BSc CSIT (TU) Science Physics (BSc CSIT, PHY113) Question Paper 2077 Nepal
This is the official BSc CSIT (TU) (Science stream) Physics (BSc CSIT, PHY113) question paper for 2077, as set in the regular annual examination. It carries 60 full marks and a time allowance of 180 minutes, across 12 questions. On Kekkei you can attempt this Physics (BSc CSIT, PHY113) past paper online with a timer, get instant AI feedback and step-by-step solutions, and track the topics where you lose marks — completely free. Whether you are revising for your BSc CSIT (TU) Physics (BSc CSIT, PHY113) exam or solving previous years' question papers, this 2077 paper is a great way to practise under real exam conditions.
Section A: Long Answer Questions
Attempt any TWO questions.
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 — AlO (sapphire) doped with about 0.05% chromium ions (Cr). The Cr 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 Cr levels form a three-level scheme:
- — ground state.
- — a broad band of pump levels (in the blue–green, –).
- — a metastable level (long lifetime ).
Steps:
- The flash lamp pumps Cr ions from to the pump band (absorption).
- The ions decay rapidly and non-radiatively from to the metastable level , giving energy to the crystal lattice.
- Because is long-lived, ions accumulate there until the population of exceeds that of — a population inversion is achieved.
- A spontaneously emitted photon (, 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: (ground), (metastable), (pump band). An upward arrow marks pumping; a wavy downward arrow marks fast non-radiative decay; a downward arrow marks the lasing transition.
Characteristics
- Output wavelength (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.
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 times the net charge enclosed by that surface:
Setup
Consider a solid sphere of radius carrying total charge distributed uniformly with volume charge density
By symmetry is radial and depends only on the distance from the centre. Choose a concentric spherical Gaussian surface of radius , so .
(a) Field OUTSIDE the sphere ()
The whole charge is enclosed:
The sphere behaves as if all its charge were concentrated at the centre (point-charge field).
(b) Field INSIDE the sphere ()
Only the charge within radius is enclosed:
Therefore
Inside, (rises linearly from zero at the centre).
(c) At the surface ()
Both results give
the maximum value; the field is continuous at the surface, rising linearly inside and falling as outside.
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 rests on a flat glass plate, enclosing a thin air film of variable thickness. Monochromatic light of wavelength 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 at normal incidence is ; an additional phase change of (path ) occurs on reflection at the denser plate. The condition for a dark ring is
Relating thickness to ring radius
For a lens of radius , the film thickness at radial distance from the point of contact is (geometry of the circle):
Substituting into the dark-ring condition (air film, ):
Using diameter :
So the diameter of the dark ring is , i.e. .
Refractive index of a liquid
Introduce the liquid of refractive index between the lens and the plate. The dark-ring condition becomes , giving
Let = diameter of the dark ring with air, and = diameter of the same ring with the liquid. Then
Dividing:
Thus by measuring the diameter of the same ring (or, better, ) 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 .
Section B: Short Answer Questions
Attempt any EIGHT questions.
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 and coherence length .
- 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.
Derive the expression for capacitance of a spherical capacitor.
Capacitance of a Spherical Capacitor
Consider two concentric conducting spheres: inner radius carrying charge , outer radius carrying (with ).
Field in the gap ()
By Gauss's law on a concentric sphere of radius ,
Potential difference
Capacitance
Special case: If the outer sphere (), — the capacitance of an isolated sphere. If a dielectric of permittivity fills the gap, replace by .
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:
where is the number of turns, 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 , the area , or the angle between them (). The faster the flux changes, the larger the induced emf. This is the basis of generators, transformers and inductors.
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 surrounded by a cladding of slightly lower index (), with a protective jacket.
Principle
When light strikes the core–cladding boundary at an angle of incidence greater than the critical angle (where ), 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
- (core denser than cladding).
- Angle of incidence at the core–cladding interface .
- Light must enter within the acceptance angle so that all internal rays exceed .
Because TIR involves no transmission across the boundary, there is essentially no energy loss at each reflection, allowing low-loss, long-distance signal transmission.
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
where is the driving frequency.
Resonance: Resonance is the condition in which the driving frequency equals (or nearly equals) the natural frequency 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 .
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
which points in the direction of energy propagation and has units W/m.
Mathematical form:
or in integral form
where the energy density is
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 ) and (ii) the work done by the field on charges/currents (, 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.
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 illuminated normally by monochromatic light of wavelength .
Formation of the pattern
Each point of the slit acts as a secondary source (Huygens). Light diffracted at angle is brought to a point on the screen. The path difference between rays from the two edges of the slit is , and the resultant amplitude obtained by integrating over the slit is
The intensity is
Maxima and minima
- Central maximum: at , — the brightest, widest fringe.
- Minima (dark fringes): with , i.e.
- Secondary maxima: approximately midway between minima, at , with rapidly decreasing intensity (relative intensities ).
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 , so a narrower slit gives broader diffraction.
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 , measured from the fiber axis) for which a ray entering the fiber will undergo total internal reflection and be guided along the core:
Derivation
Let the core index be , cladding index (), and surrounding medium (air, ). A ray enters the end face at angle and refracts to angle inside the core; it then strikes the core–cladding boundary at angle .
For guiding, must equal the critical angle , where .
Refraction at the face (Snell's law):
Now
Therefore
With :
In terms of the fractional index difference , .
Write short notes on spontaneous and stimulated emission.
Spontaneous and Stimulated Emission
An atom in an excited state can return to a lower state by emitting a photon of energy . 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 (population of upper level), governed by the Einstein coefficient .
Stimulated Emission
- An incident photon of exactly energy induces the excited atom to drop to , 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 , governed by the Einstein coefficient .
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.
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