and the wavelength λ in the medium. Take the wavelength of the light in air as that of the
wavelength of light in a vacuum. Therefore
λ0
n =
λ
633
=
474
= 1.34
The refractive index is also the ratio of the speed of the light c in a vacuum and the speed
v in the material.
c
n =
v
c
∴v =
n
3.00 × 108
=
1.34
= 2.24 × 108 m s−1
The frequency of the light is given by:
v = fλ
v
f =
λ
2.24 × 108
=
474 × 10−9
= 4.73 × 1014 Hz
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PHYS120 - OPTICS AND ATOMIC AND NUCLEAR PHYSICS: LAB TUTORIAL 3
Question 2
The refractive index of silicate flint glass is 1.66 for violet light (λ = 400 nm) and 1.61 for
red light (λ = 700 nm). Calculate the angle of minimum deviation for the colour that is
deviated the least, if both are incident on an equilateral triangular prism made of silicate
flint glass.
Red light is deviated the least (the refractive index for red light is smaller than that for
violet light).
Dmin +A
sin
2
n =
A
sin 2
Dmin +60◦
sin
2
1.61 =
60◦
sin 2
Dmin + 60◦
(1.61)(0.5) = sin
2
Dmin + 60◦
∴ 53.6◦ =
2
Dmin = 47.2◦
Question 3
Sketch on the diagram below, the position of the image formed by an object that is 3/4 of
the distance from the lens to the focal point f .
Page 2 of 9
PHYS120 - OPTICS AND ATOMIC AND NUCLEAR PHYSICS: LAB TUTORIAL 3
Question 4
A diverging lens has a focal length of 10.0 cm. Calculate the position and magnification of
the image if an object is placed 30.0 cm from the lens.
1 1
+
u v
1
1
+
30.0 v
1
v
∴v
m
=
1
f
1
10.0
3
1
−
−
3 × 10.0 30.0
−7.50 cm
v
−
u
(−7.50)
−
30.0
+0.250
= −
=
=
=
=
=
Since v is negative, the image is virtual.
m is positive and its absolute value is less than one, therefore the image is upright and
diminished.
Question 5
A double convex lens is made of glass with a refractive index of 1.50. One surface has twice
the radius of curvature of the other, and the focal length of the lens is 60 mm. Calculate the
two radii.
1
1
1
+
= (n − 1)
f
R1 R2
1
1
= (n − 1)
+
R1 2R1
2+1
= (n − 1)
2R1
3(n − 1)
=
2R1
3f (n − 1)
∴ R1 =
2
3(60 × 10−3 )(1.50 − 1)
=
2
= = 4.50 × 10−2 m = 45.0 mm
∴ R2 = 90.0 mm
Page 3 of 9
PHYS120 - OPTICS AND ATOMIC AND NUCLEAR PHYSICS: LAB TUTORIAL 3
Question 6
A screen is separated from a double-slit source by 1.20 m. The distance betweem the two slits
is 3.00 × 10−5 m. The wavelength of the light used for this experiment is 563 nm. Calculate
the distance between any two adjacent bright fringes.
∆y = ym+1 − ym
λL
λL
λL λL
λL
(m + 1) −
m=
m+
−
m
=
d
d
d
d
d
λL
=
d
(563 × 10−9 ) × (1.20)
=
3.00 × 10−5
= 2.25 × 10−2 m
Question 7
Newton’s rings are formed with yellow light between the curved surface of a plano-convex
lens and a flat glass plate. Calculate the wavelength of the light, given that the radius of the
10th dark ring from the centre is 0.250 cm and the radius of curvature of the lens is 1.00 m.
rn2 = Rnλ, n = 0, 1, 2, . . .
rn2
∴λ =
Rn
(0.250 × 10−2 )2
=
1.00 × 10
= 6.25 × 10−7 m = 625 nm
Question 8
Light of wavelength 5.80 × 102 nm is incident on a slit of width 0.300 mm, and an observing
screen is placed 2.00 m from the slit. Find y1 , the distance from the centre of the central
bright fringe and the first dark fringe.
λ
sin θn = n
a
y1
tan θ1 =
L
y1
sin θ1 ≈ tan θ1 =
(for small θ)
L
λ
∴ y1 = L
a
580 × 10−9
= (2.00)
0.300 × 10−3
= 3.87 × 10−3 m
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PHYS120 - OPTICS AND ATOMIC AND NUCLEAR PHYSICS: LAB TUTORIAL 3
Question 9
A diffraction grating has 3570 slits per cm. What is the longest wavelength for which a
fourth-order diffraction beam can be observed when the grating is illuminated normally?
What will the colour of the diffraction beam be?
The longest wavelength will occur when θ = θmax = 90◦ :
1
1
=
N
3570
2.80 × 10−4 cm
nλmax
(2.80 × 10−4 ) × sin 90◦
4
−7
7.00 × 10 m = 700 nm
d =
d sin θmax
=
=
∴ λmax =
=
The colour of the diffracted beam is red.
Question 10
Sunlight reflected off a glass sheet is totally polarized. Determine the angle of incidence of
the light on the glass sheet, taking the refractive index of the glass as 1.51.
tan θB = n
tan θB = 1.51
∴ θB = 56.5◦
Page 5 of 9
PHYS120 - OPTICS AND ATOMIC AND NUCLEAR PHYSICS: LAB TUTORIAL 3
Atomic And Nuclear Physics
Question 11
Complete the following reaction equations:
6
Li + 21.H. → 74Be + 10n
240
236
4
(b) 94 Pu → 92 U + 2 He
....
(a) 3
Question 12
Show that one atomic mass unit is equivalent to approximately 930 MeV.
1 atom of
12
6 C
carbon has mass
:
∴ 1u =
For 1 u, E =
=
=
=
12
= 12 u
6.02 × 1023
1
= 1.66 × 10−27 kg
6.02 × 1023
mc2
1.66 × 10−27 × (3.00 × 108 )2
1.49 × 10−10 J
1.49 × 10−10
≈ 930 MeV
1.60 × 10−19
Question 13
How many atoms are there in 1 kg of silver, if its atomic mass is 108 u?
108 g of silver contains
1000 g of silver contains
6.02 × 1023 atoms
1000
× 6.02 × 1023 = 5.57 × 1024 atoms
108
Question 14
11
Ordinary boron is a mixture of the 10
5 B and 5 B isotopes and has a composite atomic mass
of 10.82 u. Find the percentage of each isotope present (by mass) in ordinary boron. The
11
atomic masses of 10
5 B and 5 B are 10.01 u and 11.01 u respectively.
Suppose x is the fraction of
10
5 B
present in the mixture. Then
10.01x + 11.01(1 − x) = 10.82
∴ x = 0.19 = 19%
Therefore ordinary boron contains 19%
10
5 B
and 81%
Page 6 of 9
11
5 B.
PHYS120 - OPTICS AND ATOMIC AND NUCLEAR PHYSICS: LAB TUTORIAL 3
Question 15
For
235
92 U
calculate:
(a) the mass defect (in u),
(b) the binding energy (in MeV),
(c) the binding energy per nucleon.
(a) The mass defect is
mass of component nucleons is =
=
=
∴ mass defect =
=
92mp + 143mn + 92me
(92 × 1.0073) + (143 × 1.0087) + (92 × 0.0005)
236.9617 u
236.197 − 235.0439
1.9178 u
(b) The binding energy (in MeV) is
binding energy = 1.9178 × 930
= 1783.55 MeV
(c) The binding energy per nucleon is
1783.55
235
= 7.59 MeV/nucleon
binding energy per nucleon =
Question 16
Suppose enriched uranium fuel containing 1.70% of the fissionable isotope 235
92 U is used as fuel
5
for a ship. The water exerts an average frictional drag of 1.0 × 10 N on the ship. Assuming
an energy release of 180 MeV per fission and that the ships engine has an efficiency of 20%,
calculate how far the ship could travel on 1.0 kg of fuel.
The mass of fissionable 235
92 U is 1.70% of 1.0 kg, which is 17 g.
235
17 g of 92 U corresponds to
17 × 6.02 × 1023
= 4.355 × 1022 atoms
235
The energy produced by this many fissions is
4.355 × 1022 × 180 = 7.839 × 1024 MeV
= 1.254 × 1012 J
20% of this energy is used to overcome the frictional drag and, using W = F s, the distance
obtained is
W
0.2 × 1.254 × 1012
s=
=
= 2.5 × 106 m
F
1.0 × 105
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PHYS120 - OPTICS AND ATOMIC AND NUCLEAR PHYSICS: LAB TUTORIAL 3
Question 17
If the radius and energy associated with the first Bohr orbit of the hydrogen atom are
0.5 × 10−10 m and −13.6 MeV respectively, what are the corresponding values for the third
orbit?
h2 ε0
· n2
πme2
h2 ε0
· (1)2
2
πme
0.5 Å
h2 ε0
· (3)2
2
πme
(3)2 × r1
4.5 Å
me4
1
− 2 2· 2
8ε0 h n
1
me4
− 2 2·
8ε0 h (1)2
−13.6 MeV
me4
1
− 2 2·
8ε0 h (3)2
1
× E1
(3)2
−1.51 MeV
rn =
r1 =
=
r3 =
∴ r3 =
=
En =
E1 =
=
E3 =
∴ E3 =
=
Question 18
What energy is carried by a photon, emitted from an He-Ne laser, of wavelength 632.8 nm?
E = hf = h
c
λ
= 6.63 × 10−34 ×
3.00 × 108
632.8 × 10−9
= 3.14 × 10−19 J
Page 8 of 9
PHYS120 - OPTICS AND ATOMIC AND NUCLEAR PHYSICS: LAB TUTORIAL 3
Question 19
X-rays of wavelength 2.0×10−10 m are directed at a certain crystal and produce second order
Bragg diffraction for an incident angle of 60◦ . Calculate the spacing of the planes responsible
for this diffraction pattern.
The Bragg condition for enhancement is
nλ = 2d sin θ
where θ is the angle between the incident X-ray and the atomic plane:
θ = 90◦ − 60◦ = 30◦
Hence
nλ
2 sin 30◦
2 × 2.0 × 10−10
=
2 sin 30◦
= 4.0 × 10−10 m = 4.0 Å
d =
Question 20
A line in the X-ray spectrum of gold has wavelength 18.5 pm (18.5 × 10−12 m). The emitted
X-ray photons correspond to a transition of the gold atom between two states, the upper
one of which has energy -13.7 keV. What is the energy of the lower state?
The energy of a photon of wavelength 18.5 × 10−12 m is
E = hf = h
c
λ
= 6.63 × 10−34 ×
3.00 × 108
18.5 × 10−12
= 1.075 × 10−14 J
1.075 × 10−14
× 10−3 keV
=
1.60 × 10−19
= 67.2 keV
The lower state therefore has energy
−13.7 − 67.2 = −80.9 keV
Page 9 of 9