Chapter 2
Physics Core: Mechanics, Waves, and Thermodynamics
TU BSc past exam questions
Past questions and answers
14 questions set from this chapter. Most repeated first.
- 2081 · PHY. 201 · 10 marks
Explain construction, working and applications of Michelson interferometer. How would you use it to measure the wavelength of monochromatic light.
OR
Describe how, with the help of a Nicol prism and quarter wave plate, plane polarised light, circularly polarised light and elliptically polarised light are produced and detected.
Answer
Michelson Interferometer
Construction
The Michelson interferometer consists of the following parts:
- Extended monochromatic source S — provides a parallel beam of light (usually a sodium lamp or laser).
- Beam splitter (half-silvered glass plate) G₁ — inclined at 45° to the incident beam. Its lower surface is partially silvered so that it divides the incident light into two parts: one reflected and one transmitted.
- Two plane mirrors M₁ and M₂ — highly polished, optically flat, and silvered on the front surface. M₁ is fixed; M₂ is mounted on a micrometer screw so that it can be moved parallel to itself.
- Compensating plate G₂ — identical to G₁ in material and thickness, placed parallel to G₁. It compensates for the extra path difference introduced because the reflected beam passes through G₁ three times while the transmitted beam passes through it only once.
- Telescope T — to observe the interference fringes.
Working
Light from the source S falls on the beam splitter G₁ at 45°. A part of it is reflected toward M₁ and the other part is transmitted toward M₂. After reflection from M₁ and M₂, the two beams return to G₁ and then proceed toward the telescope T.
The beam reflected by M₁ passes through G₁ three times, while the beam reflected by M₂ passes through G₁ only once. The plate G₂ equalises these paths so that both beams travel equal distances in glass. The two beams are coherent (they come from the same source) and interfere in the telescope, producing interference fringes.
If M₁ and M₂ are not exactly perpendicular to each other, the air film between them is wedge-shaped and straight fringes are obtained. If they are exactly perpendicular, circular fringes are obtained.
Applications
- Measurement of wavelength of monochromatic light.
- Measurement of refractive index of a thin transparent plate or a gas.
- Measurement of thickness of a thin sheet.
- Determination of the wavelength difference between two nearby spectral lines (e.g., sodium D₁ and D₂ lines).
- Measurement of the coherence length of a source.
- Detection of gravitational waves (in the advanced LIGO form).
Measurement of wavelength of monochromatic light
The micrometer screw is rotated to move M₂ until a bright circular fringe is at the centre of the field of view. The reading of the micrometer is noted. The mirror M₂ is then moved slowly so that N fringes cross the centre of the field of view. The final micrometer reading is noted. The distance d moved by M₂ is the difference of the two readings.
When M₂ is moved through a distance d, the path difference changes by 2d. Each time the path difference changes by one wavelength λ, one fringe crosses the centre. Hence, for N fringes:
Observation table
| No. of fringes N | Initial reading (mm) | Final reading (mm) | Displacement d (mm) |
|---|---|---|---|
| 20 | x₁ | x₂ | d = x₂ − x₁ |
Calculation
The wavelength of the monochromatic light is thus determined.
OR
Production and detection of polarised light using a Nicol prism and a quarter wave plate
1. Plane polarised light
Production: A beam of unpolarised light is passed through a Nicol prism. The Nicol prism transmits only the extraordinary ray (ordinary ray is absorbed by Canada balsam), so the emergent light is plane polarised.
Detection: The emergent beam is passed through a second Nicol prism (analyser) and viewed through a telescope. On rotating the analyser, the intensity of the transmitted light varies from a maximum to zero (complete extinction) twice in one full rotation. Complete extinction confirms that the light is plane polarised.
2. Circularly polarised light
Production: A quarter wave plate is placed in the path of a plane polarised beam such that its optic axis makes an angle of 45° with the plane of vibration of the incident light. The incident vibration is resolved into two equal components — one along the optic axis (extraordinary) and one perpendicular to it (ordinary). Since the plate introduces a path difference of λ/4, the two components differ in phase by π/2. Two equal amplitude vibrations differing in phase by π/2 combine to give circularly polarised light.
Detection: The light is passed through a rotating Nicol prism (analyser). If the intensity remains constant for all positions of the analyser (no variation on rotation), the light is circularly polarised. Alternatively, the light is first passed through a quarter wave plate and then through a rotating analyser; if complete extinction occurs twice in one rotation, the light is circularly polarised.
3. Elliptically polarised light
Production: A quarter wave plate is placed in the path of a plane polarised beam such that its optic axis makes an angle other than 45° (but not 0° or 90°) with the plane of vibration. The incident vibration is resolved into two unequal components along and perpendicular to the optic axis. The plate introduces a phase difference of π/2 between them. Two unequal amplitude vibrations differing in phase by π/2 combine to give elliptically polarised light.
Detection: The light is passed through a rotating Nicol prism. The intensity varies from a maximum to a minimum but never becomes zero (no complete extinction) in any position of the analyser. This confirms that the light is elliptically polarised.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · PHY. 201 · 10 marks
What do you mean by Zeeman effect and Stark effect? Discuss anomalous Zeeman effect.
OR
Explain Stern-Gerlach experiment, set up and its working. Discuss the results of this experiment concerning vector atom model. How did Sommerfeld modify Bohr's model?
Answer
Zeeman Effect and Stark Effect; Anomalous Zeeman Effect
Zeeman Effect
The Zeeman effect is the splitting of a single spectral line into several closely spaced components when the light source (atom) is placed in an external magnetic field. It was discovered by P. Zeeman in 1896. The splitting arises because the magnetic field interacts with the magnetic moment of the atom, lifting the degeneracy of the magnetic sublevels (characterised by the magnetic quantum number or ).
The energy shift of a level in a magnetic field is
where is the Bohr magneton and is the Lande g-factor.
- Normal Zeeman effect: A single line splits into three components (a triplet) with spacing . It occurs for atoms with total spin (singlet states), where only orbital motion contributes.
- Anomalous Zeeman effect: A line splits into more than three components (often an even number), with unequal spacings. It occurs when , i.e. when both orbital and spin magnetic moments are present.
Stark Effect
The Stark effect is the splitting (or shifting) of spectral lines when the light source is placed in an external electric field. Discovered by J. Stark in 1913. The electric field interacts with the electric dipole moment of the atom. The energy shift is
The linear Stark effect (shift proportional to ) occurs in hydrogen-like atoms where states of different but same are degenerate. The quadratic Stark effect (shift proportional to ) occurs in atoms where such degeneracy is absent. The pattern is symmetric about the original line position.
Anomalous Zeeman Effect
When an atom has a nonzero resultant spin (), the total magnetic moment is the vector sum of the orbital and spin magnetic moments:
Because the ratio of magnetic moment to angular momentum is different for orbital () and spin () parts, the resultant is not antiparallel to the total angular momentum . The component of along is what precesses about and produces the splitting. This is described by the Lande g-factor:
The energy of a level in the field is
Since differs from level to level, the selection rule gives components whose spacings are unequal and whose number is generally more than three — this is the anomalous Zeeman effect. It could not be explained by the old Bohr model; it required the vector atom model with electron spin and the space-quantisation of .
Example: The sodium D line () splits into 4 components, and the D line () into 6 components — a classic case of anomalous Zeeman splitting.
OR
Stern-Gerlach Experiment
Principle
The experiment (O. Stern and W. Gerlach, 1922) demonstrates space quantisation of angular momentum and provides direct evidence for electron spin. A beam of neutral atoms with a magnetic moment, passing through a non-uniform magnetic field, experiences a net force and is deflected.
Set-up
- Source: A small oven produces a beam of silver (Ag) atoms by evaporation; collimating slits make it a narrow beam.
- Magnetic field: The beam passes between specially shaped pole pieces — one sharply pointed, the other grooved — producing a strongly inhomogeneous magnetic field () perpendicular to the beam.
- Detector: A cold glass plate (or photographic plate) collects the atoms; the deposit is examined after the experiment.
Working
A magnetic dipole of moment in a field has energy . In a non-uniform field the force is
Since is quantised, the beam should split into a discrete number of beams instead of spreading continuously. For orbital quantum number , the number of beams is .
Results
- A beam of silver atoms split into two distinct beams, symmetric about the undeflected direction.
- This was unexpected: silver's ground state is with , so orbital motion predicts no splitting ().
- The observed two beams correspond to and — direct proof of electron spin with .
Significance for the Vector Atom Model
- Space quantisation is real: The direction of (or ) in space is quantised, giving discrete orientations.
- Electron spin confirmed: The two beams prove the electron has an intrinsic angular momentum (spin) with two orientations.
- Vector atom model validated: The model treats , , and as vectors that precess and couple (L-S coupling), which successfully explains the observed splitting.
- Magnetic moment of electron: The deflection confirms the electron's magnetic moment of one Bohr magneton, .
Sommerfeld's Modification of Bohr's Model
Bohr's model assumed circular orbits and explained only the hydrogen spectrum. Sommerfeld (1916) extended it as follows:
- Elliptical orbits: Electrons move in ellipses with the nucleus at one focus. The major axis is fixed by the principal quantum number , and the minor axis by the azimuthal (orbital) quantum number (now ). For a given , takes values , giving possible orbits.
- Relativistic correction: Since the electron near the nucleus moves with high speed, relativistic variation of mass with velocity was included. This makes orbits of the same but different have slightly different energies — lifting the degeneracy.
- Fine structure: This energy difference explains the fine structure of spectral lines (closely spaced components), which Bohr's model could not.
- Introduction of quantum numbers: It introduced the azimuthal quantum number (later ) and the magnetic quantum number , forming the basis of the vector atom model.
- Selection rule: Transitions obey , consistent with observed spectra.
Thus Sommerfeld's model gave a more accurate picture of atomic structure and laid the foundation for the vector atom model and the later quantum mechanical treatment.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · PHY. 201 · 10 marks
Draw the circuit diagram of a class A FET amplifier and explain its operation.
OR
Discuss Thevenin's and Norton's Theorem with their applications. Write down the steps to a typical electric circuit through Norton's Theorem.
Answer
Class A FET Amplifier
Circuit Diagram
+V_DD
┌───────┴───────┐
│ │
R_1 R_D
│ ├────┤├──────── Output (v_o)
│ │ C_out
C_in │ D
──┤├───┼───────────────G (n-channel JFET)
│ S
R_G ├──────┐
│ │ │
GND R_S C_S
│ │
GND GND
Components: R_G (gate resistor to ground, sets a high input resistance; the optional R_1 is used only with voltage-divider bias), R_D (drain load), R_D (drain load), R_S (source resistor), C_S (source bypass capacitor), C_in and C_out (coupling capacitors), V_DD (supply).
Operation
-
Biasing (self-bias): The gate is held at DC ground through R_G. The drain current I_D flows through R_S, developing a voltage V_S = I_D R_S. Hence V_GS = −I_D R_S. The bias point (Q-point) is the intersection of this bias line with the transfer characteristic I_D = I_DSS(1 − V_GS/V_P)². This gives a stable Q-point in the middle of the active region.
-
Quiescent state: With no input signal, DC voltages V_GS, I_D, V_DS set the operating point. C_in and C_out block DC, so the DC bias is undisturbed.
-
With input signal: The AC input v_in is applied between gate and ground. Since the JFET is voltage-driven, v_gs varies the gate–source voltage about the Q-point. Because the JFET operates in the depletion mode, a positive half-cycle of v_in (less negative V_GS) increases I_D, and a negative half-cycle decreases I_D. Thus the drain current varies in phase with the input.
-
Output: The varying I_D flows through R_D, producing a voltage drop i_d R_D. The drain voltage is v_ds = −i_d R_D (the negative sign shows the output is 180° out of phase with the input). This amplified AC voltage is coupled to the load through C_out.
-
C_S action: C_S bypasses R_S for AC, so no AC degeneration occurs and the gain remains high. The voltage gain is A_v = −g_m (R_D ∥ r_d ∥ R_L) ≈ −g_m R_D.
-
Class A feature: The Q-point is placed at the centre of the load line, so the device conducts for the full 360° of the input cycle. The output is an exact amplified replica of the input, but the efficiency is low (maximum 25% for a resistive load).
OR
Thevenin's and Norton's Theorems
Thevenin's Theorem
Any linear bilateral network with several sources and resistances can be replaced, as seen from two terminals, by a single voltage source V_th in series with a single resistance R_th.
- V_th = open-circuit voltage across the terminals.
- R_th = resistance seen from the terminals with all independent sources killed (voltage sources shorted, current sources opened).
Norton's Theorem
Any linear bilateral network can be replaced, as seen from two terminals, by a single current source I_N in parallel with a resistance R_N.
- I_N = short-circuit current through the terminals.
- R_N = same as R_th.
Relation between the two
Applications
- Simplifies the analysis of complex networks to a simple two-terminal equivalent.
- Used to find the current through or voltage across a particular branch (load) without solving the whole network.
- Helps in maximum power transfer problems (R_L = R_th).
- Used in circuit design, matching networks, and analysis of electronic circuits such as amplifiers.
- Reduces repeated calculations when the load is varied.
Steps to Solve a Circuit by Norton's Theorem
- Remove the load resistance R_L from the terminals a–b.
- Find I_N: Short-circuit terminals a–b and calculate the current flowing through the short (using mesh, nodal, or superposition methods).
- Find R_N: Kill all independent sources (short voltage sources, open current sources) and calculate the equivalent resistance looking into terminals a–b.
- Draw the Norton equivalent: A current source I_N in parallel with R_N.
- Reconnect the load R_L across a–b.
- Find the load current by the current-divider rule:
- Find the load voltage: V_L = I_L R_L.
Answer checked by Claude Sonnet · Oct 21, 2026
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- 2081 · PHY. 201 · 8 marks
How diffraction pattern can be obtained through a single and a double slit? Discuss difference between single and double slit diffractions.
Answer
Single-slit diffraction
When monochromatic light of wavelength λ passes through a single narrow slit of width (comparable to λ) and falls on a screen at distance , the secondary wavelets from every point of the slit interfere. The path difference between wavelets from the two edges of the slit, for a ray diffracted at angle θ, is . The slit is divided into zones; at the observation point:
- Minima (dark fringes): ,
- Secondary maxima (bright):
- Central maximum: at θ = 0, of width (on screen) , much brighter and broader than the others.
The intensity falls off rapidly on either side of the central maximum.
Double-slit diffraction
Light from a narrow source is allowed to fall on two parallel slits separated by distance (each slit narrow). The wavelets from the two slits superpose on the screen:
- Bright fringes: path difference ,
- Dark fringes:
- Fringe width , all fringes equally spaced and of equal intensity (modulated by the single-slit envelope of each slit).
Difference between single- and double-slit diffraction
| Single-slit diffraction | Double-slit diffraction |
|---|---|
| Produced by one slit; wavefront from one aperture diffracts. | Produced by two coherent slits; interference of two diffracted beams. |
| Condition: (minima). | Condition: (maxima). |
| Central fringe is brightest and widest; intensity decreases rapidly. | Fringes are equally spaced and of nearly equal intensity. |
| Fringe width depends on slit width : . | Fringe width depends on slit separation : . |
| Pattern is due to division of wavefront within one slit. | Pattern is due to superposition of waves from two slits. |
| Fewer, broader fringes are observed. | Many sharp, closely spaced fringes are observed. |
Thus, a single slit gives a diffraction pattern with a broad central maximum, while two slits give an interference pattern of equally spaced bright and dark fringes within the diffraction envelope.
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- 2081 · PHY. 201 · 8 marks
Explain the laws of successive radioactive disintegration. Use this law to explain any one natural radioactive series.
Answer
Laws of Successive Radioactive Disintegration
When a radioactive nuclide decays, it produces a daughter nuclide which may itself be radioactive. The daughter then decays to a granddaughter, and so on, forming a radioactive series (also called a decay chain). The number of undecayed atoms of each member changes with time according to the laws of successive disintegration.
1. Parent nuclide (first member)
Let a parent nuclide P have decay constant and initial number of atoms at . Its decay follows the ordinary radioactive decay law:
The rate of production of the daughter nuclide is , while its rate of decay is , where is the decay constant of the daughter.
2. Daughter nuclide (second member)
Let be the number of daughter atoms present at time . The net rate of change of is
This is a linear differential equation. Solving it with the condition at gives
3. Granddaughter nuclide (third member)
Similarly, for the third member with decay constant ,
whose solution is
4. General member (nth member)
For the -th member of the series,
5. Secular equilibrium
If the parent is very long-lived compared with the daughter, i.e. , then after sufficient time and the daughter decays with the same decay constant as the parent:
This condition is called secular equilibrium, and the whole series then decays with the half-life of the parent.
Application: The Uranium (4n + 2) Series
The uranium series starts from and ends at the stable lead isotope . Every mass number in this series satisfies , so it is called the 4n + 2 series.
Decay scheme:
Explanation using the law:
- Each -decay reduces the mass number by 4 and the atomic number by 2, while each -decay leaves the mass number unchanged but increases the atomic number by 1. Hence all members keep the form .
- The parent has a very long half-life ( years) compared with its daughter ( days). Since , the series attains secular equilibrium, and the activity of each member becomes equal to that of the parent.
- The number of atoms of any member at equilibrium is given by , so a member with a short half-life has fewer atoms but the same activity.
- The final product is stable, so the chain terminates there. The total mass lost as -particles is , i.e. 8 -particles, and the corresponding 6 -particles restore the atomic number from to .
Answer checked by Claude Sonnet · Oct 21, 2026
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- 2081 · PHY. 201 · 8 marks
Draw a circuit following a stable multivibrator and explain its operation. What is the frequency of the circuit?
Answer
Astable multivibrator
An astable multivibrator has no stable state. It switches continuously between two quasi-stable states and gives a square wave without any trigger.
Circuit diagram
+Vcc ──┬──────────┬────────────┬──────────┬──
R3 R1 R2 R4
│ │ │ │
│ ├────┤├──────┼──────────┤ C1: collector Q1 -> base Q2
│ │ │ │
├────┤├────┼────────────┤ │ C2: collector Q2 -> base Q1
│ │ │ │
B1─┐ C C ┌─B2
[ Q1 ]──┘ └──[ Q2 ]
│ │
GND GND
(Wires that cross at a ┼ are not joined.)
- are the collector load resistors of .
- are the base resistors, joining the bases of to .
- couples the collector of to the base of ; couples the collector of to the base of .
Operation
- Initial state. Because of slight differences between the components, one transistor conducts faster. Say is ON and is OFF. The collector of is at about 0 V and the collector of is at about .
- held OFF. When turned ON, its collector fell by nearly . Through this pushed the base of well below 0 V, so is cut off. Now discharges through towards , so the base voltage of rises exponentially with time constant . Meanwhile charges through and the base-emitter junction of , which keeps ON.
- Switching. When the base of reaches about 0.7 V, starts to conduct. Its collector voltage falls, and through this drives the base of negative. turns OFF (regenerative action), and its collector rises to .
- Second state. Now is ON and is OFF. discharges through with time constant , while recharges through . When the base of reaches 0.7 V, turns ON again.
- The cycle repeats indefinitely, and a square wave appears at each collector, the two outputs being in antiphase.
Frequency
stays OFF for and stays OFF for . Hence
For the symmetrical circuit, and :
The duty cycle is 50%, and the frequency is changed by varying or .
Answer checked by Claude Sonnet · Oct 21, 2026
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- 2081 · PHY. 201 · 6 marks
Answer any TWO of the following.
a) Explain in brief the dispersive and resolving power of grating.
b) Explain in brief about astigmatism.
c) What is common mode rejection ratio (CMRR) in operational amplifier?
d) Distinguish Rutherford and Compton scattering.
Answer
a) Dispersive and resolving power of a grating
Dispersive power measures how widely a grating spreads light of different wavelengths. It is defined as the rate of change of the angle of diffraction with wavelength :
where is the order of the spectrum and is the grating element (slit spacing). A larger value means the wavelengths are spread farther apart. It is greatest in higher orders and at large diffraction angles.
Resolving power measures the ability of the grating to separate two wavelengths that are very close together. It is defined as
where is the smallest wavelength difference that can just be resolved and is the total number of lines on the illuminated part of the grating. Thus resolving power increases with the order and with the total number of ruled lines .
b) Astigmatism
Astigmatism is a defect of vision (or of a lens) in which the eye cannot focus both horizontal and vertical lines sharply at the same time. It arises when the refracting surfaces of the cornea or lens are not perfectly spherical but have different curvatures in different meridians — the surface is more curved in one direction than in the perpendicular direction.
Because of this, rays in the vertical plane and rays in the horizontal plane come to focus at different points (two focal lines), so a point object is imaged as a short line or a blurred patch rather than a point. A person with astigmatism sees objects distorted — lines in one direction appear sharp while lines at right angles to them appear blurred.
Correction: It is corrected by using a cylindrical lens (or a combination of cylindrical lenses) whose curvature is opposite to that of the defective surface, so that the two focal lines are brought together on the retina. Spectacle lenses with cylindrical power are prescribed after measuring the axis of the defect.
c) Common mode rejection ratio (CMRR)
An operational amplifier has two inputs. A common-mode signal is one applied equally to both inputs (e.g. noise or interference picked up by both input lines), while a differential signal is the difference between the two inputs. An ideal op-amp should amplify only the differential signal and completely reject the common-mode signal.
The common-mode rejection ratio (CMRR) measures how well an op-amp rejects the common-mode signal. It is defined as the ratio of the differential voltage gain to the common-mode voltage gain :
It is usually expressed in decibels:
A high CMRR (typically –) means the amplifier responds strongly to the wanted difference signal while suppressing unwanted common-mode noise, which is essential in instrumentation and measurement circuits.
d) Rutherford scattering vs Compton scattering
| Rutherford scattering | Compton scattering |
|---|---|
| Scattering of alpha particles by atomic nuclei. | Scattering of X-ray photons by loosely bound electrons. |
| Governed by Coulomb force between the positively charged nucleus and the alpha particle. | Governed by the photon–electron collision, treated by conservation of energy and momentum. |
| The wavelength of the particle is unchanged; only its direction changes. | The wavelength of the scattered photon increases (Compton shift), . |
| Classical treatment; explains large-angle deflection of alpha particles. | Requires quantum (photon) treatment; gives direct evidence for the particle nature of light. |
| Cross-section follows the Rutherford formula, . | Cross-section is given by the Klein–Nishina formula. |
In short, Rutherford scattering involves charged particles deflected by the nuclear Coulomb field with no change in wavelength, whereas Compton scattering involves a photon transferring part of its energy to an electron, so its wavelength increases.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · PHY. 201 · 10 marks
Answer all questions
a) What are the applications of holography?
b) Briefly describe the objectives of Large Hadron Collider (LHC).
c) What is gravitational red shift?
d) Describe X-OR gate with truth table.
Answer
(a) Applications of holography
Holography is the technique of recording and reconstructing the complete wavefront (both amplitude and phase) of light using the interference of a reference beam and an object beam. Its main applications are:
- Three-dimensional imaging (holographic display): Produces true 3-D images with full depth perception, used in art, advertising, museums and security features on currency and credit cards.
- Holographic interferometry: Detects very small surface deformations (of the order of a wavelength) and is used in non-destructive testing of tyres, turbine blades, aircraft parts and in stress–strain analysis.
- Data storage: Holographic memory stores information throughout the volume of a crystal (page-oriented, high density, fast parallel access).
- Holographic optical elements (HOEs): Used as lenses, gratings, filters and beam splitters in optical systems, scanners and head-up displays.
- Medical and biological imaging: Holographic microscopy and holographic endoscopy give 3-D images of internal organs, cells and tissues.
- Security and anti-counterfeiting: Holograms on banknotes, ID cards, passports, CDs and product labels.
- Particle size analysis and velocimetry: Holography of moving particles (e.g. in fluid flow) measures size, position and velocity.
- Acoustic holography: Used in underwater (sonar) imaging and medical ultrasound.
(b) Objectives of the Large Hadron Collider (LHC)
The LHC is the world's largest and most powerful particle accelerator, built by CERN near Geneva in a 27 km underground ring. Its main objectives are:
- To search for the Higgs boson and confirm the mechanism by which elementary particles acquire mass (the Higgs mechanism) — achieved in 2012.
- To search for new physics beyond the Standard Model, such as supersymmetric particles, extra dimensions and new gauge bosons.
- To study the nature of dark matter by looking for weakly interacting massive particles (WIMPs) produced in high-energy collisions.
- To explain the matter–antimatter asymmetry of the universe by studying differences between matter and antimatter (CP violation) in B-mesons and other particles.
- To reproduce conditions just after the Big Bang (quark–gluon plasma) and study the properties of the early universe.
- To test the Standard Model precisely at the highest energies (up to about 14 TeV in proton–proton collisions) and to search for violations of its predictions.
(c) Gravitational red shift
The gravitational red shift is the increase in the wavelength (shift towards the red end of the spectrum) of light when it travels from a region of stronger gravitational field to a region of weaker gravitational field, i.e. from a massive body outwards to a distant observer.
According to Einstein's general theory of relativity, a photon climbing out of a gravitational potential well loses energy. Since its energy is (E = h\nu), a decrease in energy means a decrease in frequency (\nu) and hence an increase in wavelength (\lambda). The fractional shift is
[ \frac{\Delta \lambda}{\lambda} = \frac{\Delta \nu}{\nu} = \frac{GM}{Rc^{2}} ]
where (M) and (R) are the mass and radius of the body, (G) is the gravitational constant and (c) is the speed of light. For the Sun this shift is about (2\times 10^{-6}), and for a white dwarf it is much larger. It has been verified experimentally by the Pound–Rebka experiment and by observations of the spectral lines of white dwarfs.
(d) X-OR gate
The X-OR (Exclusive-OR) gate is a two-input logic gate whose output is HIGH (1) only when the two inputs are different, and LOW (0) when the inputs are the same. Its Boolean expression is
[ Y = A \oplus B = A\bar{B} + \bar{A}B ]
Truth table
| (A) | (B) | (Y = A \oplus B) |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
The X-OR gate is used as a parity checker/generator, in half-adders and full-adders (sum bit), in comparators and in controlled inverters.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · PHY. 201 · 5 marks
A Ramsden's eyepiece is to have an effective focal length of 3 cm. Calculate the focal length of its lenses and their distance of separation.
Answer
For a Ramsden’s eyepiece, the two lenses have equal focal length (f), and their separation is
[ d=\frac{2f}{3} ]
The equivalent focal length (F) of two thin lenses separated by distance (d) is
[ \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2} ]
Here (f_1=f_2=f), so
[ \frac{1}{F}=\frac{2}{f}-\frac{d}{f^2} ]
Substitute (d=\frac{2f}{3}):
[ \frac{1}{F}=\frac{2}{f}-\frac{2f/3}{f^2} ]
[ \frac{1}{F}=\frac{2}{f}-\frac{2}{3f} ]
[ \frac{1}{F}=\frac{4}{3f} ]
[ F=\frac{3f}{4} ]
Given effective focal length (F=3\ \text{cm}),
[ 3=\frac{3f}{4} ]
[ f=4\ \text{cm} ]
So each lens has focal length
[ \boxed{f=4\ \text{cm}} ]
Distance of separation:
[ d=\frac{2f}{3}=\frac{2\times 4}{3} ]
[ d=\frac{8}{3}=2.67\ \text{cm} ]
[ \boxed{d=2.67\ \text{cm}} ]
Answer: Focal length of each lens (=4\ \text{cm}), and separation between the lenses (=2.67\ \text{cm}).
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- 2081 · PHY. 201 · 5 marks
How many orders will be observed by a grating having 4000 lines/cm if it is illuminated by a visible light of wave length in the range of 4000 Å to 7500 Å?
Answer
Given:
Number of lines, (N = 4000\ \text{lines/cm})
So grating spacing, [ d=\frac{1}{N}=\frac{1}{4000}\ \text{cm} =2.5\times 10^{-4}\ \text{cm} =2.5\times 10^{-6}\ \text{m} =25000\ \text{Å} ]
For a grating, [ d\sin\theta = n\lambda ]
For the (n)th order to be observed, (\sin\theta \le 1), so [ n \le \frac{d}{\lambda} ]
For the shortest wavelength, (\lambda_{\min}=4000\ \text{Å}): [ n_{\max}=\frac{25000}{4000}=6.25 ] Hence the highest possible order is [ n=6 ]
For the longest wavelength, (\lambda_{\max}=7500\ \text{Å}): [ n_{\max}=\frac{25000}{7500}=3.33 ] So up to (n=3), all wavelengths in the given range are present. Orders (4,5,6) appear only for shorter wavelengths.
Therefore, the grating will produce orders [ n=0,\ \pm1,\ \pm2,\ \pm3,\ \pm4,\ \pm5,\ \pm6 ]
So the maximum order observed is (\boxed{6}).
On each side of the central maximum there are (\boxed{6}) orders; including the central order and both sides, the total number of orders is (\boxed{13}).
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · PHY. 201 · 5 marks
Calculate the time required for 10% of a sample of thorium to disintegrate. Assume the half life of thorium to be years.
Answer
Given:
Half-life of thorium, (T_{1/2}=1.4\times 10^{10}) years
Fraction disintegrated (=10%), so fraction remaining (=90%=0.9)
Let initial amount be (N_0). After time (t), remaining amount is
[
N=0.9N_0
]
Using radioactive decay law, [ N=N_0\left(\frac12\right)^{t/T_{1/2}} ]
[ 0.9N_0=N_0\left(\frac12\right)^{t/T_{1/2}} ]
[ 0.9=\left(\frac12\right)^{t/T_{1/2}} ]
Taking natural logarithm, [ \ln(0.9)=\frac{t}{T_{1/2}}\ln\left(\frac12\right) ]
[ t=T_{1/2}\frac{\ln(0.9)}{\ln(0.5)} ]
Substituting values, [ t=1.4\times 10^{10}\times \frac{-0.10536}{-0.69315} ]
[ t=1.4\times 10^{10}\times 0.1520 ]
[ t=2.128\times 10^9\ \text{years} ]
[ \boxed{t \approx 2.13\times 10^9\ \text{years}} ]
So, about 2.13 billion years are required for 10% of thorium to disintegrate.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · PHY. 201 · 5 marks
X-rays of wavelength m are incident on free stationary electrons. What is the wavelength of the X-ray that are scattered directly backwards? What is the energy gained by the scattered electron's in this case.
Given : kg
m/s
JS
Answer
Given: m, scattering angle (directly backwards).
Wavelength of the scattered X-ray. The Compton shift is
For , , so m and
Energy gained by the electron. It equals the energy lost by the photon:
Answer checked by Claude Sonnet · Oct 21, 2026
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- 2081 · PHY. 201 · 5 marks
A dc and an ac voltmeter were used to measure the output voltage from a filter circuit. If the readings of the two voltmeters are 25 volt and 2.5 volt respectively, calculate the ripple factor of the filter circuit.
Answer
Given:
- DC voltmeter reading, (V_{\text{dc}} = 25\ \text{V})
- AC voltmeter reading (rms ripple voltage), (V_{\text{ac(rms)}} = 2.5\ \text{V})
Formula: [ \text{Ripple factor},\ r = \frac{V_{\text{ac(rms)}}}{V_{\text{dc}}} ]
Substitution: [ r = \frac{2.5}{25} = 0.1 ]
Result: [ \boxed{r = 0.1} ]
So, the ripple factor of the filter circuit is 0.1, i.e. 10%.
Answer checked by Claude Haiku · Oct 21, 2026
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- 2081 · PHY. 201 · 5 marks
Convert 110011 and 101101 into their decimal number and add them.
Answer
Binary to decimal.
Sum: .
Check in binary: .
Answer checked by Claude Sonnet · Oct 21, 2026
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Questions from TU 4 Yrs BSc 2nd Yr Linear Algebra 2082 (Bachelor Level (4 Yrs.) · MAT 201 (New Course) · 2082), TU 4 Years BSc 2nd Yr PHY 201 2081 (Bachelor (4 Yrs.) · PHY. 201 · 2081), TU 4 Years BSc 4th Yr Environmental and (Bachelor Level [4 Yrs.] · Environment and Biodiversity (Bot. 401) · 2082) and TU 4 Years BSc 4th Yr Material Science (Bachelor Level (4 Yrs.) · Material Science (Phy.405) · 2074). Answers are written for this site; check them against your class notes.