BE Computer Engineering (IOE, TU) Engineering Physics (IOE, SH 452) Question Paper 2079 Nepal

(a) Damped harmonic oscillator [6]

A mass $m$ subject to a restoring force $-kx$ and a velocity-dependent damping force $-b\dfrac{dx}{dt}$ obeys Newton's second law:

$$m\frac{d^2x}{dt^2} = -kx - b\frac{dx}{dt}$$

Dividing by $m$ and writing $2\beta = b/m$ (damping constant) and $\omega_0^2 = k/m$ (natural angular frequency):

$$\frac{d^2x}{dt^2} + 2\beta\frac{dx}{dt} + \omega_0^2 x = 0$$

Solution. Try $x = e^{\alpha t}$, giving the auxiliary equation $\alpha^2 + 2\beta\alpha + \omega_0^2 = 0$, so

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

For light (under) damping $\beta < \omega_0$, the root is complex. Writing $\omega = \sqrt{\omega_0^2 - \beta^2}$,

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

This is an oscillation of angular frequency $\omega$ whose amplitude $A_0e^{-\beta t}$ decays exponentially.

Relaxation time $\tau$: the time in which the amplitude falls to $1/e$ of its initial value. Since amplitude $\propto e^{-\beta t}$, $e^{-\beta\tau}=e^{-1}$ giving

$$\tau = \frac{1}{\beta} = \frac{2m}{b}$$

(The energy, $\propto$ amplitude$^2$, decays as $e^{-2\beta t}=e^{-t/\tau_E}$ with energy relaxation time $\tau_E = 1/2\beta$.)

Quality factor $Q$: a measure of the sharpness of resonance / number of radians of oscillation in which energy falls by factor $e$:

$$Q = \omega_0\tau = \frac{\omega_0}{2\beta} = \frac{\omega_0 m}{b}$$

A large $Q$ means weak damping and a long-lived, sharply tuned oscillation.

(b) Numerical [4]

Given $f_0 = 500$ Hz, $Q = 200$, so $\omega_0 = 2\pi f_0 = 2\pi(500) = 3141.6\ \text{rad/s}$.

Relaxation time (amplitude $1/e$):

$$\tau = \frac{Q}{\omega_0} = \frac{2Q}{\omega_0}\cdot\frac12$$

Using $Q = \omega_0\tau$ with $\tau = 1/\beta$ (amplitude decay time):

$$\tau = \frac{Q}{\omega_0} = \frac{200}{3141.6} = 0.0637\ \text{s} \approx 63.7\ \text{ms}$$

The time for the amplitude to fall to $1/e$ of its initial value is, by definition, this same relaxation time:

$$t_{1/e} = \tau \approx 6.37\times10^{-2}\ \text{s} \approx 63.7\ \text{ms}$$

(a) Maxwell's equations (integral form) [7]

Modification of Ampere's law — displacement current. The original Ampere's law $\oint \vec B\cdot d\vec l = \mu_0 I_c$ fails for a charging capacitor: an Amperian loop spanned by a surface passing through the plates encloses no conduction current, yet $\vec B$ is observed. Maxwell argued that the changing electric field between the plates is equivalent to a current. He defined the displacement current:

$$I_d = \varepsilon_0\frac{d\Phi_E}{dt}, \qquad J_d = \varepsilon_0\frac{\partial E}{\partial t}$$

Adding it makes the total current continuous and Ampere's law becomes

$$\oint \vec B\cdot d\vec l = \mu_0\left(I_c + \varepsilon_0\frac{d\Phi_E}{dt}\right)$$

This term is essential for the existence of electromagnetic waves.

(b) Numerical [3]

$E = 50\sin(10^9 t)\ \text{V/m}$. In free space the displacement current density is

$$J_d = \varepsilon_0\frac{\partial E}{\partial t} = \varepsilon_0\cdot 50\cdot 10^9\cos(10^9 t)$$

With $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$:

$$J_d = 8.85\times10^{-12}\times 50\times10^{9}\cos(10^9 t) = 0.4425\cos(10^9 t)\ \text{A/m}^2$$

Maximum (amplitude) value $\approx 0.44\ \text{A/m}^2$.

(a) Time-independent Schrodinger equation [5]

A free particle is described by the de Broglie wave $\psi(x,t)=A\,e^{i(kx-\omega t)}$ with $k = 2\pi/\lambda$ and $\lambda = h/p$. The general time-dependent equation in one dimension with potential $V(x)$ is

$$i\hbar\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2} + V(x)\psi$$

For a stationary state separate variables: $\psi(x,t)=\psi(x)\,e^{-iEt/\hbar}$, where $E=\hbar\omega$ is the total energy. Substituting and cancelling the time factor:

$$E\,\psi(x) = -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi$$

Rearranging gives the time-independent Schrodinger equation:

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

(b) Particle in a 1-D infinite well of width L [5]

Inside the well ($0<x<L$), $V=0$:

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

General solution $\psi(x) = A\sin kx + B\cos kx$.

Boundary conditions (walls infinitely high, so $\psi=0$ outside and at walls):

• $\psi(0)=0 \Rightarrow B=0$, so $\psi = A\sin kx$.
• $\psi(L)=0 \Rightarrow \sin kL = 0 \Rightarrow kL = n\pi$, $n=1,2,3,\dots$

Hence $k = n\pi/L$.

Energy eigenvalues. From $k^2 = 2mE/\hbar^2$:

$$\boxed{E_n = \frac{n^2\pi^2\hbar^2}{2mL^2} = \frac{n^2 h^2}{8mL^2}}, \quad n=1,2,3,\dots$$

Energy is quantized; the lowest (zero-point) energy is $E_1 = h^2/8mL^2 \neq 0$.

Normalization. Require $\int_0^L |\psi|^2 dx = 1$:

$$A^2\int_0^L \sin^2\!\frac{n\pi x}{L}\,dx = A^2\cdot\frac{L}{2} = 1 \Rightarrow A = \sqrt{\frac{2}{L}}$$

Normalized wave functions:

$$\boxed{\psi_n(x) = \sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L}}, \quad 0\le x\le L$$

(a) Newton's rings in reflected light [7]

When a plano-convex lens of large radius $R$ is placed on a flat glass plate, a thin air film of variable thickness forms between them, increasing radially from zero at the contact point. Monochromatic light reflected from the top and bottom surfaces of this wedge interferes, producing concentric bright and dark rings (Newton's rings) centred on the contact point.

The two reflected rays have a path difference $2\mu t$ (for air $\mu=1$) plus an extra $\lambda/2$ from the phase reversal on reflection at the denser glass plate:

$$\Delta = 2\mu t + \frac{\lambda}{2}$$

Dark ring condition (destructive): $\Delta = (2n+1)\lambda/2 \Rightarrow 2\mu t = n\lambda$.

Geometry. For a circle of radius $r$ at film thickness $t$, by the property of the circle (sagitta), $r^2 = 2Rt$, i.e. $t = r^2/2R$. Substituting (air film, $\mu=1$):

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

For diameter $D_n = 2r_n$:

$$D_n^2 = 4nR\lambda \Rightarrow \boxed{D_n = 2\sqrt{R\lambda}\,\sqrt{n}}$$

Thus the diameter of the $n$-th dark ring is proportional to $\sqrt{n}$ (the square root of natural numbers).

(b) Refractive index of liquid [3]

For an air film: $D_n^2 = 4nR\lambda$. For a liquid of index $\mu$ between lens and plate: $D_n'^2 = \dfrac{4nR\lambda}{\mu}$.

Dividing for the same ring number $n=10$:

$$\mu = \frac{D_n^2}{D_n'^2} = \frac{(1.40)^2}{(1.27)^2} = \frac{1.960}{1.6129} = 1.215$$ $$\boxed{\mu \approx 1.22}$$

Plane diffraction grating

A plane diffraction grating is an optical element consisting of a large number of equidistant, parallel, narrow slits (transparent strips) of equal width ruled on a flat transparent plate. If $a$ is the width of each slit and $b$ the width of each opaque space, then $d=(a+b)$ is the grating element (period). A grating with $N$ lines per unit length has $d = 1/N$.

Grating equation (normal incidence)

For light of wavelength $\lambda$ incident normally, wavelets from corresponding points of adjacent slits travelling at angle $\theta$ have a path difference $d\sin\theta$. Principal maxima occur when this equals a whole number of wavelengths:

$$\boxed{(a+b)\sin\theta = n\lambda \quad\text{or}\quad d\sin\theta = n\lambda}$$

where $n = 0, 1, 2,\dots$ is the order of the spectrum.

Resolving power

The resolving power of a grating is its ability to separate two spectral lines of nearly equal wavelengths $\lambda$ and $\lambda+d\lambda$. It is defined as

$$R = \frac{\lambda}{d\lambda} = nN$$

where $n$ is the order and $N$ the total number of ruled lines illuminated. Resolving power increases with order and with the number of lines.

Brewster's law

When unpolarised light is incident on the surface of a transparent medium, the reflected light becomes completely plane-polarised at one particular angle of incidence called the polarising (Brewster) angle $\theta_p$. Brewster's law states that at this angle the refractive index equals the tangent of the polarising angle:

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

Explanation. At the Brewster angle the reflected and refracted rays are mutually perpendicular ($\theta_p + \theta_r = 90^\circ$). Applying Snell's law $\mu = \sin\theta_p/\sin\theta_r = \sin\theta_p/\sin(90^\circ-\theta_p) = \tan\theta_p$. The reflected light vibrates only perpendicular to the plane of incidence (fully polarised).

Ordinary (O) ray vs Extraordinary (E) ray

In a doubly refracting (birefringent) crystal such as calcite, an incident ray splits into two refracted rays:

Along the optic axis both rays travel with the same speed and are not separated.

Definitions

Acceptance angle $\theta_a$: the maximum half-angle of the cone of light (measured from the fibre axis) that can enter the core of an optical fibre and still be guided by total internal reflection at the core–cladding boundary. Rays entering within this cone propagate; rays outside it leak into the cladding.

Numerical aperture (NA): the light-gathering ability of the fibre, equal to the sine of the acceptance angle. For a step-index fibre with core index $n_1$ and cladding index $n_2$ in a medium of index $n_0$:

$$\text{NA} = n_0\sin\theta_a = \sqrt{n_1^2 - n_2^2}$$

Numerical

Given $n_1 = 1.50$, $n_2 = 1.47$, surrounding medium air $n_0 = 1$.

$$\text{NA} = \sqrt{n_1^2 - n_2^2} = \sqrt{1.50^2 - 1.47^2} = \sqrt{2.25 - 2.1609} = \sqrt{0.0891}$$ $$\boxed{\text{NA} = 0.2985 \approx 0.30}$$

Acceptance angle:

$$\theta_a = \sin^{-1}(\text{NA}) = \sin^{-1}(0.2985) = 17.4^\circ$$ $$\boxed{\theta_a \approx 17.4^\circ}$$

Population inversion

In thermal equilibrium the number of atoms in a lower energy level $N_1$ is greater than in a higher level $N_2$ (Boltzmann distribution). Population inversion is the non-equilibrium condition in which a higher energy level has a larger population than a lower one, i.e. $N_2 > N_1$.

Why it is essential for laser action

When a photon of the right energy passes through the medium, two competing processes occur: stimulated absorption (rate $\propto N_1$) and stimulated emission (rate $\propto N_2$). Net amplification of light (gain) requires stimulated emission to dominate absorption, which happens only when $N_2 > N_1$. Without population inversion the medium absorbs more than it emits and no laser amplification is possible.

Role of metastable states

A metastable state is an excited level with an unusually long lifetime ($\sim 10^{-3}$ s instead of $\sim 10^{-8}$ s). Because atoms remain there long enough, a large population can accumulate, allowing $N_2$ to exceed $N_1$ and sustaining population inversion. It acts as the upper laser level.

Role of optical pumping

Optical pumping uses an intense external light source (e.g. a flash lamp) to excite atoms from the ground state to a higher pump band. The atoms then decay rapidly (non-radiatively) to the metastable state, building up its population. This continuous supply of energy maintains the inversion against losses, enabling continuous or repeated laser action.

Gauss's law

The total electric flux through any closed (Gaussian) surface equals $1/\varepsilon_0$ times the net charge enclosed by that surface:

$$\oint \vec E\cdot d\vec A = \frac{Q_{enc}}{\varepsilon_0}$$

Field of an infinite charged plane sheet

Consider an infinite plane sheet with uniform surface charge density $\sigma$. By symmetry the field $\vec E$ is perpendicular to the sheet and points away from it (for $\sigma>0$), with equal magnitude on both sides.

Gaussian surface: a cylinder (pillbox) of cross-sectional area $A$ piercing the sheet, with its two flat faces parallel to the sheet, one on each side.

• Flux through the curved side $=0$ (field parallel to the surface).
• Flux through each flat face $= EA$; total flux $= 2EA$.
• Charge enclosed $= \sigma A$.

Applying Gauss's law:

$$2EA = \frac{\sigma A}{\varepsilon_0}$$ $$\boxed{E = \frac{\sigma}{2\varepsilon_0}}$$

The field is uniform, independent of distance from the sheet, and directed normally away from (or toward) it depending on the sign of $\sigma$.

Capacitor partially filled with a dielectric slab

Let the plates carry surface charge density $\sigma = Q/A$, plate separation $d$, and a dielectric slab of thickness $t$ ($t<d$) and dielectric constant $K$ inserted between them. The remaining gap $(d-t)$ is air/vacuum.

Field in the air gap (free space): $E_0 = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{A\varepsilon_0}$

Field inside the dielectric (reduced by factor $K$): $E_d = \dfrac{E_0}{K} = \dfrac{\sigma}{K\varepsilon_0}$

Potential difference between the plates is the sum across the two regions:

$$V = E_0(d - t) + E_d\, t = \frac{\sigma}{\varepsilon_0}(d-t) + \frac{\sigma}{K\varepsilon_0}t$$ $$V = \frac{\sigma}{\varepsilon_0}\left[(d-t) + \frac{t}{K}\right] = \frac{Q}{A\varepsilon_0}\left[(d-t) + \frac{t}{K}\right]$$

Capacitance $C = Q/V$:

$$\boxed{C = \frac{\varepsilon_0 A}{(d - t) + \dfrac{t}{K}}}$$

Checks: if $t=0$ (no slab) $\Rightarrow C = \varepsilon_0 A/d$; if $t=d$ (slab fills gap) $\Rightarrow C = K\varepsilon_0 A/d$, as expected. Since $t/K < t$, inserting the slab always increases the capacitance.

Heisenberg's uncertainty principle

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

$$\Delta x\,\Delta p \ge \frac{\hbar}{2}, \qquad \hbar = \frac{h}{2\pi}$$

(A common exam form uses $\Delta x\,\Delta p \ge \hbar$ or $\ge h/4\pi$.)

Numerical

Confinement width $\Delta x = 1\times10^{-10}$ m. Using $\Delta p \approx \dfrac{\hbar}{2\Delta x}$ with $\hbar = 1.055\times10^{-34}$ J·s:

$$\Delta p_{min} = \frac{1.055\times10^{-34}}{2\times10^{-10}} = 5.27\times10^{-25}\ \text{kg·m/s}$$ $$\boxed{\Delta p_{min} \approx 5.3\times10^{-25}\ \text{kg·m/s}}$$

Minimum kinetic energy (taking $p \approx \Delta p$, electron mass $m = 9.11\times10^{-31}$ kg):

$$E_{min} = \frac{(\Delta p)^2}{2m} = \frac{(5.27\times10^{-25})^2}{2\times 9.11\times10^{-31}} = \frac{2.78\times10^{-49}}{1.82\times10^{-30}}$$ $$E_{min} = 1.53\times10^{-19}\ \text{J} \approx 0.95\ \text{eV}$$ $$\boxed{E_{min} \approx 1.5\times10^{-19}\ \text{J} \approx 0.95\ \text{eV}}$$

(Using $\Delta x\,\Delta p \ge \hbar$ instead gives $\Delta p \approx 1.05\times10^{-24}$ kg·m/s and $E_{min} \approx 3.8$ eV; either convention is acceptable if stated.)

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.

Cause. A current $I_x$ flows along $x$ in a slab of thickness $t$; a magnetic field $B_z$ acts along $z$. The moving charge carriers experience the Lorentz force $\vec F = q\vec v\times\vec B$, which deflects them toward one face. Charge accumulates there until the resulting transverse electric field $E_y$ balances the magnetic force.

Derivation of Hall coefficient

At equilibrium the electric and magnetic forces balance:

$$qE_y = qv_x B_z \Rightarrow E_y = v_x B_z$$

The drift velocity is related to current density $J_x = nqv_x$, where $n$ = carrier concentration, so $v_x = J_x/nq$. Thus

$$E_y = \frac{J_x B_z}{nq}$$

The Hall coefficient is defined as $R_H = \dfrac{E_y}{J_x B_z}$:

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

In terms of measurable quantities, with $V_H = E_y\,w$ and $I = J_x(wt)$:

$$R_H = \frac{V_H\, t}{I\, B}$$

$R_H$ is negative for n-type (electron) and positive for p-type (hole) conduction.

Applications

1. Determining carrier type and concentration $n = 1/(R_H q)$ and the sign of carriers (n- or p-type) in a semiconductor.
2. Measurement of magnetic field strength using Hall-effect sensors (gaussmeters); also used to measure carrier mobility ($\mu = R_H\sigma$).

Superconductivity

Superconductivity is the phenomenon in which certain materials, when cooled below a characteristic critical temperature $T_c$, lose all electrical resistance (zero resistivity) and conduct current without any energy dissipation. Below $T_c$ the material also expels magnetic flux from its interior.

Meissner effect

The Meissner effect is the complete expulsion of magnetic flux from the interior of a superconductor when it is cooled below $T_c$ in a magnetic field. The field inside becomes $B = 0$, showing that a superconductor is a perfect diamagnet ($\chi = -1$). This is a distinct property: a mere perfect conductor would only freeze in the existing flux, whereas a superconductor actively expels it. The Meissner effect persists only up to a critical field $H_c$; above it superconductivity is destroyed.

Type I vs Type II superconductors

Wave equation for a stretched string

Consider a string of linear mass density $\mu$ (mass per unit length) under uniform tension $T$, lying along the $x$-axis. A small transverse displacement $y(x,t)$ produces a small element of length $dx$ between $x$ and $x+dx$.

Forces on the element. The tension acts tangentially at both ends. For small slopes, the net upward (transverse) force arises from the difference in the vertical components of tension at the two ends:

$$F_y = T\left(\frac{\partial y}{\partial x}\right)_{x+dx} - T\left(\frac{\partial y}{\partial x}\right)_x = T\frac{\partial^2 y}{\partial x^2}\,dx$$

Newton's second law. The mass of the element is $\mu\,dx$ and its transverse acceleration is $\partial^2 y/\partial t^2$:

$$\mu\,dx\,\frac{\partial^2 y}{\partial t^2} = T\frac{\partial^2 y}{\partial x^2}\,dx$$

Dividing by $\mu\,dx$:

$$\boxed{\frac{\partial^2 y}{\partial t^2} = \frac{T}{\mu}\frac{\partial^2 y}{\partial x^2}}$$

Wave velocity

Comparing with the standard one-dimensional wave equation

$$\frac{\partial^2 y}{\partial t^2} = v^2\frac{\partial^2 y}{\partial x^2}$$

we identify $v^2 = T/\mu$, so the velocity of the transverse wave is

$$\boxed{v = \sqrt{\frac{T}{\mu}}}$$

The wave travels faster for greater tension and slower for a heavier (larger $\mu$) string, independent of frequency or amplitude.