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A

Section A: Long Answer Questions

Attempt any TWO questions.

3 questions·10 marks each
1long10 marks

Discuss Fraunhofer diffraction at a single slit and derive the condition for minima and maxima. Obtain the expression for the intensity distribution.

Fraunhofer Diffraction at a Single Slit

Setup

A plane monochromatic wavefront of wavelength λ\lambda is incident normally on a slit of width aa. A converging lens focuses the diffracted light onto a screen. In Fraunhofer diffraction, both source and screen are effectively at infinity (parallel rays).

Divide the slit into a large number of imaginary point sources (Huygens secondary wavelets). Consider rays diffracted at an angle θ\theta.

Path difference and resultant amplitude

The path difference between wavelets from the two edges of the slit is asinθa\sin\theta. Let the phase difference between extreme rays be

2α=2πλasinθ,α=πasinθλ.2\alpha = \frac{2\pi}{\lambda}\,a\sin\theta,\qquad \alpha = \frac{\pi a \sin\theta}{\lambda}.

Summing the wavelets (vector/phasor addition) gives the resultant amplitude:

E=E0sinαα.E = E_0\,\frac{\sin\alpha}{\alpha}.

Intensity distribution

I=I0(sinαα)2,α=πasinθλ\boxed{\,I = I_0\left(\frac{\sin\alpha}{\alpha}\right)^2,\qquad \alpha=\frac{\pi a\sin\theta}{\lambda}\,}

where I0I_0 is the intensity at the centre (θ=0\theta=0, where sinαα1\frac{\sin\alpha}{\alpha}\to 1).

Condition for minima

I=0I=0 when sinα=0\sin\alpha=0 but α0\alpha\neq 0, i.e. α=mπ\alpha = m\pi (m=±1,±2,m=\pm1,\pm2,\dots):

asinθ=mλ,m=1,2,3,\boxed{\,a\sin\theta = m\lambda,\quad m=1,2,3,\dots\,}

These are the dark fringes.

Condition for maxima

  • Central (principal) maximum at θ=0\theta=0 (α=0\alpha=0), with intensity I0I_0.
  • Secondary maxima occur between consecutive minima, approximately where tanα=α\tan\alpha=\alpha, giving
asinθ=(2m+1)λ2,m=1,2,3,a\sin\theta = (2m+1)\frac{\lambda}{2},\quad m=1,2,3,\dots

The secondary maxima are weak; the first has intensity I0/22  (4.5%)\approx I_0/22 \;(\approx 4.5\%), the next I0/61\approx I_0/61, decreasing rapidly.

Intensity pattern

The pattern is a broad bright central maximum (angular half-width sinθ=λ/a\sin\theta=\lambda/a) flanked by symmetric, rapidly diminishing secondary maxima separated by minima. Most of the energy is concentrated in the central maximum.

diffraction
2long10 marks

Define electric potential. Derive the expression for the electric potential due to an electric dipole at any point.

Electric Potential

The electric potential at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point against the electric field, without acceleration.

V=Wq0,(SI unit: volt, V = J/C).V = \frac{W}{q_0},\qquad \text{(SI unit: volt, V = J/C)}.

For a point charge qq at distance rr:   V=14πε0qr.\;V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}. It is a scalar quantity.

Potential due to an Electric Dipole

A dipole consists of charges +q+q and q-q separated by distance 2l2l, with dipole moment p=q(2l)p = q(2l), directed from q-q to +q+q.

Let PP be a point at distance rr from the centre OO, making angle θ\theta with the dipole axis. Let r1r_1 and r2r_2 be the distances of PP from +q+q and q-q respectively.

The net potential is

V=14πε0(qr1qr2).V = \frac{1}{4\pi\varepsilon_0}\left(\frac{q}{r_1}-\frac{q}{r_2}\right).

For rlr \gg l, using geometry (the foot of perpendiculars):

r1rlcosθ,r2r+lcosθ.r_1 \approx r - l\cos\theta,\qquad r_2 \approx r + l\cos\theta.

Then

V=q4πε0(1rlcosθ1r+lcosθ)=q4πε02lcosθr2l2cos2θ.V = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{r-l\cos\theta}-\frac{1}{r+l\cos\theta}\right) = \frac{q}{4\pi\varepsilon_0}\cdot\frac{2l\cos\theta}{r^2-l^2\cos^2\theta}.

For rlr\gg l, l2cos2θl^2\cos^2\theta is negligible and q(2l)=pq(2l)=p:

V=14πε0pcosθr2=pr^4πε0r2\boxed{\,V = \frac{1}{4\pi\varepsilon_0}\,\frac{p\cos\theta}{r^2} = \frac{\vec p\cdot\hat r}{4\pi\varepsilon_0 r^2}\,}

Special cases

  • Axial point (θ=0\theta=0):   V=14πε0pr2\;V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{p}{r^2} (maximum, positive end).
  • Equatorial point (θ=90\theta=90^\circ):   V=0\;V = 0.

Note the dipole potential falls off as 1/r21/r^2, faster than the 1/r1/r of a point charge.

electrostatics
3long10 marks

Explain the principle, construction and working of an optical fiber as a communication medium. Discuss its advantages over conventional systems.

Optical Fiber as a Communication Medium

Principle

An optical fiber guides light by total internal reflection (TIR). When light travelling in a denser medium (core, refractive index n1n_1) strikes the boundary with a rarer medium (cladding, n2<n1n_2 < n_1) at an angle greater than the critical angle θc\theta_c, it is totally reflected back into the core. Repeated TIR carries the signal along the fiber with little loss.

sinθc=n2n1.\sin\theta_c = \frac{n_2}{n_1}.

The maximum acceptance angle is given by the numerical aperture:

NA=sinθmax=n12n22.NA = \sin\theta_{max} = \sqrt{n_1^2 - n_2^2}.

Construction

  1. Core – central thin glass/silica cylinder of higher refractive index n1n_1; carries the light.
  2. Cladding – surrounding layer of lower index n2n_2; confines light to the core by TIR.
  3. Buffer/Jacket – protective polymer coating giving mechanical strength and protection from moisture and damage.

Types: step-index (sharp core–cladding boundary) and graded-index (refractive index varies gradually), and single-mode vs multi-mode depending on core diameter.

Working in communication

  • The electrical message signal modulates a light source (LED or LASER diode) at the transmitter.
  • The optical signal travels along the fiber by TIR.
  • At the receiver, a photodetector (PIN/avalanche photodiode) converts light back into an electrical signal, which is amplified and demodulated.

Advantages over conventional (metallic) systems

  • Very large bandwidth / high data rate (uses optical frequencies).
  • Low transmission loss (attenuation) → long distances between repeaters.
  • Immune to electromagnetic and radio-frequency interference; no crosstalk.
  • Electrical isolation and safety (no spark hazard, no short-circuit).
  • Light weight and small size, easy to handle.
  • High security (difficult to tap) and uses cheap, abundant raw material (silica).
optical-fiber
B

Section B: Short Answer Questions

Attempt any EIGHT questions.

9 questions·5 marks each
4short5 marks

Explain the conditions for sustained interference of light.

Conditions for Sustained Interference of Light

For a steady, observable (sustained) interference pattern the following conditions must hold:

  1. Coherent sources – The two sources must maintain a constant phase difference with time (obtained from a single source by division of wavefront or amplitude).
  2. Same frequency / wavelength – The interfering waves must be monochromatic (or of the same colour) so the fringes do not wash out.
  3. Same or comparable amplitude – For good contrast, the amplitudes should be nearly equal so that minima are (nearly) dark and contrast is maximum.
  4. Same polarization plane – The waves should be polarized in the same plane (or be unpolarized similarly) to superpose effectively.
  5. Small path/source separation – The two sources should be narrow and close together, and the screen sufficiently far, so that fringes are wide enough to be resolved.

If these are satisfied, the resultant intensity I=I1+I2+2I1I2cosϕI = I_1 + I_2 + 2\sqrt{I_1 I_2}\cos\phi gives stable bright fringes (constructive, ϕ=2mπ\phi=2m\pi) and dark fringes (destructive, ϕ=(2m+1)π\phi=(2m+1)\pi).

interference
5short5 marks

Define dielectric constant and polarization.

Dielectric Constant and Polarization

Dielectric constant (relative permittivity, εr\varepsilon_r or KK): It is the ratio of the permittivity of a medium to that of free space:

εr=εε0.\varepsilon_r = \frac{\varepsilon}{\varepsilon_0}.

Equivalently, it is the factor by which the capacitance of a capacitor increases when the gap is filled with the dielectric, εr=C/C0\varepsilon_r = C/C_0, or the factor by which the field between charges is reduced. It is a dimensionless number (εr=1\varepsilon_r=1 for vacuum, 81\approx 81 for water).

Polarization (P\vec P): When a dielectric is placed in an electric field, its molecules develop (or align their) electric dipole moments. Polarization is the electric dipole moment induced per unit volume of the dielectric:

P=pV(SI unit: C/m2).\vec P = \frac{\sum \vec p}{V}\qquad(\text{SI unit: C/m}^2).

It is related to the field by P=ε0(εr1)E=ε0χeE\vec P = \varepsilon_0(\varepsilon_r - 1)\vec E = \varepsilon_0\chi_e\vec E, where χe\chi_e is the electric susceptibility.

capacitance
6short5 marks

State Lenz's law and explain its significance.

Lenz's Law

Statement: The direction of an induced current (or induced EMF) is always such that it opposes the change in magnetic flux that produces it.

Mathematically it gives the negative sign in Faraday's law:

ε=dΦBdt.\varepsilon = -\frac{d\Phi_B}{dt}.

For example, when a north pole is pushed toward a coil, the near face of the coil becomes a north pole to repel the magnet (opposing the increase in flux); when withdrawn, it becomes a south pole to attract it (opposing the decrease).

Significance

  • It gives the direction of induced current/EMF (Faraday's law gives only the magnitude).
  • It is a direct consequence of the law of conservation of energy — the opposition means external work must be done to change the flux, and this work appears as electrical energy. Without it, energy could be created indefinitely.
  • It explains effects such as eddy-current braking and the back-EMF in motors.
electromagnetic-induction
7short5 marks

Distinguish between LED and LASER.

Distinction between LED and LASER

FeatureLED (Light Emitting Diode)LASER (Light Amplification by Stimulated Emission of Radiation)
Emission processSpontaneous emissionStimulated emission (with population inversion)
CoherenceIncoherent lightHighly coherent (single phase)
MonochromaticityBroad spectral width (multi-wavelength)Highly monochromatic (narrow line)
DirectionalityDiverges widely (non-directional)Highly directional, narrow beam
Intensity / brightnessLowVery high, intense
Optical feedbackNo resonant cavity requiredRequires optical resonant cavity (mirrors)
Threshold currentNo threshold; light rises graduallyOperates above a threshold current
Cost & complexityCheap, simpleCostlier, more complex
Use in fiber comm.Short-distance, multimodeLong-distance, high-bandwidth, single-mode

Summary: A LED emits incoherent, divergent, polychromatic light by spontaneous emission, while a LASER emits coherent, highly directional, monochromatic, intense light by stimulated emission in a resonant cavity.

laser
8short5 marks

Write the differential equation of damped oscillation and define the quality factor.

Damped Oscillation and Quality Factor

Differential equation

For a body of mass mm executing oscillation under a restoring force kx-kx and a damping (resistive) force bdxdt-b\dfrac{dx}{dt} proportional to velocity:

md2xdt2+bdxdt+kx=0,m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx = 0,

or, dividing by mm,

d2xdt2+2βdxdt+ω02x=0\boxed{\,\frac{d^2x}{dt^2} + 2\beta\frac{dx}{dt} + \omega_0^2 x = 0\,}

where β=b2m\beta = \dfrac{b}{2m} is the damping constant and ω0=k/m\omega_0 = \sqrt{k/m} is the natural angular frequency.

For light (under-)damping the solution is

x(t)=A0eβtcos(ωt+ϕ),ω=ω02β2,x(t) = A_0 e^{-\beta t}\cos(\omega t + \phi),\qquad \omega = \sqrt{\omega_0^2 - \beta^2},

an oscillation whose amplitude decays exponentially.

Quality factor (Q-factor)

The quality factor measures how lightly a system is damped — the ratio of energy stored to energy lost per radian of oscillation:

Q=2πenergy storedenergy lost per cycle=ω02β=ω0mb.Q = 2\pi\frac{\text{energy stored}}{\text{energy lost per cycle}} = \frac{\omega_0}{2\beta} = \frac{\omega_0 m}{b}.

A high QQ means low damping, a sharp resonance, and many oscillations before dying out; a low QQ means heavy damping.

oscillations
9short5 marks

State Gauss's divergence theorem.

Gauss's Divergence Theorem

Statement: The total outward flux of a vector field A\vec A through a closed surface SS equals the volume integral of the divergence of A\vec A over the volume VV enclosed by that surface.

  SAdS=V(A)dV  \boxed{\;\oint_S \vec A \cdot d\vec S = \int_V (\nabla\cdot\vec A)\,dV\;}

It converts a closed surface integral into a volume integral. Physically, it states that the net flux leaving a closed region equals the total amount of "source" of the field distributed throughout the enclosed volume. It is used, for example, to derive the differential form of Gauss's law of electrostatics, E=ρ/ε0\nabla\cdot\vec E = \rho/\varepsilon_0, from its integral form.

electromagnetism
10short5 marks

Explain the formation of colours in thin films.

Formation of Colours in Thin Films

When white light falls on a thin transparent film (e.g. a soap bubble or oil layer on water of thickness tt and refractive index μ\mu), part of the light is reflected from the top surface and part from the bottom surface. These two reflected beams are coherent (derived from the same incident beam) and superpose, producing interference.

The effective path difference for light reflected at angle rr inside the film is

Δ=2μtcosr±λ2,\Delta = 2\mu t\cos r \pm \frac{\lambda}{2},

where the extra λ/2\lambda/2 arises because reflection at the denser (top) surface causes a phase change of π\pi.

  • Constructive interference (bright) for a wavelength when 2μtcosr=(2n+1)λ22\mu t\cos r = (2n+1)\dfrac{\lambda}{2}.
  • Destructive interference (dark) when 2μtcosr=nλ2\mu t\cos r = n\lambda.

Since white light contains all wavelengths, for a given thickness tt and viewing angle, only certain colours interfere constructively while others cancel. The film therefore appears coloured, and as tt or the angle of viewing changes, the colour seen changes. This is why soap bubbles and oil films show brilliant, shifting colours.

interference
11short5 marks

Define attenuation and dispersion in optical fibers.

Attenuation and Dispersion in Optical Fibers

Attenuation (signal loss): It is the reduction in optical power (intensity) of the signal as it propagates along the fiber, expressed in decibels per kilometre (dB/km):

α=10Llog10 ⁣(PinPout)  dB/km.\alpha = \frac{10}{L}\log_{10}\!\left(\frac{P_{in}}{P_{out}}\right)\;\text{dB/km}.

It is caused mainly by absorption (by impurities and the material itself), scattering (Rayleigh scattering by density fluctuations), and bending losses. Attenuation limits the distance a signal can travel before it must be regenerated.

Dispersion (pulse broadening): It is the spreading (broadening) of light pulses as they travel along the fiber, so that pulses overlap at the output. Main types:

  • Modal (intermodal) dispersion – different modes (ray paths) take different times in multimode fiber.
  • Material (chromatic) dispersion – refractive index varies with wavelength, so different wavelengths travel at different speeds.
  • Waveguide dispersion – arises from the fiber geometry.

Dispersion limits the bandwidth and maximum data rate of the fiber, because overlapping pulses cause intersymbol interference.

optical-fiber
12short5 marks

Write short notes on Maxwell's equations.

Maxwell's Equations

Maxwell's equations are the four fundamental equations that completely describe classical electromagnetism, linking electric field E\vec E and magnetic field B\vec B to charges and currents.

#Differential formIntegral formPhysical meaning
1. Gauss's law (electricity)E=ρε0\nabla\cdot\vec E = \dfrac{\rho}{\varepsilon_0}EdS=qε0\oint \vec E\cdot d\vec S = \dfrac{q}{\varepsilon_0}Electric charges are sources of E\vec E.
2. Gauss's law (magnetism)B=0\nabla\cdot\vec B = 0BdS=0\oint \vec B\cdot d\vec S = 0No magnetic monopoles; B\vec B lines are closed.
3. Faraday's law×E=Bt\nabla\times\vec E = -\dfrac{\partial \vec B}{\partial t}Edl=dΦBdt\oint \vec E\cdot d\vec l = -\dfrac{d\Phi_B}{dt}A changing magnetic field induces an electric field.
4. Ampère–Maxwell law×B=μ0J+μ0ε0Et\nabla\times\vec B = \mu_0\vec J + \mu_0\varepsilon_0\dfrac{\partial \vec E}{\partial t}Bdl=μ0I+μ0ε0dΦEdt\oint \vec B\cdot d\vec l = \mu_0 I + \mu_0\varepsilon_0\dfrac{d\Phi_E}{dt}Currents and changing electric fields produce B\vec B.

Significance: Together they predict electromagnetic waves propagating in vacuum at speed

c=1μ0ε03×108m/s,c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} \approx 3\times10^8\,\text{m/s},

showing that light is an electromagnetic wave. The term μ0ε0E/t\mu_0\varepsilon_0\,\partial\vec E/\partial t (the displacement current) was Maxwell's key addition.

electromagnetism

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