BSc CSIT (TU) Science Physics (BSc CSIT, PHY113) Question Paper 2080 Nepal
This is the official BSc CSIT (TU) (Science stream) Physics (BSc CSIT, PHY113) question paper for 2080, 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 2080 paper is a great way to practise under real exam conditions.
Section A: Long Answer Questions
Attempt any TWO questions.
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 at near-normal incidence, the effective path difference between the two reflected rays is
The extra arises because reflection at the denser plate surface causes a phase change of . For air, and at normal incidence , so .
Conditions for fringes:
- Bright ring:
- Dark ring:
Relation between thickness and radius: If is the radius of curvature of the lens and the radius of the ring where film thickness is , then from the geometry of the circle,
so .
Radii of dark rings
For a dark ring,
Thus the radius of dark rings is proportional to , and the diameter , i.e. .
Radii of bright rings
For a bright ring,
The centre is dark because there , giving path difference (destructive interference).
Determination of wavelength
Measure the diameters of the and dark rings. Since ,
Therefore
Knowing (found independently, e.g. by a spherometer) and measuring ring diameters with a travelling microscope, the wavelength is determined. Using eliminates errors due to imperfect contact at the centre.
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 times the net charge enclosed by that surface.
Proof (from Coulomb's law)
Consider a point charge at the centre of a sphere of radius . The field at the surface is radial with magnitude
The flux through the sphere is
The result is independent of . By drawing solid-angle cones, any arbitrary closed surface intercepts the same flux (each cone of solid angle contributes , and ). A charge outside the surface contributes zero net flux (flux entering equals flux leaving). By superposition, for many charges . This proves Gauss's law.
Field due to an infinite line charge
Consider an infinitely long straight wire of uniform linear charge density (C/m). By symmetry is radial and depends only on perpendicular distance . Choose a coaxial cylindrical Gaussian surface of radius and length .
- Flux through the two flat ends (here is parallel to the ends).
- Flux through the curved surface .
- Charge enclosed .
Applying Gauss's law:
directed radially outward (for ). The field falls off as .
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 surrounded by a cladding of slightly lower index (). 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 where . The signal is thus guided along the fiber with very low loss.
Acceptance angle
Let a ray enter from a medium of index (usually air, ) at angle to the fiber axis and refract into the core at angle . At the core–cladding interface it strikes at angle . For TIR this must equal or exceed :
At the input face, Snell's law gives . Substituting the limiting condition,
The acceptance angle is
Rays entering within the cone of half-angle are guided.
Numerical aperture
In terms of the fractional index difference , . 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.
Section B: Short Answer Questions
Attempt any EIGHT questions.
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 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 ) appear bright and destructive interference (path difference ) appear dark.
Derive the energy density of an electric field.
Energy Density of an Electric Field
Consider a parallel-plate capacitor of plate area and separation , with capacitance . The energy stored when charged to voltage is
The field between the plates is uniform, , so :
Since the energy is stored in the field occupying the volume between the plates, the energy per unit volume (energy density) is
This result, though derived for a capacitor, is general: at any point where an electric field exists in vacuum, the stored electrostatic energy density is (in a dielectric, ).
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 around any closed loop equals times the total current enclosed by the loop:
Field inside a long solenoid
Consider an ideal long solenoid with turns per unit length carrying current . Inside the field is uniform and parallel to the axis; outside it is negligibly small. Choose a rectangular Amperian loop of length with side inside (parallel to the axis) and side outside.
Evaluating :
- Along (inside): .
- Along (outside): (since outside).
- Along and : ().
So .
The current enclosed by the loop is , since turns thread the loop. Applying Ampere's law:
The field inside a long solenoid is uniform and independent of position, depending only on the turn density and current .
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.
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: . By Newton's second law, , i.e.
where is the angular frequency.
Solution
This is a second-order linear homogeneous equation with auxiliary equation , giving roots . The general solution is
or equivalently
where is the amplitude and the initial phase, fixed by the initial conditions.
Verification: , which satisfies the equation.
The motion is periodic with time period
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 :
where is the volume charge density.
Differential form: Applying the divergence theorem, . Equating integrands,
This states that the divergence of the electric field at any point equals the local charge density divided by ; electric field lines originate from positive charge () 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.
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 and the opaque spacing is , the grating element (period) is . When monochromatic light of wavelength falls normally on the grating, the condition for principal maxima is
where is the order of the spectrum and 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 and . By the Rayleigh criterion it is defined as , and for a grating
where is the order of the spectrum and 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.
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 ) to a rarer medium (), 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 is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is . From Snell's law,
For a denser medium to air (), . For incidence angles greater than , TIR occurs. This is the basic principle behind light guidance in optical fibers.
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 is always more populated than a higher level (, since ). 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.
When this condition is achieved, an incoming photon of energy 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, ~ 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.
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