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Section A: Long Answer Questions

Attempt any TWO questions.

3 questions·10 marks each
1long10 marks

Explain the construction and working of a Ruby laser with an energy-level diagram.

Ruby Laser

The ruby laser, built by T. H. Maiman (1960), was the first working laser. It is a three-level, solid-state, optically-pumped pulsed laser.

Construction

  • Active medium: A cylindrical rod of synthetic ruby — Al2_2O3_3 (sapphire) doped with about 0.05% chromium ions (Cr3+^{3+}). The Cr3+^{3+} ions are the active centres; the alumina is the host.
  • The two ends of the rod are optically flat and parallel; one end is fully silvered and the other is partially silvered to form the optical resonant cavity (Fabry–Perot etalon).
  • Pumping source: A helical xenon flash lamp surrounds the rod and provides optical pumping. A power supply drives the lamp.
  • A cooling arrangement removes the heat generated.

Working and Energy Levels

The relevant Cr3+^{3+} levels form a three-level scheme:

  • E1E_1 — ground state.
  • E3E_3 — a broad band of pump levels (in the blue–green, 4000\sim 40005500A˚5500\,\text{Å}).
  • E2E_2 — a metastable level (long lifetime 3ms\sim 3\,\text{ms}).

Steps:

  1. The flash lamp pumps Cr3+^{3+} ions from E1E_1 to the pump band E3E_3 (absorption).
  2. The ions decay rapidly and non-radiatively from E3E_3 to the metastable level E2E_2, giving energy to the crystal lattice.
  3. Because E2E_2 is long-lived, ions accumulate there until the population of E2E_2 exceeds that of E1E_1 — a population inversion is achieved.
  4. A spontaneously emitted photon (λ=6943A˚\lambda = 6943\,\text{Å}, deep red) triggers stimulated emission; photons reflected back and forth by the mirrors build up an intense, coherent, monochromatic beam that emerges through the partially-silvered end.

Energy-level diagram (described)

Three horizontal lines bottom-to-top: E1E_1 (ground), E2E_2 (metastable), E3E_3 (pump band). An upward arrow E1E3E_1\to E_3 marks pumping; a wavy downward arrow E3E2E_3\to E_2 marks fast non-radiative decay; a downward arrow E2E1E_2\to E_1 marks the 6943A˚6943\,\text{Å} lasing transition.

Characteristics

  • Output wavelength 6943A˚6943\,\text{Å} (red), pulsed operation.
  • Drawback: being a three-level system, more than half the ions must be pumped out of the ground state, so a very high pump power threshold is needed.
laser
2long10 marks

State Gauss's law and use it to find the electric field due to a uniformly charged solid sphere at points inside and outside the sphere.

Gauss's Law and the Uniformly Charged Solid Sphere

Statement

Gauss's law states that the total electric flux through any closed surface equals 1/ε01/\varepsilon_0 times the net charge enclosed by that surface:

SEdA=qencε0.\oint_S \vec{E}\cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}.

Setup

Consider a solid sphere of radius RR carrying total charge QQ distributed uniformly with volume charge density

ρ=Q43πR3.\rho = \frac{Q}{\tfrac{4}{3}\pi R^3}.

By symmetry E\vec{E} is radial and depends only on the distance rr from the centre. Choose a concentric spherical Gaussian surface of radius rr, so EdA=E(4πr2)\oint \vec{E}\cdot d\vec{A} = E\,(4\pi r^2).

(a) Field OUTSIDE the sphere (r>Rr > R)

The whole charge QQ is enclosed:

E(4πr2)=Qε0    E=14πε0Qr2.E\,(4\pi r^2) = \frac{Q}{\varepsilon_0} \;\Rightarrow\; \boxed{E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}}.

The sphere behaves as if all its charge were concentrated at the centre (point-charge field).

(b) Field INSIDE the sphere (r<Rr < R)

Only the charge within radius rr is enclosed:

qenc=ρ43πr3=Qr3R3.q_{\text{enc}} = \rho\cdot\tfrac{4}{3}\pi r^3 = Q\frac{r^3}{R^3}.

Therefore

E(4πr2)=Qr3ε0R3    E=14πε0QrR3.E\,(4\pi r^2) = \frac{Q r^3}{\varepsilon_0 R^3} \;\Rightarrow\; \boxed{E = \frac{1}{4\pi\varepsilon_0}\frac{Q r}{R^3}}.

Inside, ErE \propto r (rises linearly from zero at the centre).

(c) At the surface (r=Rr = R)

Both results give

E=14πε0QR2,E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2},

the maximum value; the field is continuous at the surface, rising linearly inside and falling as 1/r21/r^2 outside.

electrostatics
3long10 marks

Derive the expression for the diameter of dark rings in Newton's rings experiment and explain how it is used to measure the refractive index of a liquid.

Newton's Rings — Diameter of Dark Rings and Refractive Index of a Liquid

Arrangement

A plano-convex lens of large radius of curvature RR rests on a flat glass plate, enclosing a thin air film of variable thickness. Monochromatic light of wavelength λ\lambda falls normally and is reflected; interference between rays reflected from the top and bottom of the air film produces concentric bright and dark rings (Newton's rings) viewed by reflection.

Path difference and condition for dark rings

The geometric path difference for a film of thickness tt at normal incidence is 2μt2\mu t; an additional phase change of π\pi (path λ/2\lambda/2) occurs on reflection at the denser plate. The condition for a dark ring is

2μt=nλ,n=0,1,2,2\mu t = n\lambda, \qquad n = 0,1,2,\dots

Relating thickness to ring radius

For a lens of radius RR, the film thickness at radial distance rr from the point of contact is (geometry of the circle):

t=r22R.t = \frac{r^2}{2R}.

Substituting into the dark-ring condition (air film, μ=1\mu = 1):

2r22R=nλ    rn2=nλR.2\cdot\frac{r^2}{2R} = n\lambda \;\Rightarrow\; r_n^2 = n\lambda R.

Using diameter Dn=2rnD_n = 2r_n:

Dn2=4nλR.\boxed{D_n^2 = 4n\lambda R}.

So the diameter of the nthn^{\text{th}} dark ring is Dn=2nλRD_n = 2\sqrt{n\lambda R}, i.e. DnnD_n \propto \sqrt{n}.

Refractive index of a liquid

Introduce the liquid of refractive index μ\mu between the lens and the plate. The dark-ring condition becomes 2μt=nλ2\mu t = n\lambda, giving

Dn2=4nλRμ.D_n^2 = \frac{4n\lambda R}{\mu}.

Let DnD_n = diameter of the nthn^{\text{th}} dark ring with air, and DnD_n' = diameter of the same ring with the liquid. Then

Dn2=4nλR(air),Dn2=4nλRμ(liquid).D_n^2 = 4n\lambda R \quad\text{(air)}, \qquad D_n'^2 = \frac{4n\lambda R}{\mu} \quad\text{(liquid)}.

Dividing:

μ=Dn2Dn2.\boxed{\mu = \frac{D_n^2}{D_n'^2}}.

Thus by measuring the diameter of the same ring (or, better, Dn+p2Dn2D_{n+p}^2 - D_n^2) with air and with the liquid present, the refractive index of the liquid is obtained. The liquid rings are smaller, since they contract by the factor 1/μ1/\sqrt{\mu}.

interference
B

Section B: Short Answer Questions

Attempt any EIGHT questions.

9 questions·5 marks each
4short5 marks

Define coherence and explain its importance in interference.

Coherence is the property of two or more light waves having a constant phase relationship over time and space (and the same frequency). Two sources are coherent if the phase difference between them does not change with time.

  • Temporal (longitudinal) coherence relates to monochromaticity — how long the wave maintains a fixed phase; measured by coherence time τc\tau_c and coherence length Lc=cτcL_c = c\tau_c.
  • Spatial (transverse) coherence relates to the phase correlation across the wavefront, set by the source size.

Importance in interference: A stationary, observable interference pattern is produced only when the superposing waves are coherent. If the phase difference varied randomly with time (incoherent sources), the bright and dark fringes would shift rapidly and average out, giving uniform illumination with no visible fringes. Hence coherence is essential to obtain stable, high-contrast fringes — which is why interference uses a single source split into two (e.g. Young's slits, Lloyd's mirror) or a laser.

interference
5short5 marks

Derive the expression for capacitance of a spherical capacitor.

Capacitance of a Spherical Capacitor

Consider two concentric conducting spheres: inner radius aa carrying charge +Q+Q, outer radius bb carrying Q-Q (with b>ab > a).

Field in the gap (a<r<ba < r < b)

By Gauss's law on a concentric sphere of radius rr,

E=14πε0Qr2.E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}.

Potential difference

V=baEdr=Q4πε0abdrr2=Q4πε0(1a1b)=Q4πε0baab.V = -\int_b^a E\,dr = \frac{Q}{4\pi\varepsilon_0}\int_a^b \frac{dr}{r^2} = \frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{a}-\frac{1}{b}\right) = \frac{Q}{4\pi\varepsilon_0}\,\frac{b-a}{ab}.

Capacitance

C=QV=4πε0abba.\boxed{C = \frac{Q}{V} = 4\pi\varepsilon_0\,\frac{ab}{b-a}}.

Special case: If the outer sphere \to\infty (bb\to\infty), C=4πε0aC = 4\pi\varepsilon_0 a — the capacitance of an isolated sphere. If a dielectric of permittivity ε=εrε0\varepsilon = \varepsilon_r\varepsilon_0 fills the gap, replace ε0\varepsilon_0 by ε\varepsilon.

capacitance
6short5 marks

State Faraday's laws of electromagnetic induction.

Faraday's Laws of Electromagnetic Induction

First law: Whenever the magnetic flux linked with a closed circuit changes, an electromotive force (emf) is induced in the circuit; the induced emf lasts only as long as the change in flux continues.

Second law: The magnitude of the induced emf is directly proportional to the rate of change of magnetic flux linkage:

ε=NdΦBdt,\varepsilon = -N\frac{d\Phi_B}{dt},

where NN is the number of turns, ΦB\Phi_B the flux through one turn, and the negative sign expresses Lenz's law — the induced emf opposes the change in flux that produces it (a consequence of conservation of energy).

Explanation: Flux can change by varying the field BB, the area AA, or the angle between them (ΦB=BAcosθ\Phi_B = BA\cos\theta). The faster the flux changes, the larger the induced emf. This is the basis of generators, transformers and inductors.

electromagnetic-induction
7short5 marks

Explain the principle of light propagation through an optical fiber.

Light Propagation Through an Optical Fiber

An optical fiber guides light by total internal reflection (TIR). It consists of a central core of refractive index n1n_1 surrounded by a cladding of slightly lower index n2n_2 (n1>n2n_1 > n_2), with a protective jacket.

Principle

When light strikes the core–cladding boundary at an angle of incidence greater than the critical angle θc\theta_c (where sinθc=n2/n1\sin\theta_c = n_2/n_1), it is totally internally reflected back into the core. Rays that enter the fiber within the acceptance cone repeatedly undergo TIR at the boundary and zig-zag down the length of the fiber with negligible loss, even around bends, until they emerge at the far end.

Conditions

  • n1>n2n_1 > n_2 (core denser than cladding).
  • Angle of incidence at the core–cladding interface >θc> \theta_c.
  • Light must enter within the acceptance angle so that all internal rays exceed θc\theta_c.

Because TIR involves no transmission across the boundary, there is essentially no energy loss at each reflection, allowing low-loss, long-distance signal transmission.

optical-fiber
8short5 marks

Define forced oscillation and resonance.

Forced oscillation: When a body capable of oscillating is acted upon by an external periodic (driving) force, it eventually vibrates with the frequency of the applied force (not its own natural frequency). Such sustained oscillation under an external periodic force is called a forced oscillation. The equation of motion is

mx¨+bx˙+kx=F0cosωt,m\ddot{x} + b\dot{x} + kx = F_0\cos\omega t,

where ω\omega is the driving frequency.

Resonance: Resonance is the condition in which the driving frequency equals (or nearly equals) the natural frequency ω0=k/m\omega_0 = \sqrt{k/m} of the system. At resonance the system absorbs maximum energy from the driver and the amplitude of oscillation becomes maximum (limited only by damping). Example: a swing pushed in time with its natural period, or an LCR circuit at ω=1/LC\omega = 1/\sqrt{LC}.

oscillations
9short5 marks

State and explain Poynting's theorem.

Poynting's Theorem

Statement: Poynting's theorem expresses conservation of energy for the electromagnetic field. The rate of flow of electromagnetic energy per unit area is given by the Poynting vector

S=1μ0E×B=E×H,\vec{S} = \frac{1}{\mu_0}\,\vec{E}\times\vec{B} = \vec{E}\times\vec{H},

which points in the direction of energy propagation and has units W/m2^2.

Mathematical form:

ut=S+JE,-\frac{\partial u}{\partial t} = \nabla\cdot\vec{S} + \vec{J}\cdot\vec{E},

or in integral form

ddtVudV=SSdA+VJEdV,-\frac{d}{dt}\int_V u\,dV = \oint_S \vec{S}\cdot d\vec{A} + \int_V \vec{J}\cdot\vec{E}\,dV,

where the energy density is

u=12(ε0E2+B2μ0).u = \frac{1}{2}\left(\varepsilon_0 E^2 + \frac{B^2}{\mu_0}\right).

Explanation: The theorem states that the rate of decrease of electromagnetic energy stored in a volume equals the sum of (i) the energy flowing out through the bounding surface (flux of S\vec{S}) and (ii) the work done by the field on charges/currents (JE\vec{J}\cdot\vec{E}, the ohmic dissipation). It thus accounts for where the field energy goes — it is either radiated away or converted into other forms — confirming energy conservation in electromagnetism.

electromagnetism
10short5 marks

Explain Fraunhofer diffraction at a single slit.

Fraunhofer Diffraction at a Single Slit

In Fraunhofer diffraction the source and screen are effectively at infinity (parallel light obtained with a collimating lens, pattern focused by a second lens). Consider a slit of width aa illuminated normally by monochromatic light of wavelength λ\lambda.

Formation of the pattern

Each point of the slit acts as a secondary source (Huygens). Light diffracted at angle θ\theta is brought to a point on the screen. The path difference between rays from the two edges of the slit is asinθa\sin\theta, and the resultant amplitude obtained by integrating over the slit is

A=A0sinββ,β=πasinθλ.A = A_0\frac{\sin\beta}{\beta}, \qquad \beta = \frac{\pi a\sin\theta}{\lambda}.

The intensity is

I=I0(sinββ)2.I = I_0\left(\frac{\sin\beta}{\beta}\right)^2.

Maxima and minima

  • Central maximum: at θ=0\theta = 0, I=I0I = I_0 — the brightest, widest fringe.
  • Minima (dark fringes): sinβ=0\sin\beta = 0 with β0\beta\neq 0, i.e.
asinθ=nλ,n=±1,±2,a\sin\theta = n\lambda, \quad n = \pm1, \pm2,\dots
  • Secondary maxima: approximately midway between minima, at asinθ=(n+12)λa\sin\theta = (n+\tfrac12)\lambda, with rapidly decreasing intensity (relative intensities 1,0.045,0.016,\approx 1, 0.045, 0.016,\dots).

Features

The pattern is a bright central band twice as wide as the others, flanked by progressively fainter secondary maxima. The angular half-width of the central maximum is sinθλ/a\sin\theta \approx \lambda/a, so a narrower slit gives broader diffraction.

diffraction
11short5 marks

Define numerical aperture and derive its expression.

Numerical Aperture of an Optical Fiber

Definition: The numerical aperture (NA) is the light-gathering capacity of an optical fiber. It is defined as the sine of the maximum angle of incidence (the acceptance angle θa\theta_a, measured from the fiber axis) for which a ray entering the fiber will undergo total internal reflection and be guided along the core:

NA=sinθa.\text{NA} = \sin\theta_a.

Derivation

Let the core index be n1n_1, cladding index n2n_2 (n1>n2n_1 > n_2), and surrounding medium n0n_0 (air, n0=1n_0 = 1). A ray enters the end face at angle θa\theta_a and refracts to angle θr\theta_r inside the core; it then strikes the core–cladding boundary at angle ϕ=90θr\phi = 90^\circ - \theta_r.

For guiding, ϕ\phi must equal the critical angle θc\theta_c, where sinθc=n2/n1\sin\theta_c = n_2/n_1.

Refraction at the face (Snell's law):

n0sinθa=n1sinθr=n1sin(90θc)=n1cosθc.n_0\sin\theta_a = n_1\sin\theta_r = n_1\sin(90^\circ-\theta_c) = n_1\cos\theta_c.

Now

cosθc=1sin2θc=1n22n12=n12n22n1.\cos\theta_c = \sqrt{1-\sin^2\theta_c} = \sqrt{1-\frac{n_2^2}{n_1^2}} = \frac{\sqrt{n_1^2 - n_2^2}}{n_1}.

Therefore

n0sinθa=n1n12n22n1=n12n22.n_0\sin\theta_a = n_1\cdot\frac{\sqrt{n_1^2-n_2^2}}{n_1} = \sqrt{n_1^2 - n_2^2}.

With n0=1n_0 = 1:

NA=sinθa=n12n22.\boxed{\text{NA} = \sin\theta_a = \sqrt{n_1^2 - n_2^2}}.

In terms of the fractional index difference Δ=(n1n2)/n1\Delta = (n_1 - n_2)/n_1, NAn12Δ\text{NA} \approx n_1\sqrt{2\Delta}.

optical-fiber
12short5 marks

Write short notes on spontaneous and stimulated emission.

Spontaneous and Stimulated Emission

An atom in an excited state E2E_2 can return to a lower state E1E_1 by emitting a photon of energy hν=E2E1h\nu = E_2 - E_1. This happens in two ways.

Spontaneous Emission

  • The excited atom decays to the lower state on its own, without any external trigger, after an average lifetime in the excited state.
  • The emitted photons have random phase, random direction and random polarization.
  • The light produced is incoherent and non-directional (e.g. ordinary lamps, fluorescent tubes).
  • Rate N2\propto N_2 (population of upper level), governed by the Einstein coefficient A21A_{21}.

Stimulated Emission

  • An incident photon of exactly energy hν=E2E1h\nu = E_2-E_1 induces the excited atom to drop to E1E_1, emitting a second photon.
  • The emitted photon is identical to the incident one — same frequency, phase, direction and polarization.
  • This gives coherent, monochromatic, directional, amplified light, and is the basis of laser action (Light Amplification by Stimulated Emission of Radiation).
  • Rate N2ρ(ν)\propto N_2\,\rho(\nu), governed by the Einstein coefficient B21B_{21}.

Key difference: Spontaneous emission gives one random photon and incoherent light; stimulated emission turns one photon into two identical photons, enabling coherent amplification (lasing) — provided a population inversion exists.

laser

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