Perfect Circular Objective - Intermediate Physics - Exam, Exams for Physics. Baddi University of Emerging Sciences and Technologies

Physics

Description: This is the Exam of Intermediate Physics which includes Width of Central Maximum, Intensity Distribution, Horizontal Position Axis, Orders of Diffraction, Fabry-Perot Interferometer, Beam of Light, Transmission Directions etc. Key important points are: Perfect Circular Objective, Earth’s Atmosphere, Chromatic Resolving Power, Transparent Model, Supporting Beam, Spacing of Lines in Grating, Second Order Light, Fabry-Perot Interferometer
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9322
THE UNIVERSITY OF SYDNEY
FACULTY OF SCIENCE
INTERMEDIATE PHYSICS
PHYS 2011 PHYSICS 2A
JUNE 2008 TIME ALLOWED: 2 HOURS
ALL QUESTIONS HAVE THE VALUE SHOWN
INSTRUCTIONS:
This paper consists of 2 sections.
Section A Optics 50 marks
Section B Nuclear Physics 40 marks
Candidates should attempt all questions.
USE A SEPARATE ANSWER BOOK FOR EACH SECTION.
In answering the questions in this paper, it is particularly important to give rea-
sons for your answer. Only partial marks will be awarded for correct answers
with inadequate reasons.
No written material of any kind may be taken into the examination room. Calcu-
lators are permitted.
9322 Semester 1 2008 Page 2 of 10
Table of constants
Avogadro’s number NA=6.022 ×1023 mole1
speed of light c=2.998 ×108m.s1
electronic charge e=1.602 ×1019 C
electron rest mass me=9.110 ×1031 kg
electron rest energy mec2= 511 keV
electron volt 1 eV = 1.602 ×1019 J
proton rest mass mp=1.673 ×1027 kg
neutron rest mass mn=1.675 ×1027 kg
Planck’s constant h=6.626 ×1034 J.s
Planck’s constant (reduced) ¯h=1.055 ×1034 J.s
Boltzmann’s constant kB=1.380 ×1023 J.K1
Universal gas constant R=8.315 J.mol1K1
Stefan’s constant σ=5.670 ×108W.m2.K4
permittivity of free space ε0=8.854 ×1012 C2.N1.m2
gravitational constant G=6.673 ×1011 N.m2.kg2
atomic mass constant u=1.660 ×1027 kg
degrees/radian 180'57.2958
9322 Semester 1 2008 Page 3 of 10
SECTION A
OPTICS
FORMULAS
n=c/v v =fλ
nasin θa=nbsin θb
Iav =1
20cE2
max
dsin θ=
dsin θ= (m+1
2)λ
E2
tot =E2
1+E2
2+ 2E1E2cos(φ2φ1)
I=I0cos2 πd sin θ
λ!
β=2πa sin θ
λ
I=I0
sin(β/2)
(β/2)
2
sin θ=
a
φ=2πd sin θ
λ
I=I0cos2(φ/2)
sin(β/2)
(β/2)
2
`=λ2
λ
I=I0
sin(β/2)
(β/2)
2
sin(Nφ/2)
sin(φ/2)
2
sin θsin θi=
dsin θ=
d
R=λ
λ
R=mN
9322 Semester 1 2008 Page 4 of 10
Itrans =Iincid cos2φ
tan θp=nb
na
θ=1.22λ
D
φ= 2π 2n2t
λvac !+φ12 +φ23
φ= 2π 2n2tcos θ2
λvac !+φ12 +φ23
It
Ii
=1
1 + Fsin2(δ/2) where δ= 2π 2tcos θ
λ!
F=4R
(1 − R)2
F=πF
2
2tcos θ=
R=mπ
2F
λF SR =λ
m'λ2
2t
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