BE Computer Engineering (Pokhara University) Applied Physics (PU, PHY 110) Question Paper 2078 Nepal

(a) Damped harmonic oscillator

A particle of mass $m$ experiences a restoring force $-kx$ and a damping force $-b\dfrac{dx}{dt}$ proportional to velocity. 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).

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, $\alpha = -\beta \pm i\omega'$ with $\omega' = \sqrt{\omega_0^2 - \beta^2}$. The solution is

$$\boxed{x(t) = a_0\, e^{-\beta t}\cos(\omega' t + \phi)}$$

This is an oscillation of angular frequency $\omega'$ whose amplitude $a_0 e^{-\beta t}$ decays exponentially with time.

Logarithmic decrement ($\lambda$): the natural logarithm of the ratio of successive amplitudes one period $T = 2\pi/\omega'$ apart:

$$\lambda = \ln\frac{a_n}{a_{n+1}} = \beta T = \frac{2\pi\beta}{\omega'}$$

It measures how rapidly the amplitude dies away.

Quality factor ($Q$): $Q = 2\pi \times \dfrac{\text{energy stored}}{\text{energy lost per cycle}} = \dfrac{\omega'}{2\beta} \approx \dfrac{\omega_0}{2\beta}$ for light damping. A high $Q$ means a sharply resonant, slowly decaying oscillator.

(b) Numerical

Given $m = 0.2$ kg, $k = 80$ N/m, $b = 4$ kg/s.

$$\omega_0^2 = \frac{k}{m} = \frac{80}{0.2} = 400\ \text{s}^{-2}, \qquad \beta = \frac{b}{2m} = \frac{4}{0.4} = 10\ \text{s}^{-1}$$

(i) Angular frequency of damped oscillation:

$$\omega' = \sqrt{\omega_0^2 - \beta^2} = \sqrt{400 - 100} = \sqrt{300} = 17.32\ \text{rad/s}$$

(ii) The amplitude factor is $e^{-\beta t}$. It falls to $1/e$ when $\beta t = 1$:

$$t = \frac{1}{\beta} = \frac{1}{10} = 0.1\ \text{s}$$

Results: $\omega' \approx 17.3$ rad/s and $t = 0.1$ s.

(a) Newton's rings by reflected light

When a plano-convex lens of large radius of curvature $R$ is placed on a flat glass plate, a thin wedge-shaped air film of variable thickness $t$ is formed between them. Monochromatic light incident normally is partly reflected from the top and bottom of the air film; these two beams interfere, producing concentric bright and dark circular fringes centred at the point of contact — Newton's rings.

For reflection at the lower (denser) surface there is an additional path change of $\lambda/2$ (phase change of $\pi$). The effective path difference is

$$\Delta = 2\mu t + \frac{\lambda}{2} \quad (\mu = 1 \text{ for air})$$

Bright rings: $2t + \dfrac{\lambda}{2} = n\lambda \Rightarrow 2t = (2n-1)\dfrac{\lambda}{2}$

Dark rings: $2t = n\lambda,\quad n = 0,1,2,\dots$

Geometry of the air film: if $r$ is the radius of a ring, $r^2 = (2R - t)t \approx 2Rt$ (since $t \ll R$), so $t = \dfrac{r^2}{2R}$.

Diameter of dark rings ($D = 2r$): $2\cdot\dfrac{r^2}{2R} = n\lambda \Rightarrow r^2 = nR\lambda$

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

Thus $D_n \propto \sqrt{n}$ — the dark-ring diameter is proportional to the square root of the natural numbers $\sqrt{1},\sqrt{2},\sqrt{3},\dots$

Diameter of bright rings: $r^2 = \dfrac{(2n-1)R\lambda}{2} \Rightarrow D_n = \sqrt{2(2n-1)R\lambda}$, i.e. $D_n \propto \sqrt{2n-1}$.

(b) Refractive index of the liquid

In air: $D_n^2 = 4nR\lambda$. In a liquid of index $\mu$: $D_n'^2 = \dfrac{4nR\lambda}{\mu}$.

Dividing,

$$\mu = \frac{D_n^2}{D_n'^2} = \frac{(1.40)^2}{(1.27)^2} = \frac{1.960}{1.613} = 1.215$$

Refractive index of the liquid $\mu \approx 1.22$.

(a) Maxwell's equations (charge-free, non-conducting medium)

For a region with $\rho = 0$ and $\mathbf{J} = 0$ (free space, permittivity $\varepsilon_0$, permeability $\mu_0$):

Displacement current: Maxwell noticed Ampère's law $\nabla\times\mathbf{B} = \mu_0\mathbf{J}$ fails for a charging capacitor (no conduction current between plates). He added a term $\varepsilon_0\dfrac{\partial \mathbf{E}}{\partial t}$, the displacement current density $\mathbf{J}_d = \varepsilon_0\dfrac{\partial \mathbf{E}}{\partial t}$. It is not a flow of charge but a changing electric flux that produces a magnetic field exactly like a real current, making the equations consistent and predicting EM waves.

(b) Wave equation and wave speed

Take the curl of Faraday's law:

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

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

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

Comparing with the standard wave equation $\nabla^2\mathbf{E} = \dfrac{1}{v^2}\dfrac{\partial^2\mathbf{E}}{\partial t^2}$, the wave speed is

$$v = \frac{1}{\sqrt{\mu_0\varepsilon_0}}$$

Substituting $\mu_0 = 4\pi\times10^{-7}$ H/m and $\varepsilon_0 = 8.854\times10^{-12}$ F/m:

$$v = \frac{1}{\sqrt{(4\pi\times10^{-7})(8.854\times10^{-12})}} = 3\times10^{8}\ \text{m/s} = c$$

This equals the measured speed of light, establishing that light is an electromagnetic wave.

(a) Time-independent Schrödinger equation (1-D)

A particle of mass $m$ is described by a wavefunction. For a plane matter wave the time-dependent wavefunction is

$$\Psi(x,t) = \psi(x)\,e^{-iEt/\hbar}$$

Start from the classical energy relation $E = \dfrac{p^2}{2m} + V(x)$. Using the de Broglie relation and the operator correspondences $E \to i\hbar\dfrac{\partial}{\partial t}$ and $p \to -i\hbar\dfrac{\partial}{\partial x}$, the time-dependent Schrödinger equation is

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

Substituting $\Psi = \psi(x)e^{-iEt/\hbar}$, the time part gives $E\psi$ on the left, and dividing through by $e^{-iEt/\hbar}$:

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

This is the time-independent Schrödinger equation.

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

Let $V = 0$ for $0 < x < L$ and $V = \infty$ outside. Inside the well:

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

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

Boundary conditions: $\psi = 0$ at $x=0$ and $x=L$ (walls are impenetrable).

• At $x=0$: $B = 0 \Rightarrow \psi = A\sin kx$.
• At $x=L$: $A\sin kL = 0 \Rightarrow kL = n\pi,\ n = 1,2,3,\dots$

So $k = \dfrac{n\pi}{L}$.

Energy levels: from $E = \dfrac{\hbar^2 k^2}{2m}$,

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

The energy is quantized; the lowest (zero-point) energy is $E_1 = \dfrac{h^2}{8mL^2} \neq 0$.

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

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

Sketches (described):

• $n=1$: $\psi_1$ is a single half-sine loop with a maximum at $x=L/2$; $|\psi_1|^2$ has one hump peaked at the centre.
• $n=2$: $\psi_2$ has one full sine wave with a node at $x=L/2$; $|\psi_2|^2$ shows two humps with zero probability at the centre and at the walls.
• $n=3$: $\psi_3$ has three half-loops (nodes at $x=L/3, 2L/3$); $|\psi_3|^2$ shows three humps.

In every case the wavefunction and probability density vanish at the walls, and the number of nodes increases with $n$.

(a) Fresnel vs Fraunhofer diffraction

(b) Grating: second-order maximum

Grating element (spacing) $d = \dfrac{1}{N} = \dfrac{1}{5000}\ \text{cm} = 2\times10^{-6}\ \text{m} = 2000$ nm.

Grating equation: $d\sin\theta = m\lambda$ with $m = 2$, $\lambda = 5890\ \text{\AA} = 5.89\times10^{-7}$ m.

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

Angle of diffraction for the second order $\theta \approx 36.1^{\circ}$.

(a) Brewster's law

When unpolarized light is incident on a transparent dielectric, the reflected light is completely plane-polarized at one particular angle of incidence called the polarizing (Brewster) angle $\theta_p$. Brewster's law states that the refractive index of the medium equals the tangent of the polarizing angle:

$$\mu = \tan\theta_p$$

Reflected and refracted rays are perpendicular: By Snell's law $\mu = \dfrac{\sin\theta_p}{\sin r}$. By Brewster's law $\mu = \dfrac{\sin\theta_p}{\cos\theta_p}$. Comparing,

$$\sin r = \cos\theta_p = \sin(90^\circ - \theta_p) \Rightarrow r = 90^\circ - \theta_p$$

Therefore $\theta_p + r = 90^\circ$. Since the reflected ray makes angle $\theta_p$ and the refracted ray makes angle $r$ with the normal on opposite sides, the angle between them is $180^\circ - (\theta_p + r) = 90^\circ$. Hence the reflected and refracted rays are mutually perpendicular at the polarizing angle.

(b) Polarizing angle for glass ($\mu = 1.54$)

$$\theta_p = \tan^{-1}(\mu) = \tan^{-1}(1.54) = 57.0^{\circ}$$

Polarizing angle $\theta_p \approx 57^{\circ}$.

Key terms

Spontaneous emission: An atom in an excited state $E_2$ spontaneously decays to a lower state $E_1$ emitting a photon of energy $h\nu = E_2 - E_1$ at a random time and in a random direction. The emitted photons are incoherent.

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 that is identical to the incident one — same frequency, phase, direction and polarization. This produces coherent amplification of light (the basis of laser action).

Population inversion: The non-equilibrium condition in which more atoms occupy the higher energy level than the lower one ($N_2 > N_1$). It is essential for stimulated emission to dominate over absorption so that light is amplified. It is achieved by pumping (optical or electrical).

He–Ne laser

The active medium is a mixture of helium and neon (about 10:1) in a discharge tube; an electric discharge excites the He atoms.

Energy-level scheme (described):

1. Electrons in the discharge excite He atoms to metastable levels $F_2$ (≈20.61 eV) and $F_3$ (≈19.81 eV).
2. These He levels are nearly equal in energy to the Ne levels $E_6$ and $E_4$. By resonant collision the excited He atoms transfer their energy to Ne atoms (He returns to ground state), populating Ne levels $E_6$ and $E_4$. This creates population inversion between these Ne levels and the lower Ne levels.
3. Stimulated transitions in neon give the laser output, the principal one being

$$E_6 \to E_3:\quad \lambda = 6328\ \text{\AA (632.8 nm, red)}$$

(other lines at 1.15 µm and 3.39 µm). 4. From $E_3$ the Ne atoms decay rapidly to $E_2$ and return to the ground state through collisions with the tube walls, keeping the lower laser level empty and sustaining the inversion.

A pair of mirrors (one fully reflecting, one partially reflecting) forms an optical resonator that builds up the coherent beam, which emerges as a continuous, highly monochromatic, coherent red beam.

(a) Acceptance angle and numerical aperture

Acceptance angle ($\theta_a$): the maximum angle (measured from the fibre axis) at which light entering the core end face is guided down the fibre by total internal reflection. Rays entering within this angle propagate; rays outside it escape into the cladding.

Numerical aperture (NA): the light-gathering ability of the fibre, defined as $\text{NA} = \sin\theta_a$ (in air).

Derivation: Let the core index be $n_1$, cladding index $n_2$ ($n_1 > n_2$) and surrounding medium $n_0$. For a ray to be guided, it must strike the core–cladding boundary at the critical angle $\theta_c$, where $\sin\theta_c = n_2/n_1$.

For the limiting ray entering at $\theta_a$, Snell's law at the end face gives $n_0\sin\theta_a = n_1\sin\theta_r$, where $\theta_r = 90^\circ - \theta_c$. Thus $\sin\theta_r = \cos\theta_c = \sqrt{1 - \sin^2\theta_c} = \sqrt{1 - n_2^2/n_1^2}$.

$$n_0\sin\theta_a = n_1\sqrt{1 - \frac{n_2^2}{n_1^2}} = \sqrt{n_1^2 - n_2^2}$$

Taking $n_0 = 1$ (air):

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

(b) Numerical

$n_1 = 1.50,\ n_2 = 1.47$.

$$\text{NA} = \sqrt{1.50^2 - 1.47^2} = \sqrt{2.25 - 2.1609} = \sqrt{0.0891} = 0.2985 \approx 0.30$$ $$\theta_a = \sin^{-1}(0.2985) = 17.4^{\circ}$$

NA $\approx 0.30$, acceptance angle $\theta_a \approx 17.4^{\circ}$.

(a) Hall effect and Hall coefficient

When a current-carrying conductor or semiconductor is placed in a magnetic field perpendicular to the current, a transverse voltage (the Hall voltage $V_H$) appears across the specimen, perpendicular to both the current and the field. This is the Hall effect.

Let current density $J_x$ flow along $x$ and field $B_z$ along $z$. Moving charges experience the magnetic (Lorentz) force $qv_x B_z$, which deflects them sideways and builds up a transverse electric field $E_y$ until it balances the magnetic force:

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

With $J_x = nqv_x$ (carrier density $n$, charge $q$), $v_x = J_x/(nq)$, so

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

The Hall coefficient is defined as

$$\boxed{R_H = \frac{E_y}{J_x B_z} = \frac{1}{nq}}$$

Its sign gives the sign of the charge carriers: $R_H > 0$ for p-type (holes), $R_H < 0$ for n-type (electrons).

Applications: (i) determination of the type, sign and concentration of charge carriers; (ii) measurement of magnetic field (Hall probe / gaussmeter); also measurement of carrier mobility.

(b) Numerical

$R_H = +3.66\times10^{-4}\ \text{m}^3/\text{C}$. The positive sign indicates p-type material (holes are the majority carriers).

$$R_H = \frac{1}{nq} \Rightarrow n = \frac{1}{R_H\,q} = \frac{1}{(3.66\times10^{-4})(1.6\times10^{-19})}$$ $$n = \frac{1}{5.856\times10^{-23}} = 1.71\times10^{22}\ \text{m}^{-3}$$

Carriers are holes (p-type), concentration $n \approx 1.71\times10^{22}\ \text{m}^{-3}$.

(a) Polar vs non-polar dielectrics; types of polarization

Non-polar dielectrics: molecules in which the centres of positive and negative charge coincide, so they have no permanent dipole moment in the absence of a field (e.g. $H_2$, $N_2$, $O_2$, $CO_2$). A dipole is induced only when a field is applied.

Polar dielectrics: molecules with an asymmetric structure so the charge centres do not coincide, giving a permanent dipole moment even without a field (e.g. $H_2O$, $HCl$, $NH_3$). Without a field the dipoles are randomly oriented; a field tends to align them.

Types of polarization:

1. Electronic polarization – displacement of the electron cloud relative to the nucleus by the field; present in all dielectrics, very fast.
2. Ionic (atomic) polarization – relative displacement of positive and negative ions in ionic solids.
3. Orientational (dipolar) polarization – partial alignment of permanent dipoles in polar dielectrics; temperature dependent.
4. Space-charge (interfacial) polarization – accumulation of charges at interfaces/grain boundaries in inhomogeneous materials.

(b) Clausius–Mossotti relation

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

where

• $\varepsilon_r$ = relative permittivity (dielectric constant) of the material,
• $N$ = number of molecules (polarizable units) per unit volume,
• $\alpha$ = molecular polarizability (dipole moment induced per unit local field),
• $\varepsilon_0$ = permittivity of free space.

It relates the macroscopic dielectric constant $\varepsilon_r$ to the microscopic polarizability $\alpha$, accounting for the local (internal) field acting on each molecule, and is valid for non-polar dielectrics.

(a) Diamagnetic, paramagnetic and ferromagnetic materials

(b) B–H hysteresis loop

When a ferromagnetic specimen is taken through a complete cycle of magnetization, the plot of flux density $B$ against magnetizing field $H$ does not retrace itself but forms a closed loop — hysteresis (B lags behind H).

Description of the loop (described in words):

• Starting unmagnetized, increasing $H$ raises $B$ along the initial curve $O\to a$ up to saturation.
• Reducing $H$ to zero leaves a residual flux density $Ob$ = retentivity (remanence) $B_r$ — the material stays magnetized.
• Reversing $H$ to a value $Oc$ reduces $B$ to zero; this reverse field is the coercivity $H_c$ — the field needed to fully demagnetize the specimen.
• Continuing the cycle traces the full symmetric loop $a\!-\!b\!-\!c\!-\!d\!-\!e\!-\!f\!-\!a$.

Significance of the loop area: the area enclosed by the B–H loop equals the energy dissipated as heat (hysteresis loss) per unit volume per cycle of magnetization. Therefore:

• Soft magnetic materials (e.g. soft iron) have a thin loop (small area, low loss, low $H_c$) → suitable for transformer cores and electromagnets.
• Hard magnetic materials (e.g. steel) have a wide loop (large area, high retentivity and coercivity) → suitable for permanent magnets.

(a) Heisenberg's uncertainty principle

It is impossible to determine simultaneously and with unlimited precision both the position and the momentum of a particle. The product of the uncertainties satisfies

$$\Delta x\,\Delta p \gtrsim \frac{\hbar}{2}\quad\left(\hbar = \frac{h}{2\pi}\right)$$

Electron cannot exist inside the nucleus: A nucleus has radius $\sim 10^{-14}$ m, so if an electron were confined inside it, $\Delta x \le 10^{-14}$ m. Then

$$\Delta p \ge \frac{\hbar}{2\Delta x} = \frac{1.055\times10^{-34}}{2\times10^{-14}} \approx 5.3\times10^{-21}\ \text{kg·m/s}$$

The minimum momentum $p \approx \Delta p$, giving a (relativistic) energy $E \approx pc \approx (5.3\times10^{-21})(3\times10^8) \approx 1.6\times10^{-12}$ J $\approx 10$ MeV.

An electron bound in the nucleus would thus need an energy of the order of tens of MeV. But electrons emitted in $\beta$-decay have energies of only a few MeV, and no nuclear binding can supply ~tens of MeV to hold an electron. Hence an electron cannot exist inside the nucleus.

(b) de Broglie wavelength of an electron accelerated through 100 V

The kinetic energy gained is $eV$, and the wavelength is

$$\lambda = \frac{h}{\sqrt{2meV}} = \frac{12.27}{\sqrt{V}}\ \text{\AA} \quad(V \text{ in volts})$$

Substituting $V = 100$ V:

$$\lambda = \frac{12.27}{\sqrt{100}} = \frac{12.27}{10} = 1.227\ \text{\AA} = 1.227\times10^{-10}\ \text{m}$$

de Broglie wavelength $\lambda \approx 1.23\ \text{\AA} = 0.123$ nm.