BE Computer Engineering (Pokhara University) Applied Physics (PU, PHY 110) Question Paper 2079 Nepal

(a) Damped harmonic oscillator (9 marks)

A particle of mass $m$ experiences a restoring force $-kx$ and a velocity-dependent damping force $-b\dfrac{dx}{dt}$. By Newton's second law:

$$m\frac{d^2x}{dt^2} = -kx - b\frac{dx}{dt}$$ $$\frac{d^2x}{dt^2} + 2\beta\frac{dx}{dt} + \omega_0^2 x = 0, \qquad 2\beta=\frac{b}{m},\;\; \omega_0^2=\frac{k}{m}$$

Trying $x = e^{\alpha t}$ gives the auxiliary equation $\alpha^2 + 2\beta\alpha + \omega_0^2 = 0$, so

$$\alpha = -\beta \pm \sqrt{\beta^2 - \omega_0^2}$$

Under-damped case ($\beta < \omega_0$): the root is complex, $\alpha=-\beta\pm i\omega_d$ with $\omega_d=\sqrt{\omega_0^2-\beta^2}$. The solution is an exponentially decaying oscillation:

$$\boxed{x(t) = A_0\,e^{-\beta t}\cos(\omega_d t + \phi)}$$

The amplitude $A_0 e^{-\beta t}$ decays slowly while the system oscillates at the reduced frequency $\omega_d$.

Three cases

• Under-damping ($\beta<\omega_0$): oscillatory motion with amplitude falling exponentially; the displacement crosses zero many times. Sketch: a cosine curve bounded by a decaying $\pm A_0e^{-\beta t}$ envelope.
• Critical damping ($\beta=\omega_0$): equal roots; $x=(A+Bt)e^{-\beta t}$. The system returns to equilibrium in the shortest time without oscillating. Sketch: a single smooth curve rising/falling to zero with no overshoot.
• Over-damping ($\beta>\omega_0$): two real negative roots; motion is a slow non-oscillatory return to equilibrium, slower than critical. Sketch: a sluggish exponential decay to zero, lying above the critical curve.

(b) Numerical (5 marks)

Given $m=0.2\,\text{kg}$, $k=80\,\text{N/m}$, $b=0.16\,\text{kg/s}$.

$$\omega_0^2=\frac{k}{m}=\frac{80}{0.2}=400\,\text{s}^{-2}, \qquad \beta=\frac{b}{2m}=\frac{0.16}{0.4}=0.4\,\text{s}^{-1}$$

Damped angular frequency:

$$\omega_d=\sqrt{\omega_0^2-\beta^2}=\sqrt{400-0.16}=\sqrt{399.84}\approx 20.0\,\text{rad/s}$$

Time for amplitude to halve: amplitude $\propto e^{-\beta t}$, so $e^{-\beta t}=\tfrac12$:

$$t_{1/2}=\frac{\ln 2}{\beta}=\frac{0.693}{0.4}\approx 1.73\,\text{s}$$

(a) Maxwell's equations and displacement current (8 marks)

In differential form, in a medium of permittivity $\varepsilon$ and permeability $\mu$:

Displacement current. The original Ampère's law $\nabla\times\mathbf{B}=\mu_0\mathbf{J}$ gives $\nabla\cdot(\nabla\times\mathbf{B})=\mu_0\nabla\cdot\mathbf{J}=0$, but the continuity equation requires $\nabla\cdot\mathbf{J}=-\partial\rho/\partial t$, which is non-zero when charge accumulates (e.g. between capacitor plates while charging). Maxwell removed the inconsistency by adding a term. Using $\rho=\varepsilon_0\nabla\cdot\mathbf{E}$,

$$\nabla\cdot\mathbf{J} + \frac{\partial}{\partial t}(\varepsilon_0\nabla\cdot\mathbf{E})=0 \;\Rightarrow\; \nabla\cdot\!\left(\mathbf{J}+\varepsilon_0\frac{\partial\mathbf{E}}{\partial t}\right)=0$$

The extra term $\mathbf{J}_d=\varepsilon_0\dfrac{\partial\mathbf{E}}{\partial t}$ is the displacement current density, giving the corrected law $\nabla\times\mathbf{B}=\mu_0(\mathbf{J}+\mathbf{J}_d)$.

(b) Wave equation and speed of EM waves (6 marks)

In free space $\rho=0,\ \mathbf{J}=0$. Take the curl of Faraday's law:

$$\nabla\times(\nabla\times\mathbf{E})=-\frac{\partial}{\partial t}(\nabla\times\mathbf{B})=-\mu_0\varepsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}$$

Using the identity $\nabla\times(\nabla\times\mathbf{E})=\nabla(\nabla\cdot\mathbf{E})-\nabla^2\mathbf{E}$ with $\nabla\cdot\mathbf{E}=0$:

$$\boxed{\nabla^2\mathbf{E}=\mu_0\varepsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}}$$

Comparing with the standard wave equation $\nabla^2\mathbf{E}=\dfrac{1}{c^2}\dfrac{\partial^2\mathbf{E}}{\partial t^2}$ gives the speed of EM waves in vacuum:

$$c=\frac{1}{\sqrt{\mu_0\varepsilon_0}}=\frac{1}{\sqrt{(4\pi\times10^{-7})(8.85\times10^{-12})}}\approx 3\times10^{8}\,\text{m/s}$$

which matches the measured speed of light, establishing that light is an electromagnetic wave.

(a) Time-independent Schrödinger equation (6 marks)

A particle of mass $m$ is described by a wave $\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$. Starting from the classical plane wave $\Psi=Ae^{i(kx-\omega t)}$ and using the de Broglie relations $p=\hbar k$ and $E=\hbar\omega$:

$$\frac{\partial^2\Psi}{\partial x^2}=-k^2\Psi=-\frac{p^2}{\hbar^2}\Psi, \qquad i\hbar\frac{\partial\Psi}{\partial t}=E\Psi$$

Total energy $E=\dfrac{p^2}{2m}+V$, hence $p^2=2m(E-V)$. Substituting:

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V\psi=E\psi \;\Rightarrow\; \boxed{\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}(E-V)\psi=0}$$

This is the time-independent Schrödinger equation (3-D form: $\nabla^2\psi+\tfrac{2m}{\hbar^2}(E-V)\psi=0$).

(b) Particle in a 1-D infinite well of width $L$ (6 marks)

Inside the box $V=0$ ($0<x<L$); outside $V=\infty$ so $\psi=0$. The equation becomes

$$\frac{d^2\psi}{dx^2}+k^2\psi=0,\qquad k^2=\frac{2mE}{\hbar^2}$$

General solution $\psi=A\sin kx + B\cos kx$. Boundary condition $\psi(0)=0\Rightarrow B=0$. Condition $\psi(L)=0\Rightarrow \sin kL=0\Rightarrow kL=n\pi,\ n=1,2,3,\dots$

Quantized energy:

$$E_n=\frac{\hbar^2 k^2}{2m}=\boxed{\frac{n^2\pi^2\hbar^2}{2mL^2}=\frac{n^2 h^2}{8mL^2}}$$

Energy is discrete (quantized) and the lowest state ($n=1$) has non-zero zero-point energy.

Normalization: $\displaystyle\int_0^L A^2\sin^2\!\frac{n\pi x}{L}\,dx=1\Rightarrow A=\sqrt{\tfrac{2}{L}}$, so

$$\psi_n(x)=\sqrt{\frac{2}{L}}\,\sin\!\frac{n\pi x}{L}$$

Sketch of first three eigenstates: $\psi_1$ is a single half-sine (one antinode, no node inside); $\psi_2$ is a full sine (one node at $x=L/2$); $\psi_3$ has two interior nodes. All vanish at $x=0$ and $x=L$.

(a) Light propagation, acceptance angle and numerical aperture (8 marks)

An optical fibre has a high-index core ($n_1$) surrounded by a lower-index cladding ($n_2<n_1$). A ray entering the core strikes the core–cladding boundary; if the angle of incidence there exceeds the critical angle $\theta_c$ (where $\sin\theta_c=n_2/n_1$), the ray undergoes total internal reflection and is guided down the fibre with negligible loss, repeatedly reflecting along its length.

Acceptance angle ($\theta_a$): the maximum half-angle of the entrance cone for which an incident ray is still totally internally reflected inside the core. Rays within this cone are guided; rays outside it escape into the cladding.

Numerical aperture (NA): the light-gathering ability of the fibre, $NA=n_0\sin\theta_a$ (with $n_0$ the index of the medium outside).

Derivation. Let a ray enter from a medium of index $n_0$ at angle $\theta_a$, refracting to angle $\theta_r$ inside the core: $n_0\sin\theta_a=n_1\sin\theta_r$. To be guided it must hit the wall at exactly the critical angle, so the angle with the wall normal is $90^\circ-\theta_r$ and

$$\sin(90^\circ-\theta_r)=\cos\theta_r=\frac{n_2}{n_1}\;\Rightarrow\;\sin\theta_r=\sqrt{1-\frac{n_2^2}{n_1^2}}$$

Therefore

$$n_0\sin\theta_a=n_1\sin\theta_r=n_1\sqrt{1-\frac{n_2^2}{n_1^2}}=\sqrt{n_1^2-n_2^2}$$ $$\boxed{NA=n_0\sin\theta_a=\sqrt{n_1^2-n_2^2}}\quad(\text{in air, }n_0=1)$$

(b) Numerical (4 marks)

Given $n_1=1.50,\ n_2=1.46$, fibre in air ($n_0=1$).

$$NA=\sqrt{1.50^2-1.46^2}=\sqrt{2.25-2.1316}=\sqrt{0.1184}\approx 0.344$$ $$\theta_a=\sin^{-1}(0.344)\approx 20.1^\circ$$

Numerical aperture $\approx 0.344$; acceptance angle $\approx 20.1^\circ$.

Newton's rings in reflected light

When a plano-convex lens of large radius $R$ rests on a flat glass plate, a thin air film of gradually increasing thickness forms between them. Monochromatic light incident from above is partly reflected at the top and bottom of the air film; these two beams interfere. Because of the circular symmetry of the film, loci of constant thickness are circles, so a pattern of alternately bright and dark concentric rings with a dark central spot is seen in reflected light.

For near-normal incidence the path difference between the two reflected rays is $2\mu t$, plus an extra $\lambda/2$ from the phase change on reflection at the glass surface (rarer-to-denser). For an air film $\mu=1$:

• Dark ring: $2t=n\lambda$
• Bright ring: $2t=(2n-1)\tfrac{\lambda}{2}$

Diameter of the $n$-th dark ring

From the geometry of the lens, the film thickness at radius $r$ is $t=\dfrac{r^2}{2R}$. For the $n$-th dark ring $2t=n\lambda$:

$$2\cdot\frac{r_n^2}{2R}=n\lambda \;\Rightarrow\; r_n^2=nR\lambda$$

With diameter $D_n=2r_n$:

$$\boxed{D_n^2=4nR\lambda}$$

The diameter is proportional to $\sqrt{n}$, so rings get closer together outward.

Determination of wavelength

Measure the diameters of the $n$-th and $m$-th dark rings. Then

$$D_n^2=4nR\lambda,\quad D_m^2=4mR\lambda \;\Rightarrow\; D_m^2-D_n^2=4(m-n)R\lambda$$ $$\boxed{\lambda=\frac{D_m^2-D_n^2}{4(m-n)R}}$$

This subtraction eliminates errors from imperfect contact at the centre. Knowing $R$ (from a spherometer) and the measured diameters, $\lambda$ is found.

Plane diffraction grating

A plane diffraction grating is an optical component consisting of a large number of equally spaced, identical, parallel slits (rulings) on a flat transparent surface. If the slit width is $a$ and the opaque space between slits is $b$, the grating element is $d=a+b$, equal to the reciprocal of the number of lines per unit length.

Grating equation (principal maxima)

When light of wavelength $\lambda$ falls normally on the grating, the secondary wavelets from corresponding points of adjacent slits travel an extra path $d\sin\theta$ in a direction making angle $\theta$ with the normal. Constructive interference (a principal maximum) occurs when this path difference is a whole number of wavelengths:

$$\boxed{d\sin\theta = n\lambda}, \qquad n=0,1,2,\dots$$

where $n$ is the order of the spectrum.

Numerical

Grating has $N=5000$ lines/cm, so

$$d=\frac{1}{5000}\,\text{cm}=2\times10^{-4}\,\text{cm}=2\times10^{-6}\,\text{m}$$

For $n=2$, $\lambda=589\,\text{nm}=589\times10^{-9}\,\text{m}$:

$$\sin\theta=\frac{n\lambda}{d}=\frac{2\times589\times10^{-9}}{2\times10^{-6}}=0.589$$ $$\theta=\sin^{-1}(0.589)\approx 36.1^\circ$$

Angle of diffraction for the second-order maximum $\approx 36.1^\circ$.

(a) Brewster's law and polarizing angle (4 marks)

Brewster's law: When unpolarized light is incident on a transparent dielectric, there is a particular angle of incidence, the polarizing (Brewster) angle $\theta_p$, at which the reflected light is completely plane-polarized (vibrations perpendicular to the plane of incidence). At this angle the refractive index equals the tangent of the polarizing angle:

$$\boxed{\mu=\tan\theta_p}$$

At $\theta_p$ the reflected and refracted rays are mutually perpendicular ($\theta_p+\theta_r=90^\circ$). Proof: by Snell's law $\mu=\dfrac{\sin\theta_p}{\sin\theta_r}=\dfrac{\sin\theta_p}{\sin(90^\circ-\theta_p)}=\dfrac{\sin\theta_p}{\cos\theta_p}=\tan\theta_p$. The polarizing angle is thus the angle of incidence for which the reflected beam is fully polarized.

(b) Quarter-wave vs half-wave plate (3 marks)

Emission processes and population inversion

Spontaneous emission: An atom in an excited state $E_2$ decays on its own to a lower state $E_1$, emitting a photon of energy $h\nu=E_2-E_1$. The photons are emitted in random directions, with random phase — the light is incoherent (ordinary light).

Stimulated emission: An incident photon of energy $h\nu=E_2-E_1$ induces an excited atom to drop to $E_1$, emitting a second photon identical in frequency, phase, direction and polarization to the incident one. This produces coherent light amplification — the basis of laser action.

Population inversion: A non-equilibrium condition in which the number of atoms in the higher energy state exceeds that in the lower state ($N_2>N_1$). In thermal equilibrium the Boltzmann distribution gives $N_2<N_1$, so inversion must be created artificially.

Why population inversion is essential

Stimulated emission (gain) competes with absorption. The net rate of stimulated emission $\propto (N_2-N_1)$, so a photon beam is amplified only if $N_2>N_1$; otherwise absorption dominates and there is no light amplification. Hence population inversion is a necessary condition for laser action.

Three-level system

Atoms are pumped (optically) from the ground state $E_1$ to a high state $E_3$. From $E_3$ they decay rapidly and non-radiatively to a metastable state $E_2$ which has a long lifetime, so atoms accumulate there. When the population of $E_2$ exceeds that of the ground state $E_1$, inversion between $E_2$ and $E_1$ is achieved, and lasing occurs on the $E_2\to E_1$ transition (e.g. the ruby laser). A drawback is that the lower laser level is the heavily populated ground state, so very strong pumping is required.

Formation of the depletion region

When p-type and n-type semiconductors are joined, the large concentration gradient causes majority electrons to diffuse from n to p and holes from p to n. Near the junction electrons fill holes; the n-side is left with fixed positive donor ions and the p-side with fixed negative acceptor ions. This region, depleted of mobile charge carriers, is the depletion region. The exposed immobile ions set up an internal electric field (and a built-in potential barrier $V_0$, about 0.7 V for Si, 0.3 V for Ge) directed from n to p, which opposes further diffusion. At equilibrium the diffusion current is exactly balanced by the drift current and no net current flows.

In the band picture, the Fermi levels of the two sides align; the conduction and valence bands of the p-side bend upward relative to the n-side, the offset $eV_0$ being the barrier height.

Forward bias

The p-side is connected to the positive terminal and the n-side to the negative terminal. The external field opposes the built-in field, reducing the barrier height to $e(V_0-V)$ and narrowing the depletion region. The bands shift so the offset decreases. Majority carriers cross the junction easily, and a large current flows that rises steeply once $V$ exceeds the knee voltage.

Reverse bias

The p-side is made negative and the n-side positive. The external field aids the built-in field, increasing the barrier to $e(V_0+V)$ and widening the depletion region. The bands bend further apart. Majority-carrier flow is blocked; only a tiny reverse saturation current due to thermally generated minority carriers flows, nearly independent of voltage until breakdown.

Hall effect

When a current-carrying conductor (or semiconductor) is placed in a magnetic field perpendicular to the current, a transverse potential difference (the Hall voltage $V_H$) develops across the specimen, perpendicular to both the current and the field. This is the Hall effect, and it arises from the magnetic Lorentz force on the moving charge carriers.

Derivation of the Hall coefficient

Let current $I$ flow along $x$ in a slab of width $w$ and thickness $t$, with magnetic field $B$ along $z$. Carriers of charge $q$ drift with velocity $v_d$. The Lorentz force $qv_dB$ pushes them sideways (along $y$) until the resulting transverse electric field $E_H$ balances it:

$$qE_H=qv_dB \;\Rightarrow\; E_H=v_dB$$

The current density is $J=nqv_d$, where $n$ is the carrier concentration, so $v_d=\dfrac{J}{nq}$ and

$$E_H=\frac{JB}{nq}$$

The Hall coefficient is defined as $R_H=\dfrac{E_H}{JB}$:

$$\boxed{R_H=\frac{1}{nq}}$$

In terms of measurable quantities, $V_H=E_H w$ and $J=\dfrac{I}{wt}$, giving $V_H=\dfrac{IB}{nqt}$ and $R_H=\dfrac{V_H t}{IB}$. The sign of $R_H$ is negative for n-type (electrons) and positive for p-type (holes).

Applications

1. Determination of carrier type and concentration — the sign of $R_H$ gives the majority-carrier type and its magnitude gives $n=1/(R_H q)$.
2. Measurement of carrier mobility — combining $R_H$ with the conductivity $\sigma$ gives the Hall mobility $\mu_H=R_H\sigma$. (Also used in Hall-effect magnetic-field sensors/probes.)

Polarization mechanisms in dielectrics

When a dielectric is placed in an electric field, bound charges shift slightly, giving each unit volume a dipole moment (polarization $P$). The main mechanisms are:

1. Electronic polarization: displacement of the electron cloud relative to the nucleus of each atom. Present in all dielectrics, very fast (up to optical frequencies).
2. Ionic (atomic) polarization: in ionic solids, positive and negative ions displace in opposite directions, changing inter-ionic distances.
3. Orientational (dipolar) polarization: in polar molecules (with permanent dipoles, e.g. water), the field tends to align the dipoles against thermal randomization; strongly temperature-dependent.
4. Space-charge (interfacial) polarization: accumulation of mobile charges at grain boundaries/interfaces; significant only at low frequencies.

Clausius–Mossotti relation

For a non-polar dielectric, the local field acting on an atom (Lorentz field) is

$$E_{loc}=E+\frac{P}{3\varepsilon_0}$$

The induced dipole moment per atom is $p=\alpha E_{loc}$, and with $N$ atoms per unit volume the polarization is

$$P=N\alpha E_{loc}=N\alpha\left(E+\frac{P}{3\varepsilon_0}\right)$$

Solving for $P$:

$$P=\frac{N\alpha E}{1-\dfrac{N\alpha}{3\varepsilon_0}}$$

Using $P=\varepsilon_0(\varepsilon_r-1)E$ to eliminate $P/E$ and simplifying gives the Clausius–Mossotti relation:

$$\boxed{\frac{\varepsilon_r-1}{\varepsilon_r+2}=\frac{N\alpha}{3\varepsilon_0}}$$

This links the macroscopic dielectric constant $\varepsilon_r$ to the microscopic atomic polarizability $\alpha$ and the number density $N$.

Classification of magnetic materials

Summary: Diamagnetics oppose the applied field weakly, paramagnetics enhance it weakly, and ferromagnetics enhance it enormously and retain magnetization.

Heisenberg's uncertainty principle

It is impossible to determine simultaneously, with unlimited precision, both the position and the momentum of a particle. The product of the uncertainties is at least of the order of $\hbar$:

$$\Delta x\,\Delta p \ge \frac{\hbar}{2}, \qquad \hbar=\frac{h}{2\pi}=1.055\times10^{-34}\,\text{J·s}$$

Minimum energy of an electron confined in a nucleus

If the electron is confined within a nucleus, $\Delta x \approx 10^{-14}\,\text{m}$. Taking the minimum momentum $p\approx\Delta p\approx\dfrac{\hbar}{\Delta x}$:

$$p\approx\frac{1.055\times10^{-34}}{10^{-14}}=1.055\times10^{-20}\,\text{kg·m/s}$$

Since $pc=(1.055\times10^{-20})(3\times10^{8})=3.16\times10^{-12}\,\text{J}$ is far larger than the electron rest energy $m_0c^2=8.2\times10^{-14}\,\text{J}$, the electron must be treated relativistically, $E\approx pc$:

$$E\approx pc=3.16\times10^{-12}\,\text{J}=\frac{3.16\times10^{-12}}{1.6\times10^{-19}}\,\text{eV}\approx 1.98\times10^{7}\,\text{eV}\approx 20\,\text{MeV}$$

Comment

An electron bound inside the nucleus would need a minimum energy of about 20 MeV. However, electrons emitted in beta decay have energies of only a few MeV, and no nuclear binding mechanism can hold an electron with 20 MeV. Therefore electrons cannot exist inside the nucleus; the beta-decay electron is created at the instant of decay.