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Section A: Long Answer Questions

Attempt any TWO questions.

3 questions·10 marks each
1long10 marks

Discuss the theory of Newton's rings. Derive expressions for the radii of bright and dark rings and explain the experimental determination of the wavelength of light.

Newton's Rings

Principle: Newton's rings are a system of concentric circular interference fringes produced when a plano-convex lens of large radius of curvature is placed on a flat glass plate. A thin air film of gradually increasing thickness is formed between the curved surface of the lens and the plate. Monochromatic light incident normally is partly reflected from the top and bottom of this air film, and the two reflected beams interfere, producing alternate bright and dark rings (interference by division of amplitude).

Path difference: For a film of thickness tt at near-normal incidence, the effective path difference between the two reflected rays is

Δ=2μtcosr+λ2\Delta = 2\mu t \cos r + \frac{\lambda}{2}

The extra λ/2\lambda/2 arises because reflection at the denser plate surface causes a phase change of π\pi. For air, μ=1\mu = 1 and at normal incidence cosr=1\cos r = 1, so Δ=2t+λ/2\Delta = 2t + \lambda/2.

Conditions for fringes:

  • Bright ring: 2t=(2n1)λ22t = (2n-1)\dfrac{\lambda}{2}
  • Dark ring: 2t=nλ2t = n\lambda

Relation between thickness and radius: If RR is the radius of curvature of the lens and rnr_n the radius of the ring where film thickness is tt, then from the geometry of the circle,

rn2=(2Rt)t2Rt(since tR)r_n^2 = (2R - t)t \approx 2Rt \quad (\text{since } t \ll R)

so t=rn22Rt = \dfrac{r_n^2}{2R}.

Radii of dark rings

For a dark ring, 2t=nλ2rn22R=nλ2t = n\lambda \Rightarrow 2\cdot\dfrac{r_n^2}{2R} = n\lambda

rn=nλR\boxed{r_n = \sqrt{n\lambda R}}

Thus the radius of dark rings is proportional to n\sqrt{n}, and the diameter Dn=2nλRD_n = 2\sqrt{n\lambda R}, i.e. DnnD_n \propto \sqrt{n}.

Radii of bright rings

For a bright ring, 2t=(2n1)λ2rn2R=(2n1)λ22t = (2n-1)\dfrac{\lambda}{2} \Rightarrow \dfrac{r_n^2}{R} = (2n-1)\dfrac{\lambda}{2}

rn=(2n1)λR2\boxed{r_n = \sqrt{\frac{(2n-1)\lambda R}{2}}}

The centre is dark because there t0t \to 0, giving path difference λ/2\lambda/2 (destructive interference).

Determination of wavelength

Measure the diameters of the nthn^{th} and mthm^{th} dark rings. Since Dn2=4nλRD_n^2 = 4n\lambda R,

Dm2Dn2=4(mn)λRD_m^2 - D_n^2 = 4(m-n)\lambda R

Therefore

λ=Dm2Dn24(mn)R\boxed{\lambda = \frac{D_m^2 - D_n^2}{4(m-n)R}}

Knowing RR (found independently, e.g. by a spherometer) and measuring ring diameters with a travelling microscope, the wavelength λ\lambda is determined. Using Dm2Dn2D_m^2-D_n^2 eliminates errors due to imperfect contact at the centre.

interference
2long10 marks

State and prove Gauss's law in electrostatics. Apply it to find the field due to an infinitely long line charge.

Gauss's Law in Electrostatics

Statement: The total electric flux through any closed surface is equal to 1/ε01/\varepsilon_0 times the net charge enclosed by that surface.

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

Proof (from Coulomb's law)

Consider a point charge qq at the centre of a sphere of radius rr. The field at the surface is radial with magnitude

E=14πε0qr2E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}

The flux through the sphere is

Φ=EdA=E4πr2=14πε0qr24πr2=qε0\Phi = \oint \vec{E}\cdot d\vec{A} = E \cdot 4\pi r^2 = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\cdot 4\pi r^2 = \frac{q}{\varepsilon_0}

The result is independent of rr. By drawing solid-angle cones, any arbitrary closed surface intercepts the same flux q/ε0q/\varepsilon_0 (each cone of solid angle dΩd\Omega contributes q4πε0dΩ\frac{q}{4\pi\varepsilon_0}d\Omega, and dΩ=4π\oint d\Omega = 4\pi). A charge outside the surface contributes zero net flux (flux entering equals flux leaving). By superposition, for many charges Φ=qi/ε0=qenc/ε0\Phi = \sum q_i/\varepsilon_0 = q_{enc}/\varepsilon_0. This proves Gauss's law.

Field due to an infinite line charge

Consider an infinitely long straight wire of uniform linear charge density λ\lambda (C/m). By symmetry E\vec{E} is radial and depends only on perpendicular distance rr. Choose a coaxial cylindrical Gaussian surface of radius rr and length LL.

  • Flux through the two flat ends =0= 0 (here E\vec{E} is parallel to the ends).
  • Flux through the curved surface =E(2πrL)= E\cdot(2\pi r L).
  • Charge enclosed =λL= \lambda L.

Applying Gauss's law:

E(2πrL)=λLε0E\,(2\pi r L) = \frac{\lambda L}{\varepsilon_0} E=λ2πε0r\boxed{E = \frac{\lambda}{2\pi\varepsilon_0 r}}

directed radially outward (for λ>0\lambda>0). The field falls off as 1/r1/r.

electrostatics
3long10 marks

Explain the propagation of light through an optical fiber. Derive expressions for the acceptance angle and numerical aperture, and discuss types of fibers.

Light Propagation in an Optical Fiber

An optical fiber has a cylindrical core of refractive index n1n_1 surrounded by a cladding of slightly lower index n2n_2 (n1>n2n_1 > n_2). Light launched into the core travels by repeated total internal reflection (TIR) at the core–cladding boundary, since the ray strikes the interface at an angle greater than the critical angle θc\theta_c where sinθc=n2/n1\sin\theta_c = n_2/n_1. The signal is thus guided along the fiber with very low loss.

Acceptance angle

Let a ray enter from a medium of index n0n_0 (usually air, n0=1n_0=1) at angle θ0\theta_0 to the fiber axis and refract into the core at angle θr\theta_r. At the core–cladding interface it strikes at angle (90θr)(90^\circ-\theta_r). For TIR this must equal or exceed θc\theta_c:

sinθrcosθc=1sin2θc=1n22n12\sin\theta_r \le \cos\theta_c = \sqrt{1-\sin^2\theta_c} = \sqrt{1-\frac{n_2^2}{n_1^2}}

At the input face, Snell's law gives n0sinθ0=n1sinθrn_0\sin\theta_0 = n_1\sin\theta_r. Substituting the limiting condition,

n0sinθ0,max=n11n22n12=n12n22n_0\sin\theta_{0,\max} = n_1\sqrt{1-\frac{n_2^2}{n_1^2}} = \sqrt{n_1^2 - n_2^2}

The acceptance angle is

θ0,max=sin1 ⁣(n12n22n0)\boxed{\theta_{0,\max} = \sin^{-1}\!\left(\frac{\sqrt{n_1^2-n_2^2}}{n_0}\right)}

Rays entering within the cone of half-angle θ0,max\theta_{0,\max} are guided.

Numerical aperture

NA=n0sinθ0,max=n12n22\boxed{NA = n_0\sin\theta_{0,\max} = \sqrt{n_1^2 - n_2^2}}

In terms of the fractional index difference Δ=(n1n2)/n1\Delta = (n_1-n_2)/n_1, NAn12ΔNA \approx n_1\sqrt{2\Delta}. NANA measures the light-gathering capacity of the fiber.

Types of fibers

  • Step-index single-mode fiber: very small core (~8–10 µm); supports only one mode; negligible intermodal dispersion; used for long-haul, high-bandwidth links.
  • Step-index multimode fiber: larger core (~50–62.5 µm); supports many modes; suffers intermodal dispersion; used for short distances.
  • Graded-index multimode fiber: core index decreases gradually from axis to edge, so rays follow curved paths and arrive nearly together, greatly reducing intermodal dispersion.
optical-fiber
B

Section B: Short Answer Questions

Attempt any EIGHT questions.

9 questions·5 marks each
4short5 marks

Explain the conditions for observing interference fringes.

Conditions for Sustained Interference

For steady, observable interference fringes the following conditions must be satisfied:

  • Coherence: The two sources must be coherent, i.e. emit waves of a constant (or zero) phase difference that does not change with time. This is normally achieved by deriving both beams from a single source (division of wavefront or amplitude).
  • Same frequency / wavelength: The two waves must be monochromatic and of the same frequency, otherwise the phase difference varies continuously and fringes wash out.
  • Comparable amplitude: The amplitudes (intensities) should be equal or nearly equal so that destructive interference gives near-zero intensity, producing good contrast between bright and dark fringes.
  • Small separation of sources and narrow source: The sources must be close together and the source slit narrow, so that fringe width β=λD/d\beta = \lambda D/d is large enough to be resolved.
  • Same polarization: For transverse light waves, the interfering waves should be in the same plane of polarization.

When these are met, regions of constructive interference (path difference =nλ= n\lambda) appear bright and destructive interference (path difference =(2n+1)λ/2= (2n+1)\lambda/2) appear dark.

interference
5short5 marks

Derive the energy density of an electric field.

Energy Density of an Electric Field

Consider a parallel-plate capacitor of plate area AA and separation dd, with capacitance C=ε0AdC = \dfrac{\varepsilon_0 A}{d}. The energy stored when charged to voltage VV is

U=12CV2=12ε0AdV2U = \frac{1}{2}CV^2 = \frac{1}{2}\frac{\varepsilon_0 A}{d}V^2

The field between the plates is uniform, E=V/dE = V/d, so V=EdV = Ed:

U=12ε0Ad(Ed)2=12ε0E2(Ad)U = \frac{1}{2}\frac{\varepsilon_0 A}{d}(Ed)^2 = \frac{1}{2}\varepsilon_0 E^2 (Ad)

Since the energy is stored in the field occupying the volume AdAd between the plates, the energy per unit volume (energy density) is

u=UAd=12ε0E2\boxed{u = \frac{U}{Ad} = \frac{1}{2}\varepsilon_0 E^2}

This result, though derived for a capacitor, is general: at any point where an electric field EE exists in vacuum, the stored electrostatic energy density is u=12ε0E2u = \tfrac{1}{2}\varepsilon_0 E^2 (in a dielectric, u=12εE2=12DEu = \tfrac{1}{2}\varepsilon E^2 = \tfrac{1}{2}\vec{D}\cdot\vec{E}).

capacitance
6short5 marks

State Ampere's law and obtain the magnetic field inside a solenoid.

Ampere's Law and the Solenoid

Ampere's circuital law: The line integral of the magnetic field B\vec{B} around any closed loop equals μ0\mu_0 times the total current enclosed by the loop:

Bdl=μ0Ienc\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc}

Field inside a long solenoid

Consider an ideal long solenoid with nn turns per unit length carrying current II. Inside the field is uniform and parallel to the axis; outside it is negligibly small. Choose a rectangular Amperian loop abcdabcd of length LL with side abab inside (parallel to the axis) and side cdcd outside.

Evaluating Bdl\oint \vec{B}\cdot d\vec{l}:

  • Along abab (inside): BLBL.
  • Along cdcd (outside): 00 (since B0B\approx 0 outside).
  • Along bcbc and dada: 00 (Bdl\vec{B}\perp d\vec{l}).

So Bdl=BL\oint \vec{B}\cdot d\vec{l} = BL.

The current enclosed by the loop is Ienc=(nL)II_{enc} = (nL)I, since nLnL turns thread the loop. Applying Ampere's law:

BL=μ0(nL)IBL = \mu_0 (nL) I B=μ0nI\boxed{B = \mu_0 n I}

The field inside a long solenoid is uniform and independent of position, depending only on the turn density nn and current II.

magnetism
7short5 marks

What are the characteristics of laser light?

Characteristics of Laser Light

Laser (Light Amplification by Stimulated Emission of Radiation) light differs from ordinary light in four main properties:

  • Monochromaticity: Laser light is almost perfectly single-wavelength, with an extremely narrow spectral linewidth, because all photons arise from the same atomic transition.
  • Coherence: Laser light is highly coherent, both temporally (constant phase over long time) and spatially (constant phase across the wavefront), since stimulated emission produces photons identical in phase.
  • Directionality: The beam is highly directional with very little angular spread (divergence), because only photons travelling along the cavity axis are amplified.
  • High intensity / brightness: Energy is concentrated into a narrow beam and very small bandwidth, giving very high intensity and brightness compared with conventional sources.

These properties make lasers useful in communication, holography, surgery, metrology, and material processing.

laser
8short5 marks

Write and solve the differential equation of simple harmonic motion.

Differential Equation of SHM and Its Solution

In simple harmonic motion the restoring force is proportional to displacement and oppositely directed: F=kxF = -kx. By Newton's second law, md2xdt2=kxm\dfrac{d^2x}{dt^2} = -kx, i.e.

d2xdt2+kmx=0d2xdt2+ω2x=0\frac{d^2x}{dt^2} + \frac{k}{m}x = 0 \quad\Rightarrow\quad \boxed{\frac{d^2x}{dt^2} + \omega^2 x = 0}

where ω=k/m\omega = \sqrt{k/m} is the angular frequency.

Solution

This is a second-order linear homogeneous equation with auxiliary equation D2+ω2=0D^2 + \omega^2 = 0, giving roots D=±iωD = \pm i\omega. The general solution is

x(t)=Acosωt+Bsinωtx(t) = A\cos\omega t + B\sin\omega t

or equivalently

x(t)=asin(ωt+ϕ)\boxed{x(t) = a\sin(\omega t + \phi)}

where a=A2+B2a = \sqrt{A^2+B^2} is the amplitude and ϕ\phi the initial phase, fixed by the initial conditions.

Verification: d2xdt2=aω2sin(ωt+ϕ)=ω2x\dfrac{d^2x}{dt^2} = -a\omega^2\sin(\omega t+\phi) = -\omega^2 x, which satisfies the equation.

The motion is periodic with time period

T=2πω=2πmk.T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}.
oscillations
9short5 marks

State the differential and integral forms of Gauss's law.

Gauss's Law: Integral and Differential Forms

Integral form: The total electric flux through any closed surface equals the net enclosed charge divided by ε0\varepsilon_0:

SEdA=qencε0=1ε0VρdV\oint_S \vec{E}\cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0} = \frac{1}{\varepsilon_0}\int_V \rho\,dV

where ρ\rho is the volume charge density.

Differential form: Applying the divergence theorem, SEdA=V(E)dV\oint_S \vec{E}\cdot d\vec{A} = \int_V (\nabla\cdot\vec{E})\,dV. Equating integrands,

E=ρε0\boxed{\nabla\cdot\vec{E} = \frac{\rho}{\varepsilon_0}}

This states that the divergence of the electric field at any point equals the local charge density divided by ε0\varepsilon_0; electric field lines originate from positive charge (ρ>0\rho>0) and terminate on negative charge. The two forms are equivalent — the differential form is the point (local) statement and the integral form the global statement of the same law.

electromagnetism
10short5 marks

Explain the diffraction grating and its resolving power.

Diffraction Grating and Resolving Power

Diffraction grating: A diffraction grating is an optical element consisting of a large number of equally spaced, identical parallel slits (lines). If the slit width is aa and the opaque spacing is bb, the grating element (period) is d=a+bd = a+b. When monochromatic light of wavelength λ\lambda falls normally on the grating, the condition for principal maxima is

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

where nn is the order of the spectrum and θ\theta the diffraction angle. Each order disperses different wavelengths to different angles, so the grating acts as a spectroscope.

Resolving power: The resolving power is the ability of the grating to separate two spectral lines of nearly equal wavelengths λ\lambda and λ+dλ\lambda+d\lambda. By the Rayleigh criterion it is defined as R=λ/dλR = \lambda/d\lambda, and for a grating

R=λdλ=nN\boxed{R = \frac{\lambda}{d\lambda} = nN}

where nn is the order of the spectrum and NN is the total number of illuminated lines on the grating. Thus the resolving power increases with the order and with the number of ruled lines; it is independent of the grating spacing.

diffraction
11short5 marks

Define total internal reflection and critical angle.

Total Internal Reflection and Critical Angle

Total internal reflection (TIR): When light travels from an optically denser medium (refractive index n1n_1) to a rarer medium (n2<n1n_2 < n_1), and the angle of incidence exceeds a certain value, the light is no longer refracted into the second medium but is completely reflected back into the denser medium. This phenomenon is called total internal reflection. It occurs only when (i) light goes from a denser to a rarer medium and (ii) the angle of incidence is greater than the critical angle.

Critical angle: The critical angle θc\theta_c is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 9090^\circ. From Snell's law,

n1sinθc=n2sin90n_1\sin\theta_c = n_2\sin 90^\circ sinθc=n2n1\boxed{\sin\theta_c = \frac{n_2}{n_1}}

For a denser medium to air (n2=1n_2 = 1), sinθc=1/n1\sin\theta_c = 1/n_1. For incidence angles greater than θc\theta_c, TIR occurs. This is the basic principle behind light guidance in optical fibers.

optical-fiber
12short5 marks

Write short notes on population inversion.

Population Inversion

Under normal thermal equilibrium, the population of atoms in energy levels follows the Boltzmann distribution, so a lower energy level E1E_1 is always more populated than a higher level E2E_2 (N1>N2N_1 > N_2, since N2/N1=e(E2E1)/kTN_2/N_1 = e^{-(E_2-E_1)/kT}). In such a situation absorption dominates over stimulated emission and no light amplification is possible.

Population inversion is the non-equilibrium condition in which the number of atoms in a higher (metastable) energy level exceeds that in a lower level, i.e.

N2>N1.N_2 > N_1.

When this condition is achieved, an incoming photon of energy hν=E2E1h\nu = E_2 - E_1 triggers more stimulated emission than absorption, so the light beam is amplified — the essential requirement for laser action.

Key points:

  • It is achieved by pumping (optical, electrical, or chemical) which raises atoms to higher levels.
  • A metastable state (relatively long lifetime, ~10310^{-3} s) is needed so that atoms accumulate there long enough to create the inversion.
  • A simple two-level system cannot produce a steady inversion; three-level (e.g. ruby) or four-level (e.g. He–Ne) systems are used.

Thus population inversion is the precondition that makes stimulated emission, and hence lasing, possible.

laser

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