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Section A: Long Answer Questions

Attempt any TWO questions.

3 questions·10 marks each
1long10 marks

What is meant by polarization of light? Explain the production of plane-, circularly- and elliptically-polarized light, and state the law of Malus.

Polarization of Light

Polarization is the property of a transverse wave by which the vibrations of the electric field vector E\vec{E} are confined to a single plane (or follow a definite pattern) perpendicular to the direction of propagation. Ordinary (unpolarized) light has E\vec{E} vibrating randomly in all directions perpendicular to the ray; in polarized light the vibrations are restricted. Polarization confirms that light is a transverse wave.

Plane (Linearly) Polarized Light

When the tip of the electric vector traces a straight line as the wave advances, the light is plane-polarized. It is produced by passing unpolarized light through a polaroid or by reflection at the Brewster angle (tanθB=n\tan\theta_B = n). A single linear vibration can be written as:

Ey=acos(ωtkz),Ex=0E_y = a\cos(\omega t - kz), \qquad E_x = 0

Circularly and Elliptically Polarized Light

These are produced by superposing two mutually perpendicular plane-polarized waves of the same frequency with a phase difference δ\delta, e.g. by passing plane-polarized light through a birefringent crystal plate (quarter- or half-wave plate):

Ex=acos(ωt),Ey=bcos(ωtδ)E_x = a\cos(\omega t), \qquad E_y = b\cos(\omega t - \delta)
  • Circularly polarized: equal amplitudes (a=ba=b) and phase difference δ=π/2\delta = \pi/2 (a quarter-wave plate). The tip of E\vec{E} traces a circle.
  • Elliptically polarized: unequal amplitudes (aba \neq b), or δ0,π/2,π\delta \neq 0,\pi/2,\pi. The tip of E\vec{E} traces an ellipse. (Plane and circular polarization are special cases of the general ellipse.)

Law of Malus

When plane-polarized light of intensity I0I_0 passes through an analyser whose transmission axis makes an angle θ\theta with the plane of polarization, the transmitted intensity is:

I=I0cos2θ\boxed{I = I_0\cos^2\theta}

Thus II is maximum when θ=0\theta = 0^\circ (axes parallel) and zero when θ=90\theta = 90^\circ (axes crossed). This follows because only the component E0cosθE_0\cos\theta of the amplitude is transmitted, and intensity \propto amplitude2^2.

polarization
2long10 marks

Derive Maxwell's electromagnetic wave equation in free space and show that electromagnetic waves travel with the speed of light.

Maxwell's Equations in Free Space

In free space (ρ=0\rho = 0, J=0\vec{J}=0) Maxwell's equations are:

E=0(1)B=0(2)\nabla\cdot\vec{E}=0 \quad (1) \qquad \nabla\cdot\vec{B}=0 \quad (2) ×E=Bt(3)×B=μ0ε0Et(4)\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t} \quad (3) \qquad \nabla\times\vec{B}=\mu_0\varepsilon_0\frac{\partial\vec{E}}{\partial t} \quad (4)

Wave Equation for E\vec{E}

Take the curl of equation (3):

×(×E)=t(×B)\nabla\times(\nabla\times\vec{E}) = -\frac{\partial}{\partial t}(\nabla\times\vec{B})

Using the vector identity ×(×E)=(E)2E\nabla\times(\nabla\times\vec{E}) = \nabla(\nabla\cdot\vec{E}) - \nabla^2\vec{E} and substituting (1) (E=0\nabla\cdot\vec{E}=0) and (4):

2E=t(μ0ε0Et)-\nabla^2\vec{E} = -\frac{\partial}{\partial t}\left(\mu_0\varepsilon_0\frac{\partial\vec{E}}{\partial t}\right) 2E=μ0ε02Et2\boxed{\nabla^2\vec{E} = \mu_0\varepsilon_0\frac{\partial^2\vec{E}}{\partial t^2}}

Similarly, taking the curl of (4) and using (2) gives:

2B=μ0ε02Bt2\nabla^2\vec{B} = \mu_0\varepsilon_0\frac{\partial^2\vec{B}}{\partial t^2}

Speed of the Wave

These are standard three-dimensional wave equations of the form 2ψ=1v22ψt2\nabla^2\psi = \dfrac{1}{v^2}\dfrac{\partial^2\psi}{\partial t^2}. Comparing coefficients:

1v2=μ0ε0v=1μ0ε0\frac{1}{v^2} = \mu_0\varepsilon_0 \quad\Rightarrow\quad v = \frac{1}{\sqrt{\mu_0\varepsilon_0}}

Substituting μ0=4π×107H/m\mu_0 = 4\pi\times10^{-7}\,\text{H/m} and ε0=8.854×1012F/m\varepsilon_0 = 8.854\times10^{-12}\,\text{F/m}:

v=1(4π×107)(8.854×1012)3×108 m/s=cv = \frac{1}{\sqrt{(4\pi\times10^{-7})(8.854\times10^{-12})}} \approx 3\times10^{8}\ \text{m/s} = c

Since this equals the measured speed of light cc, Maxwell concluded that light is an electromagnetic wave and that all electromagnetic waves travel at speed cc in vacuum.

electromagnetism
3long10 marks

Explain the construction and working of a He-Ne laser. Compare it with a semiconductor (diode) laser.

He-Ne Laser

Construction

A He-Ne laser consists of a long, narrow discharge tube (typically 30-50 cm long, ~few mm bore) filled with a mixture of helium and neon gases in the ratio about 10:1 at low pressure (~1 torr). A high-voltage DC supply drives a discharge between electrodes. The tube is closed by an optical resonant cavity formed by two mirrors: one fully reflecting and one partially reflecting (output mirror), aligned parallel and often with Brewster-angle windows to give polarized output.

Working (Population Inversion)

  1. The electric discharge excites He atoms to the metastable states 21S2^1S and 23S2^3S by electron collisions.
  2. These He atoms transfer their energy to Ne atoms by resonant (collisional) energy transfer, because Ne has energy levels almost coincident with the He metastable levels. This selectively populates the Ne upper laser levels, producing population inversion.
  3. Stimulated emission between Ne levels gives laser output, the strongest line being 632.8 nm (red) (others at 1.15 µm and 3.39 µm).
  4. The cavity mirrors provide optical feedback, amplifying the beam; a continuous (CW), highly monochromatic, coherent beam emerges through the partial mirror.

He-Ne is a four-level gas laser giving continuous, low-power (~mW), highly coherent and monochromatic output.

Comparison with Semiconductor (Diode) Laser

FeatureHe-Ne LaserSemiconductor (Diode) Laser
Active mediumGas mixture (He + Ne)Doped semiconductor p-n junction (e.g. GaAs)
PumpingElectrical discharge (electron + atom collisions)Direct electric current injection (forward bias)
WavelengthFixed, 632.8 nm (red)Tunable by material/composition (IR to visible)
SizeLarge (tens of cm)Very small (sub-mm chip)
EfficiencyLow (~0.1%)High (tens of %)
Output powerLow (mW, CW)Low to moderate, can be pulsed/CW
Beam qualityExcellent coherence and monochromaticityPoorer; larger divergence, broader linewidth
Cost / lifetimeCostlier, bulkyCheap, compact, long-lived
ApplicationsHolography, interferometry, alignmentOptical fiber communication, CD/DVD, barcode, pointers
laser
B

Section B: Short Answer Questions

Attempt any EIGHT questions.

9 questions·5 marks each
4short5 marks

Distinguish between constructive and destructive interference.

Constructive vs Destructive Interference

When two coherent waves superpose, the resultant depends on their phase/path difference.

Constructive InterferenceDestructive Interference
Phase differenceδ=2nπ\delta = 2n\pi (even multiple of π\pi)δ=(2n+1)π\delta = (2n+1)\pi (odd multiple of π\pi)
Path differenceΔ=nλ\Delta = n\lambdaΔ=(2n+1)λ2\Delta = (2n+1)\frac{\lambda}{2}
ConditionWaves arrive in phase (crest meets crest)Waves arrive out of phase (crest meets trough)
Resultant amplitudeMaximum, A=a1+a2A = a_1 + a_2Minimum, A=a1a2A = \lvert a_1 - a_2\rvert
Resultant intensityMaximum, Imax=(I1+I2)2I_{max} = (\sqrt{I_1}+\sqrt{I_2})^2Minimum, Imin=(I1I2)2I_{min} = (\sqrt{I_1}-\sqrt{I_2})^2
AppearanceBright fringeDark fringe

For equal amplitudes aa, constructive gives A=2aA=2a (intensity 4a24a^2) while destructive gives A=0A=0 (intensity 00). Energy is not destroyed but redistributed from dark to bright regions.

interference
5short5 marks

Derive the capacitance of a cylindrical capacitor.

Capacitance of a Cylindrical Capacitor

Consider two coaxial conducting cylinders of length LL, inner radius aa and outer radius bb, with LbL \gg b (fringing neglected). Let the inner cylinder carry charge +Q+Q and the outer Q-Q, so the linear charge density is λ=Q/L\lambda = Q/L.

Field between the cylinders (Gauss's law)

Using a coaxial Gaussian cylinder of radius rr (a<r<ba < r < b) and length LL:

E(2πrL)=Qε0E=Q2πε0rLE\,(2\pi r L) = \frac{Q}{\varepsilon_0} \quad\Rightarrow\quad E = \frac{Q}{2\pi\varepsilon_0 r L}

Potential difference

V=baEdr=abQ2πε0Lrdr=Q2πε0Lln ⁣(ba)V = -\int_b^a E\,dr = \int_a^b \frac{Q}{2\pi\varepsilon_0 L r}\,dr = \frac{Q}{2\pi\varepsilon_0 L}\ln\!\left(\frac{b}{a}\right)

Capacitance

C=QV=2πε0Lln(b/a)\boxed{C = \frac{Q}{V} = \frac{2\pi\varepsilon_0 L}{\ln(b/a)}}

The capacitance depends only on the geometry (LL, aa, bb). If a dielectric of permittivity εr\varepsilon_r fills the gap, replace ε0\varepsilon_0 by εrε0\varepsilon_r\varepsilon_0.

capacitance
6short5 marks

State and explain the Biot-Savart law.

Biot-Savart Law

The Biot-Savart law gives the magnetic field produced by a small current element. A current element IdlI\,d\vec{l} produces, at a point P located at displacement r\vec{r} from the element, a magnetic field dBd\vec{B} given by:

dB=μ04πIdl×r^r2dB=μ04πIdlsinθr2d\vec{B} = \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{l}\times\hat{r}}{r^2} \qquad\Rightarrow\qquad dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\,\sin\theta}{r^2}

where θ\theta is the angle between dld\vec{l} and r\vec{r}, and μ0=4π×107T\cdotpm/A\mu_0 = 4\pi\times10^{-7}\,\text{T·m/A} is the permeability of free space.

Key points

  • dBIdB \propto I, dBdldB \propto dl, dBsinθdB \propto \sin\theta, and dB1/r2dB \propto 1/r^2.
  • The direction of dBd\vec{B} is perpendicular to the plane containing dld\vec{l} and r\vec{r}, given by the right-hand rule (the cross product dl×r^d\vec{l}\times\hat{r}).
  • dB=0dB = 0 along the axis of the element (θ=0\theta=0 or π\pi) and is maximum at θ=90\theta = 90^\circ.

The total field of an extended conductor is obtained by integration: B=μ0I4πdl×r^r2\vec{B} = \dfrac{\mu_0 I}{4\pi}\displaystyle\int \dfrac{d\vec{l}\times\hat{r}}{r^2}. For example, for an infinite straight wire it gives B=μ0I2πrB = \dfrac{\mu_0 I}{2\pi r}.

magnetism
7short5 marks

Explain the V-number of an optical fiber.

V-Number (Normalized Frequency) of an Optical Fiber

The V-number (or normalized frequency / VV-parameter) is a dimensionless quantity that determines the number of modes an optical fiber can support. It is defined as:

V=2πaλNA=2πaλn12n22V = \frac{2\pi a}{\lambda}\,\text{NA} = \frac{2\pi a}{\lambda}\sqrt{n_1^2 - n_2^2}

where aa is the core radius, λ\lambda is the free-space wavelength, n1n_1 and n2n_2 are the refractive indices of core and cladding, and NA=n12n22\text{NA} = \sqrt{n_1^2-n_2^2} is the numerical aperture.

Significance

  • If V2.405V \le 2.405, the fiber is single-mode (only the fundamental LM01\text{LM}_{01} mode propagates). V=2.405V=2.405 is the cut-off (first zero of the Bessel function J0J_0).
  • If V>2.405V > 2.405, the fiber is multimode.
  • For a multimode step-index fiber the number of supported modes is approximately:
NV22N \approx \frac{V^2}{2}

Thus VV increases with larger core radius and numerical aperture, and decreases with longer wavelength.

optical-fiber
8short5 marks

Define the time period and frequency of SHM.

Time Period and Frequency of SHM

Simple harmonic motion (SHM) is periodic motion in which the restoring force is proportional to displacement and directed toward the mean position, described by x=Asin(ωt+ϕ)x = A\sin(\omega t + \phi) with angular frequency ω=k/m\omega = \sqrt{k/m}.

Time Period (TT)

The time period is the time taken to complete one full oscillation (one complete cycle). It is given by:

T=2πω=2πmkT = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}

SI unit: second (s).

Frequency (ff)

The frequency is the number of complete oscillations per unit time. It is the reciprocal of the time period:

f=1T=ω2π=12πkmf = \frac{1}{T} = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

SI unit: hertz (Hz), i.e. cycles per second.

The two are related by ω=2πf\omega = 2\pi f and T=1/fT = 1/f. Both depend only on mm and kk, not on amplitude (isochronism).

oscillations
9short5 marks

State Stokes' theorem.

Stokes' Theorem

Stokes' theorem relates the surface integral of the curl of a vector field over an open surface SS to the line integral of the field around the closed boundary curve CC of that surface:

CFdl=S(×F)dS\boxed{\oint_C \vec{F}\cdot d\vec{l} = \iint_S (\nabla\times\vec{F})\cdot d\vec{S}}

where:

  • F\vec{F} is a continuously differentiable vector field,
  • dld\vec{l} is the line element along the closed curve CC,
  • dS=n^dSd\vec{S} = \hat{n}\,dS is the outward area element of the surface SS bounded by CC,
  • the direction of traversal of CC and the normal n^\hat{n} follow the right-hand rule.

In words: the circulation of a vector field around a closed loop equals the flux of its curl through any surface bounded by that loop. It is used in electromagnetism to convert Ampère's and Faraday's laws between integral and differential forms.

electromagnetism
10short5 marks

Explain the missing orders in a double-slit diffraction pattern.

Missing Orders in Double-Slit Diffraction

In a double-slit pattern, the resultant intensity is the product of the single-slit diffraction envelope and the double-slit interference pattern. Two conditions operate simultaneously (slit width ee, opaque gap so spacing between slit centres =e+d= e + d, often written (a+b)(a+b)):

Interference maxima:

(e+d)sinθ=nλ(n=0,1,2,)(e + d)\sin\theta = n\lambda \qquad (n = 0, 1, 2, \dots)

Diffraction minima (single slit):

esinθ=mλ(m=1,2,3,)e\sin\theta = m\lambda \qquad (m = 1, 2, 3, \dots)

Missing orders

A missing order occurs when the direction of an interference maximum coincides with a diffraction minimum — the diffraction envelope is zero there, so although interference predicts a bright fringe, no light reaches that point and the order is absent. Dividing the two equations:

e+de=nm\frac{e + d}{e} = \frac{n}{m}

So the orders nn that are missing depend on the ratio (e+d)/e(e+d)/e. For example:

  • If e+d=2ee + d = 2e (i.e. d=ed = e): orders n=2,4,6,n = 2, 4, 6, \dots are missing.
  • If e+d=3ee + d = 3e (i.e. d=2ed = 2e): orders n=3,6,9,n = 3, 6, 9, \dots are missing.

Thus missing orders arise purely from the geometric ratio of slit spacing to slit width.

diffraction
11short5 marks

Distinguish between step-index and graded-index fibers.

Step-Index vs Graded-Index Fiber

FeatureStep-Index FiberGraded-Index Fiber
Refractive index profileCore has uniform n1n_1; abrupt step down to cladding n2n_2Core index varies gradually, maximum at axis, decreasing toward cladding (parabolic)
Ray pathRays travel in zig-zag straight lines, reflecting at core-cladding boundary (total internal reflection)Rays follow smooth curved (sinusoidal) paths, continuously refracted toward axis
Modal dispersionHigh (different rays travel different path lengths)Low (off-axis rays travel faster in lower-index region, compensating path difference)
BandwidthLowerHigher
Pulse broadeningLargeSmall
Core diameter50-200 µm (multimode)~50 µm (multimode)
Index relationn(r)=n1n(r) = n_1 (constant)n(r)=n112Δ(r/a)αn(r) = n_1\sqrt{1 - 2\Delta(r/a)^\alpha}
Manufacture / costSimpler, cheaperMore complex, costlier
UseShort-distance / single-mode linksMedium-distance, higher data-rate links

In short, the key distinction is the index profile: abrupt (step) versus gradual (graded), which makes graded-index fibers superior in reducing intermodal dispersion.

optical-fiber
12short5 marks

Write short notes on the displacement current.

Displacement Current

Displacement current is the term introduced by Maxwell to account for a changing electric field acting as a source of magnetic field, just like a conduction current. It is not a flow of charge but arises from a time-varying electric flux.

Need for it

The original Ampère's law, Bdl=μ0I\oint\vec{B}\cdot d\vec{l} = \mu_0 I, fails for a charging capacitor: between the plates no conduction current flows, yet a magnetic field exists. Maxwell resolved this by adding a displacement current IdI_d.

Expression

The displacement current is defined through the rate of change of electric flux ΦE\Phi_E:

Id=ε0dΦEdt,Jd=ε0EtI_d = \varepsilon_0\frac{d\Phi_E}{dt}, \qquad J_d = \varepsilon_0\frac{\partial \vec{E}}{\partial t}

Modified Ampère-Maxwell law

Bdl=μ0(I+ε0dΦEdt)=μ0(I+Id)\oint \vec{B}\cdot d\vec{l} = \mu_0\left(I + \varepsilon_0\frac{d\Phi_E}{dt}\right) = \mu_0(I + I_d)

or in differential form ×B=μ0J+μ0ε0Et\nabla\times\vec{B} = \mu_0\vec{J} + \mu_0\varepsilon_0\dfrac{\partial\vec{E}}{\partial t}.

Significance

  • It makes the set of Maxwell's equations consistent (continuity of current).
  • It predicts the existence of electromagnetic waves, since a changing E\vec{E} generates B\vec{B} and vice versa.
electromagnetism

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