### Optics - Diffraction Grating

Consider a single-slit diffraction pattern for a slit of width d. It is observed that for a light of wavelength 400 nanometers, the angle between the first minimum and the central maximum is 4 × 10−3 radians. The value of d is

A. 1 × 10−5 m
B. 5 × 10−5  m
C. 1 × 10−4  m
D. 2 × 10−4 m
E. 1 × 10−3 m
(GR9677 #57)
Solution:

Single slit diffraction: d sin θ =

λ = 400 nm = 4 × 10−7 m
θ = 4 × 10−3 radians
sin θ ≈ θ for small θ

=
d =  / θ = 1 × (4 × 10−7) / (4 × 10−3) radians = 10−4 m

### Optics - Telescope

A collimated laser beam emerging from a commercial HeNe laser has a diameter of about 1 millimeter. In order to convert this beam into a well-collimated beam of diameter 10 millimeter, two convex lenses are to be used. The first lens if of focal length 1.5 centimeters and is to be mounted at the output of the laser. What is the focal length, f of the second lens and how far from the first lens should it be placed?

A. f  = 4.5 cm, Distance = 6.0 cm
B. f  = 10 cm, Distance = 10 cm
C. f  = 10 cm, Distance = 11.5 cm
D. f  = 15 cm, Distance = 15 cm
E. f  = 15 cm, Distance = 16.5 cm
(GR9677 #58)
Solution:

Since two convex lenses are used, we can treat this system as a telescope.

Magnification, M =  fobject /  feye = 10
fobject = 10 feye  = 10 × 1.5 cm = 15 cm

D =  fobject +  feye  = 10 + 1.5 = 16.5 cm

### Quantum Mechanics - Laser

The approximate number of photons in a femtosecond (10−15 s) pulse of 600 nanometers wavelength light from a 10-kilowatt peak-power dye laser is

A. 103
B. 107
C. 1011
D. 1015
E. 1018
(GR9677 #59)
Solution:

Number of photon, n = Etotal / Esingle

Energy of a single photon,  Esingle hc/λ

= 6.63 × 10−34 Js
= 3 × 108 m/s
λ  =  600 nanometers = 600 × 10−9 m

→ Esingle = 3 × 10−19  J

From Power, P = W/tEtotal/t

→ Total energy, Etotal =  Pt

= 10 kW = 104 W
t = 10−15 s

Etotal = 10−11 J

n = Etotal / Esingle = 10−11  / (3 × 10−19)  = 0.3 × 108 = 3 × 10 J

### Special Relativity - Doppler Effect

The Lyman alpha spectral line of Hydrogen (λ = 122 nanometers) differs by 1.8 × 10−12 meter in spectra taken at opposite ends of the Sun’s equator. What is the speed of a particle on the equator due to the Sun’s rotation, in kilometers per second?

A. 0.22
B. 2.2
C. 22
D. 220
E. 2200
(GR9677 #60)
Solution:

From the answers provided, we can see that the problem deals with v ≪ c since the biggest velocity (answer E) is 2200 km/s = 2.2 × 106 m/s.

→ Redshift parameter, z = Δλ / λ v / c  for v ≪ c

=  cΔλ / λ

Δλ 1.8 × 10−12 m
λ 122 nanometers = 1.22 × 10−7 m
c × 108  m/s

→ = (5.4/1.22) × 10m/s  = (5.4/1.22) km/s ≈ 2.2 km/s

### Electromagnetism - Gauss’ Law, Electric Field

A sphere of radius R carries charge density proportional to the square of the distance from the center: ρ = Ar2, where A is a positive constant. At a distance of R/2 from the center, the magnitude of the electric field is:

A. A/4πɛ0
B. AR3/40ɛ0
C. AR3/24ɛ0
D. AR3/5ɛ0
E. AR3/3ɛ0
(GR9677 #61)
Solution:

### Electromagnetism - Capacitor

Two capacitors of capacitance 1.0 microfarad and 2.0 microfarad are each charged by being connected across a 5.0-Volt battery. They are disconnected from the battery and then connected to each other with resistive wires so that plates of opposite charge are connected together. What will be the magnitude of the final voltage across the 2.0 microfarad capacitor?

A. 0 V
B. 0.6 V
C. 1.7 V
D. 3.3 V
E. 5.0 V
(GR9677 #62)
Solution:

The charges stored at each capacitor:

Q1 = C× Vt = 1 × 5 = 5 μF
Q2 = C× Vt = 2 × 5 = μF

The plates of opposite charge are connected together (series connection):

Q1
=  Q2' = Qt
1/C = 1/C +  1/C= 1  +  1/2 = 3/2
C= 2/3
Since  Vt = 5
Q2' = Qt  =  C× Vt = 2/3 × 5 = 10/3

→ V2 = Q2' / C= 10/3 × 1/2 =  10/6 = 1.66 V

### Nuclear & Particle Physics - Standard Model

According to the Standard Model of elementary particles, which of the following is NOT a composite object?

A. Muon
B. Pi-meson
C. Neutron
D. Deuteron
E. Alpha particle
(GR9677 #63)
Solution:

A.TRUE
Muon is a lepton. Leptons, along with quarks, are considered the fundamental particles.

B. FALSE
Pi-Meson consists of a quark and its antiparticle. Moreover, a pi-meson is a hadron. Hadrons interact with the strong-force, and all of them are composed of combinations of quarks. (The fundamental particles are classified as quarks and leptons.)

C. FALSE
A neutron is made up of 3 quarks.

D. FALSE
A deuteron consists of a proton and a neutron. (tritium is two neutrons and a proton, while regular Hydrogen is just an electron and proton).

E. FALSE
An alpha particle consists of electrons and protons and neutrons.

### Nuclear & Particle Physics - Binding Energy

The binding energy of a heavy nucleus is about 7 million electron volts per nucleon, whereas the binding energy of a medium-weight nucleus is about 8 million electron volts per nucleon. Therefore, the total kinetic energy liberated when a heavy nucleus undergoes symmetric fission is most nearly

A. 1876 MeV
B. 938 MeV
C. 200 MeV
D. 8 MeV
E. 7 MeV
(GR9677 #64)
Solution:

KEEf   Ei

Symmetric fission: the splitting of the nucleus into two fragments of approximately equal mass.

AX  →  A1Y  + A2Z
with A1 =  A= A/2 and Y = Z  (for symmetric fission)

A Ei  → A1 Ef 1 + AEf = 2(A/2) Ef  = A Ef

KE = A Ef  − A E = A (8  − 7) MeV/nucleon = A MeV/nucleon

For heavy nucleus, A ≈ 200 nucleons. Example: 238U → A ≈ 238 nucleons

→ KE ≈ 200  MeV

### Classical Mechanics - Conservation of Momentum

A man of mass m on an initially stationary boat gets off the boat by leaping to the left in an exactly horizontal direction. Immediately after the leap, the boat of mass M is observed to be moving to the right at speed v. How much work did the man do during the leap (both on his own body and on the boat)?

A. ½ Mv²
B. ½ mv²
C. ½ (m)v²
D. ½ (M²/m)v²
E. ½ [Mm/(m)]v²
(GR9677 #65)
Solution:

Conservation of momentum:

$mv_0=Mv&space;\rightarrow&space;v_0=\frac{M}{m}v$

The man does work on both himself and the boat:

\begin{aligned}&space;\\W&=KE_{\text{total}}&space;\\&=\frac{1}{2}mv_0^2+\frac{1}{2}Mv^2&space;\\&=\frac{1}{2}m\left&space;(&space;\frac{M}{m}v&space;\right&space;)^2+\frac{1}{2}Mv^2&space;\\&=\frac{1}{2}\left&space;(&space;\frac{M^2}{m}+M&space;\right&space;)v^2&space;\end{aligned}

### Classical Mechanics - Orbital Path

When it is about the same distance from the Sun as is Jupiter, a spacecraft on a mission to the outer planets has a speed that is 1.5 times the speed of Jupiter in its orbit. Which of the following describes the orbit of the spacecraft about the Sun?

A. Spiral
B. Circle
C. Ellipse
D. Parabola
E. Hyperbola
(GR9677 #66)
Solution:

The mission is to outer planets, so the path should not be bounded (no longer a close path): A, B, C are FALSE.

If vescape = vcircular → Parabola
If ve $>$ √2 v→ Hyperbola
Since ve = 1.5 $>$ √2 → Hyperbola