BE Civil Engineering (IOE, TU) Engineering Physics (IOE, SH 402) Question Paper 2080 Nepal

Differential equation

For a body of mass $m$ acted on by a restoring force $-kx$ and a velocity-proportional damping force $-b\,\dfrac{dx}{dt}$, Newton's second law gives

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

where $2\beta = \dfrac{b}{m}$ (damping coefficient) and $\omega_0^2 = \dfrac{k}{m}$ (natural angular frequency).

Solution (under-damped, $\beta < \omega_0$)

Trying $x = e^{\alpha t}$ gives $\alpha = -\beta \pm \sqrt{\beta^2 - \omega_0^2}$. For $\beta<\omega_0$ the root is complex, $\alpha = -\beta \pm i\omega$ with $\omega = \sqrt{\omega_0^2 - \beta^2}$, so

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

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

Relaxation time $\tau$ is the time for the amplitude to fall to $1/e$ of its initial value: $e^{-\beta\tau}=e^{-1}\Rightarrow \tau = 1/\beta$.

Logarithmic decrement $\delta$ is the natural log of the ratio of successive amplitudes one period $T$ apart: $\delta = \ln\dfrac{A(t)}{A(t+T)} = \beta T$.

Numerical part

Given $m=0.20\ \text{kg}$, $k=80\ \text{N/m}$, $b=0.40\ \text{kg/s}$.

$\omega_0^2 = k/m = 80/0.20 = 400\ \text{s}^{-2}$, so $\omega_0 = 20\ \text{rad/s}$.

$\beta = b/(2m) = 0.40/(2\times0.20) = 1.0\ \text{s}^{-1}$.

(i) $\omega = \sqrt{\omega_0^2 - \beta^2} = \sqrt{400 - 1} = \sqrt{399} = 19.975\ \text{rad/s}$.

The damped angular frequency is $\boxed{\omega \approx 19.97\ \text{rad/s}}$.

(ii) Amplitude falls to $1/e$ when $t = \tau = 1/\beta = 1/1.0 = 1.0\ \text{s}$.

The relaxation time is $\boxed{\tau = 1.0\ \text{s}}$.

Formation

A plano-convex lens resting on a flat glass plate encloses a thin air film of gradually increasing thickness. Monochromatic light reflected from the top and bottom surfaces of the air film interferes. Because of the circular symmetry of the film thickness, the interference fringes are concentric circles centred at the point of contact — Newton's rings.

Condition and diameter of dark rings

For the air film of thickness $t$, including the $\pi$ phase change at the glass–air–glass reflection, dark rings occur when $2t = n\lambda$. Geometry of the lens gives $t = \dfrac{r^2}{2R}$, where $r$ is the ring radius and $R$ the radius of curvature. Hence

$$r_n^2 = nR\lambda \;\Rightarrow\; D_n^2 = 4nR\lambda \quad(D_n = 2r_n).$$

Determining $\lambda$

For two rings $n$ and $m$:

$$D_m^2 - D_n^2 = 4(m-n)R\lambda \;\Rightarrow\; \lambda = \frac{D_m^2 - D_n^2}{4(m-n)R}.$$

Numerical part

$D_{10} = 4.00\ \text{mm}=4.00\times10^{-3}\,\text{m}$, $D_{20}=6.00\ \text{mm}=6.00\times10^{-3}\,\text{m}$, $R=1.00\ \text{m}$, $m-n = 20-10 = 10$.

$D_{20}^2 = 36.0\times10^{-6}\ \text{m}^2$, $D_{10}^2 = 16.0\times10^{-6}\ \text{m}^2$.

$D_{20}^2 - D_{10}^2 = 20.0\times10^{-6}\ \text{m}^2$.

$$\lambda = \frac{20.0\times10^{-6}}{4\times10\times1.00} = \frac{20.0\times10^{-6}}{40} = 5.0\times10^{-7}\ \text{m}.$$

The wavelength is $\boxed{\lambda = 5.0\times10^{-7}\ \text{m} = 500\ \text{nm}}$.

Maxwell's equations (free space, no charges/currents)

$$\nabla\cdot\mathbf{E}=0,\quad \nabla\cdot\mathbf{B}=0,\quad \nabla\times\mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t},\quad \nabla\times\mathbf{B}=\mu_0\varepsilon_0\frac{\partial \mathbf{E}}{\partial t}.$$

• $\nabla\cdot\mathbf{E}=0$ (Gauss law): no net charge, field lines have no sources.
• $\nabla\cdot\mathbf{B}=0$: no magnetic monopoles; $\mathbf{B}$ lines are closed.
• $\nabla\times\mathbf{E}=-\partial\mathbf{B}/\partial t$ (Faraday): changing $\mathbf{B}$ induces a curling $\mathbf{E}$.
• $\nabla\times\mathbf{B}=\mu_0\varepsilon_0\,\partial\mathbf{E}/\partial t$ (Ampère–Maxwell): changing $\mathbf{E}$ (displacement current) produces $\mathbf{B}$.

Wave equation

Take the curl of Faraday's law:

$$\nabla\times(\nabla\times\mathbf{E}) = -\frac{\partial}{\partial t}(\nabla\times\mathbf{B}).$$

Use the identity $\nabla\times(\nabla\times\mathbf{E}) = \nabla(\nabla\cdot\mathbf{E}) - \nabla^2\mathbf{E} = -\nabla^2\mathbf{E}$ (since $\nabla\cdot\mathbf{E}=0$), and substitute Ampère–Maxwell:

$$-\nabla^2\mathbf{E} = -\frac{\partial}{\partial t}\left(\mu_0\varepsilon_0\frac{\partial\mathbf{E}}{\partial t}\right) \;\Rightarrow\; \nabla^2\mathbf{E} = \mu_0\varepsilon_0\frac{\partial^2\mathbf{E}}{\partial t^2}.$$

This is a wave equation $\nabla^2\mathbf{E} = \dfrac{1}{c^2}\dfrac{\partial^2\mathbf{E}}{\partial t^2}$ with

$$c = \frac{1}{\sqrt{\mu_0\varepsilon_0}}.$$

An identical equation holds for $\mathbf{B}$.

Numerical value

$$\mu_0\varepsilon_0 = (4\pi\times10^{-7})(8.85\times10^{-12}) = 1.2566\times10^{-6}\times8.85\times10^{-12} = 1.1121\times10^{-17}\ \text{s}^2/\text{m}^2.$$ $$c = \frac{1}{\sqrt{1.1121\times10^{-17}}} = \frac{1}{3.335\times10^{-9}} = 2.998\times10^{8}\ \text{m/s}.$$

The speed of EM waves in free space is $\boxed{c \approx 3.0\times10^{8}\ \text{m/s}}$.

Schrödinger equation inside the well

Inside the well $V=0$ for $0<x<L$ and $V=\infty$ outside. The time-independent equation is

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi \;\Rightarrow\; \frac{d^2\psi}{dx^2} + k^2\psi = 0,\quad k^2 = \frac{2mE}{\hbar^2}.$$

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

Boundary conditions

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

Thus $k_n = \dfrac{n\pi}{L}$.

Normalization

$\int_0^L A^2\sin^2(k_n x)\,dx = 1 \Rightarrow A^2\dfrac{L}{2}=1 \Rightarrow A=\sqrt{\dfrac{2}{L}}$. So

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

Energy levels

$E_n = \dfrac{\hbar^2 k_n^2}{2m} = \dfrac{n^2\pi^2\hbar^2}{2mL^2} = \dfrac{n^2 h^2}{8mL^2}$.

Numerical part

$L = 0.10\ \text{nm} = 1.0\times10^{-10}\ \text{m}$.

Ground state ($n=1$):

$$E_1 = \frac{h^2}{8mL^2} = \frac{(6.63\times10^{-34})^2}{8(9.11\times10^{-31})(1.0\times10^{-10})^2}.$$

Numerator: $(6.63\times10^{-34})^2 = 4.396\times10^{-67}$. Denominator: $8\times9.11\times10^{-31}\times1.0\times10^{-20} = 7.288\times10^{-50}$.

$$E_1 = \frac{4.396\times10^{-67}}{7.288\times10^{-50}} = 6.03\times10^{-18}\ \text{J} = \frac{6.03\times10^{-18}}{1.60\times10^{-19}} = 37.7\ \text{eV}.$$

First excited state ($n=2$): $E_2 = 4E_1 = 4\times37.7 = 150.7\ \text{eV}$.

$\boxed{E_1 \approx 37.7\ \text{eV},\quad E_2 \approx 150.7\ \text{eV}}$.

Definition

Capacitance is the charge stored per unit potential difference, $C = Q/V$ (unit: farad).

Derivation (partly filled)

Let free charge density on the plates be $\sigma = Q/A$. The field in the air gap (thickness $d-t$) is $E_0 = \sigma/\varepsilon_0$; inside the dielectric (thickness $t$) the field is reduced to $E = E_0/K = \sigma/(K\varepsilon_0)$.

Total potential difference:

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

With $\sigma = Q/A$:

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

Numerical part

$A = 100\ \text{cm}^2 = 100\times10^{-4}\ \text{m}^2 = 1.0\times10^{-2}\ \text{m}^2$. $d = 5.0\times10^{-3}\ \text{m}$, $t = 3.0\times10^{-3}\ \text{m}$, $K=4.0$.

Denominator: $(d-t) + t/K = (5.0-3.0)\times10^{-3} + \dfrac{3.0\times10^{-3}}{4.0} = 2.0\times10^{-3} + 0.75\times10^{-3} = 2.75\times10^{-3}\ \text{m}$.

$$C = \frac{(8.85\times10^{-12})(1.0\times10^{-2})}{2.75\times10^{-3}} = \frac{8.85\times10^{-14}}{2.75\times10^{-3}} = 3.22\times10^{-11}\ \text{F}.$$

The capacitance is $\boxed{C \approx 32.2\ \text{pF}}$.

Sabine's formula

$$T = \frac{0.161\,V}{\sum a_i S_i} = \frac{0.161\,V}{A},$$

where $T$ = reverberation time (s), $V$ = volume of the hall (m³), $A=\sum a_i S_i$ = total absorption (m²·open-window-unit, the sabine). Reverberation time increases with volume and decreases with greater total absorption (i.e. larger absorption coefficients and surface areas, more occupants/furnishings).

Numerical part

Volume $V = 20\times15\times8 = 2400\ \text{m}^3$. Total absorption $A = 560\ \text{sabine}$.

$$T = \frac{0.161\times2400}{560} = \frac{386.4}{560} = 0.69\ \text{s}.$$

The reverberation time is $\boxed{T \approx 0.69\ \text{s}}$.

Grating action

A plane transmission grating is a large number $N$ of equally spaced parallel slits of grating element (spacing) $d$. For light incident normally, secondary wavelets from successive slits interfere; principal maxima (bright) occur where the path difference between adjacent slits equals an integral number of wavelengths:

$$d\sin\theta = m\lambda,\qquad m = 0,1,2,\dots$$

Large $N$ makes these maxima sharp, allowing high-resolution spectroscopy.

Numerical part

Grating element: $d = \dfrac{1\ \text{cm}}{5000} = \dfrac{1.0\times10^{-2}\ \text{m}}{5000} = 2.0\times10^{-6}\ \text{m}$.

$\lambda = 589\ \text{nm} = 5.89\times10^{-7}\ \text{m}$, order $m=2$.

$$\sin\theta = \frac{m\lambda}{d} = \frac{2\times5.89\times10^{-7}}{2.0\times10^{-6}} = \frac{1.178\times10^{-6}}{2.0\times10^{-6}} = 0.589.$$ $$\theta = \sin^{-1}(0.589) = 36.1^{\circ}.$$

The second-order maximum is at $\boxed{\theta \approx 36.1^{\circ}}$.

Propagation

An optical fibre consists of a high-index core ($n_1$) surrounded by a lower-index cladding ($n_2$). Light launched within a certain cone strikes the core–cladding boundary at an angle greater than the critical angle and undergoes repeated total internal reflection, guiding it along the fibre with low loss.

Acceptance angle $\theta_a$: the maximum half-angle of the entrance cone for which rays are guided by total internal reflection.

Numerical aperture $\text{NA} = \sin\theta_a = \sqrt{n_1^2 - n_2^2}$ (for a fibre in air), a measure of the light-gathering ability.

Numerical part

$$\text{NA} = \sqrt{n_1^2 - n_2^2} = \sqrt{1.50^2 - 1.46^2} = \sqrt{2.2500 - 2.1316} = \sqrt{0.1184} = 0.344.$$

Acceptance angle:

$$\theta_a = \sin^{-1}(0.344) = 20.1^{\circ}.$$

$\boxed{\text{NA} \approx 0.344,\quad \theta_a \approx 20.1^{\circ}}$.

Spontaneous vs stimulated emission

Population inversion

In thermal equilibrium more atoms occupy the lower level than the upper ($N_1 > N_2$), so absorption dominates. Population inversion is the non-equilibrium condition $N_2 > N_1$. Only then does stimulated emission exceed absorption, so a passing photon produces net amplification (optical gain) — the prerequisite for laser action.

Metastable states and pumping

A metastable state has an unusually long lifetime, allowing atoms to accumulate there and sustain the inversion. Pumping (optical, electrical, etc.) supplies energy to raise atoms from the ground state to higher levels from which they decay into the metastable level, maintaining $N_2 > N_1$.

Hall effect

When a current-carrying conductor is placed in a transverse magnetic field, the moving charges experience a Lorentz force $q\mathbf{v}\times\mathbf{B}$ that deflects them sideways. Charge accumulates on one face until the transverse electric field balances the magnetic force, producing a measurable transverse Hall voltage $V_H$.

Derivation

At equilibrium: $eE_H = e v_d B \Rightarrow E_H = v_d B$. With drift velocity $v_d = \dfrac{I}{n e A} = \dfrac{I}{n e (w t)}$ and $V_H = E_H w$:

$$V_H = v_d B w = \frac{I B}{n e t} \;\Rightarrow\; n = \frac{I B}{e t V_H}.$$

The Hall coefficient is $R_H = \dfrac{1}{ne} = \dfrac{V_H t}{I B}$.

Numerical part

$I=5.0\ \text{A}$, $B=0.40\ \text{T}$, $t=1.0\times10^{-3}\ \text{m}$, $V_H=2.0\times10^{-3}\ \text{V}$.

$$n = \frac{I B}{e t V_H} = \frac{(5.0)(0.40)}{(1.60\times10^{-19})(1.0\times10^{-3})(2.0\times10^{-3})}.$$

Numerator $= 2.0$. Denominator $= 1.60\times10^{-19}\times2.0\times10^{-6} = 3.20\times10^{-25}$.

$$n = \frac{2.0}{3.20\times10^{-25}} = 6.25\times10^{24}\ \text{m}^{-3}.$$

The carrier concentration is $\boxed{n \approx 6.25\times10^{24}\ \text{m}^{-3}}$.

Superconductivity

Below a critical temperature $T_c$ certain materials lose all electrical resistance, conducting current without dissipation.

Meissner effect

A superconductor cooled below $T_c$ in a weak magnetic field expels the magnetic flux from its interior ($\mathbf{B}=0$ inside), behaving as a perfect diamagnet. This active flux expulsion is distinct from mere zero resistance and is the defining magnetic signature of superconductivity.

Type-I vs Type-II

• Type-I: single critical field $H_c$; complete Meissner expulsion until $H_c$, then abrupt loss of superconductivity. Usually pure metals, low $H_c$.
• Type-II: two critical fields $H_{c1}, H_{c2}$; between them flux partially penetrates as quantized vortices (mixed state) while superconductivity persists. High $H_{c2}$, used in high-field magnets.

Numerical part

$H_0 = 6.0\times10^4\ \text{A/m}$, $T_c = 7.0\ \text{K}$, $T = 4.0\ \text{K}$.

$$\left(\frac{T}{T_c}\right)^2 = \left(\frac{4.0}{7.0}\right)^2 = (0.5714)^2 = 0.3265.$$ $$H_c = 6.0\times10^4\,[1 - 0.3265] = 6.0\times10^4\times0.6735 = 4.04\times10^{4}\ \text{A/m}.$$

The critical field at $4.0\ \text{K}$ is $\boxed{H_c \approx 4.04\times10^{4}\ \text{A/m}}$.