Interference by Division of Amplitude and Division of Wavefront; Spatial and Temporal Coherence

1. Interference by Division of Wavefront

In this method, a single wavefront from a primary source is divided into two (or more) parts by allowing it to pass through two closely spaced apertures or by reflection from two inclined mirrors. The divided parts travel different paths and are then superposed to produce interference.

Examples:

• Young's double slit experiment: A cylindrical wavefront from a narrow slit S falls on two slits S₁ and S₂. The secondary wavelets from S₁ and S₂ are coherent and interfere on a screen.
• Fresnel's biprism: Light from a slit is divided into two virtual images by two prisms, which act as two coherent sources.
• Lloyd's mirror: The direct wave and the wave reflected from a mirror interfere.

Condition: The two divided parts must originate from the same wavefront, so the path difference must be less than the coherence length of the source. The two sources must be very close together (separation comparable to wavelength) for the fringes to be visible.

2. Interference by Division of Amplitude

Here the amplitude (intensity) of a wave is divided into two or more parts by partial reflection and partial transmission at the surfaces of a thin film or a beam splitter. The divided beams travel different paths and recombine to give interference.

Examples:

• Thin film interference: Light reflected from the upper and lower surfaces of a soap film or oil film interferes, giving coloured fringes.
• Michelson interferometer: A beam splitter divides the beam into two, which are reflected back and recombined.
• Newton's rings: Interference between light reflected from the curved surface of a lens and the flat glass plate.

Condition: The two beams must have a constant phase difference; this requires the source to be sufficiently monochromatic and the path difference to be within the coherence length.

3. Spatial Coherence

Spatial coherence describes the correlation between the waves at two different points in space at the same instant, i.e., the extent to which the wavefront is uniform in phase across its cross-section.

• A point source or a very small source gives high spatial coherence because the wavefront is nearly spherical and the phase difference between any two points on it is well defined.
• An extended source gives low spatial coherence because different points on the source emit independently, so the wavefront is irregular.
• Measure: The spatial coherence is measured by the coherence width (or transverse coherence length), which is the maximum separation of two points on the wavefront for which the phase difference remains constant. For a source of angular diameter θ, the coherence width is approximately

$l_c \approx \frac{\lambda}{\theta}$

4. Temporal Coherence

Temporal coherence describes the correlation between the waves at the same point at two different instants of time, i.e., the extent to which the wave maintains a constant phase over time.

• It depends on the monochromaticity of the source. A perfectly monochromatic source has infinite temporal coherence; a source with bandwidth Δλ has finite coherence time.
• Measure: The temporal coherence is measured by the coherence time τ_c and the coherence length L_c:

$\tau_c = \frac{1}{\Delta\nu}, \qquad L_c = c\,\tau_c = \frac{\lambda^2}{\Delta\lambda}$

where Δν is the frequency bandwidth and Δλ the wavelength bandwidth.

• For interference to occur, the path difference between the two waves must be less than the coherence length L_c.

Summary Table

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OR

Plane Transmission Grating: Working, Theory, Number of Maxima and Resolving Power

1. Construction

A plane transmission grating consists of a large number of equidistant parallel slits of equal width ruled on a transparent plate (glass or quartz). The ruling is done by a diamond point on a dividing engine. The region between two consecutive slits is opaque. If N lines are ruled per unit length, the grating element (grating constant) is

$d = \frac{1}{N}$

For a good grating, N is of the order of 5000 to 6000 lines per cm, so d ≈ 2 × 10⁻⁴ cm, comparable to the wavelength of visible light.

2. Theory of the Grating

Let a plane wavefront of monochromatic light of wavelength λ fall normally on the grating. Each slit acts as a secondary source of coherent waves (Huygens' principle). The waves from all slits superpose in the direction making an angle θ with the normal.

Path difference between waves from two consecutive slits:

$\Delta = d\sin\theta$

Condition for principal maxima (bright fringes):

$d\sin\theta = n\lambda, \qquad n = 0, 1, 2, 3, \dots$

where n is the order of the spectrum. This is the grating equation.

Condition for minima (dark fringes):

$d\sin\theta = \left(m + \frac{1}{N}\right)\lambda$

where N is the total number of lines on the grating and m is an integer.

3. Number of Diffraction Maxima

From the grating equation,

$\sin\theta = \frac{n\lambda}{d}$

Since sin θ cannot exceed 1, the maximum possible order is

$n_{\max} \le \frac{d}{\lambda}$

The total number of principal maxima on one side of the central maximum is n_max, and the total number of maxima (including the central one) is

$\text{Total maxima} = 2n_{\max} + 1$

For example, if d = 2 × 10⁻⁴ cm and λ = 5 × 10⁻⁵ cm, then n_max = 4, so the total number of maxima is 2(4) + 1 = 9.

3. Resolving Power of the Grating

The resolving power of a grating is its ability to separate two closely spaced wavelengths λ and λ + dλ in the same order of the spectrum. It is defined as

$R = \frac{\lambda}{d\lambda}$

Expression: By Rayleigh's criterion, two wavelengths are just resolved when the principal maximum of one falls on the first minimum of the other. The condition gives

$R = \frac{\lambda}{d\lambda} = nN$

where n is the order of the spectrum and N is the total number of lines on the grating.

Thus the resolving power increases with the order and with the total number of lines illuminated. A grating with a larger number of lines and used in a higher order gives better resolution.

Franck–Hertz experiment

Set-up: A tube containing mercury vapour at low pressure has a heated filament (cathode), a grid and a collector plate. Electrons emitted from the filament are accelerated towards the grid by a voltage $V$; a small retarding potential (about 0.5 V) between grid and plate stops electrons that have lost most of their energy.

Observation: The plate current rises with $V$, then falls sharply at $V = 4.9$ V, rises again, and falls again at 9.8 V, 14.7 V, ... (multiples of 4.9 V).

Explanation: Below 4.9 eV the collisions are elastic and electrons keep their energy. At 4.9 eV an electron transfers all its energy to a mercury atom in an inelastic collision, exciting it from the ground state to the first excited state, so it can no longer cross the retarding field and the current drops. At 9.8 V the electron can excite two atoms, and so on. The excited atom emits a photon of

$\lambda=\frac{hc}{4.9\,\text{eV}}\approx 2537\ \text{\AA}$

which is observed. This proves that atomic energy levels are discrete, as Bohr postulated.

Advantages of Bohr's model

• Explains the stability of the atom through stationary orbits.
• Gives the correct energy levels $E_n=-13.6/n^2$ eV and the Rydberg formula for hydrogen and hydrogen-like ions (He$^+$, Li$^{2+}$).
• Quantisation of angular momentum, $L=nh/2\pi$, gives the Bohr radius $0.529$ \AA.
• Explains the Balmer, Lyman and Paschen series, ionisation energy, and (via Franck–Hertz) excitation potentials.

Limitations of Bohr's model

• Fails for multi-electron atoms such as He.
• Cannot explain fine structure, Zeeman and Stark effects, or the relative intensities of spectral lines.
• Postulates are ad hoc: classical mechanics mixed with arbitrary quantum conditions.
• Violates the uncertainty principle by assigning definite orbits and momenta.
• Gives no account of chemical bonding or of the wave nature of the electron.

Schrödinger's modification

Schrödinger replaced fixed orbits by a wave function $\psi$ obeying

$\nabla^2\psi+\frac{2m}{\hbar^2}(E-V)\psi=0 .$

• For the Coulomb potential, acceptable solutions exist only for $E_n=-13.6/n^2$ eV, so quantisation follows naturally from boundary conditions, with no extra postulate.
• Three quantum numbers $n,\ l,\ m_l$ arise (with spin $m_s$ added), explaining orbital shapes, degeneracy and space quantisation. Angular momentum is $\sqrt{l(l+1)}\,\hbar$, with $l=0$ allowed, unlike Bohr.
• $|\psi|^2dV$ gives the probability of finding the electron, replacing definite orbits (consistent with the uncertainty principle).
• It applies to multi-electron atoms, molecules and fine structure (with relativistic corrections), and reduces to classical results for large $n$ (correspondence principle).

Current gain in CE, CB and CC configurations

Common Emitter (CE): Input is base current $I_B$, output is collector current $I_C$. The dc current gain is

$\beta_{dc}=\frac{I_C}{I_B}$

Since $I_C$ is much larger than $I_B$, $\beta$ is large (typically 20–500). The CE stage gives both current gain and voltage gain, but the output is $180^\circ$ out of phase with the input.

Common Base (CB): Input is emitter current $I_E$, output is collector current $I_C$. The dc current gain is

$\alpha_{dc}=\frac{I_C}{I_E}$

Because $I_C$ is slightly less than $I_E$, $\alpha$ is slightly less than 1 (typically 0.95–0.99). The CB stage gives no current gain but good voltage gain and high frequency response.

Common Collector (CC): Input is base current $I_B$, output is emitter current $I_E$. The dc current gain is

$\gamma_{dc}=\frac{I_E}{I_B}$

Since $I_E=I_C+I_B$, we get $\gamma=1+\beta$, so $\gamma$ is the largest of the three (typically 21–501). The CC stage (emitter follower) gives current gain but voltage gain slightly less than unity.

Relation between $\alpha$ and $\beta$

By KCL in the transistor,

$I_E=I_C+I_B$

Dividing throughout by $I_C$:

$\frac{I_E}{I_C}=\frac{I_C}{I_C}+\frac{I_B}{I_C}$

$\frac{1}{\alpha}=1+\frac{1}{\beta}=\frac{\beta+1}{\beta}$

$\boxed{\alpha=\frac{\beta}{\beta+1}}$

Inverting this relation gives

$\boxed{\beta=\frac{\alpha}{1-\alpha}}$

Check: If $\alpha=0.98$, then $\beta=\dfrac{0.98}{0.02}=49$. Also $\gamma=1+\beta=50$, consistent with $\gamma=\dfrac{1}{1-\alpha}$.

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OR

Inverting and non-inverting modes of an Op-Amp

An ideal op-amp has infinite open-loop gain $A_{OL}$, infinite input impedance, zero output impedance and infinite bandwidth. In practice it is always used with negative feedback, which fixes the closed-loop gain by external resistors.

(a) Inverting amplifier

The input $V_{in}$ is applied to the inverting terminal through $R_1$, and $R_f$ feeds back from output to the inverting terminal. The non-inverting terminal is grounded.

Because of the virtual ground (the differential input is driven to zero by feedback), the inverting terminal is at $0$ V. Hence the current through $R_1$ is

$I_1=\frac{V_{in}}{R_1}$

Since the op-amp input draws no current, all of $I_1$ flows through $R_f$:

$V_{out}=-I_1R_f=-\frac{R_f}{R_1}V_{in}$

$\boxed{A_v=-\frac{R_f}{R_1}}$

The output is inverted (phase shift of $180^\circ$) and the gain is set by the resistor ratio. Input impedance is $R_1$.

(b) Non-inverting amplifier

$V_{in}$ is applied to the non-inverting terminal, and $R_1$ is grounded while $R_f$ feeds back to the inverting terminal. The feedback drives the inverting terminal to $V_{in}$ (virtual short). The current through $R_1$ is $V_{in}/R_1$, and the same current flows in $R_f$, so

$V_{out}=V_{in}+I R_f=V_{in}\left(1+\frac{R_f}{R_1}\right)$

$\boxed{A_v=1+\frac{R_f}{R_1}}$

The output is in phase with the input. Input impedance is very high (equal to the op-amp's own input impedance), so it is preferred for weak signal sources. When $R_f=0$ and $R_1=\infty$, $A_v=1$ and the circuit becomes a voltage follower.

Differential and common mode gain

An op-amp is a differential amplifier with two inputs $V_1$ and $V_2$. Define

$V_d=V_1-V_2 \quad\text{(differential input)}$ $V_c=\frac{V_1+V_2}{2} \quad\text{(common mode input)}$

The output can be written as

$V_{out}=A_dV_d+A_cV_c$

Differential mode gain $A_d$: the amplification given to the difference of the two inputs, with the common part removed. For an ideal op-amp $A_d\to\infty$; practically it is $10^5$–$10^6$.

Common mode gain $A_c$: the unwanted amplification of the signal common to both inputs. For an ideal op-amp $A_c=0$, because a perfectly balanced differential stage cancels it.

Common Mode Rejection Ratio (CMRR):

$\text{CMRR}=\frac{A_d}{A_c}$

It is usually expressed in decibels, $\text{CMRR(dB)}=20\log_{10}\dfrac{A_d}{A_c}$. A high CMRR means the amplifier rejects noise and interference picked up equally by both inputs — an essential property for instrumentation amplifiers.

Double refraction

When a ray of light enters an anisotropic crystal such as calcite (Iceland spar) or quartz, it splits into two refracted rays that travel with different speeds and are plane-polarised in mutually perpendicular planes. This phenomenon is called double refraction (birefringence).

Outline sketch

            Unpolarised ray
                  |
                  v
        ┌───────────────────┐
        │   Calcite crystal │
        │      /     \      │
        └─────/───────\─────┘
             /         \
         E-ray         O-ray
   (vibrations in   (vibrations ⊥ to
  principal section)  principal section)

For normal incidence the O-ray goes straight through while the E-ray is deviated sideways. On emerging, the two rays are parallel and separated, and are polarised at 90° to each other.

Example: Calcite has $\mu_o = 1.658$ and $\mu_e = 1.486$. A dot or line viewed through a calcite crystal appears as a double image. Along the optic axis both rays travel with the same speed, so there is no double refraction in that direction.

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Nicol prism

A Nicol prism is a device for producing and analysing plane-polarised light, based on double refraction.

Construction

   Canada balsam layer (μ = 1.55)
          ╲
   ┌───────╲──────────┐
   │ calcite╲  calcite │   Optic axis makes ~48° with the
   │ half 1  ╲ half 2  │   prism's long axis; ends cut to
   └──────────╲───────┘   68° (instead of the natural 71°)

• A calcite crystal about three times as long as it is wide has its end faces ground so that the angles become $68^\circ$ instead of the natural $71^\circ$.
• It is cut diagonally into two halves perpendicular to the new end faces and the halves are cemented with Canada balsam ($\mu = 1.55$).
• The long side faces are blackened.

Working principle

1. An unpolarised ray entering the prism splits into an O-ray and an E-ray.
2. For the O-ray, $\mu_o = 1.658 >$ $\mu_{balsam} = 1.55$, so it passes from a denser to a rarer medium. It meets the balsam layer at an angle greater than the critical angle ($\sin C = 1.55/1.658$, $C \approx 69^\circ$) and suffers total internal reflection. It is thrown to the blackened side face and absorbed.
3. For the E-ray, $\mu_e = 1.486 <$ $\mu_{balsam} = 1.55$, so it passes into a slightly denser medium, is not totally reflected and is transmitted through the balsam.
4. The emergent light is therefore plane-polarised, with vibrations in the principal section.

Uses

• As a polariser to produce plane-polarised light.
• As an analyser to detect and analyse polarised light.
• To determine the plane of vibration of polarised light.

Stern–Gerlach Experiment

Introduction

The Stern–Gerlach experiment (performed by Otto Stern and Walther Gerlach in 1922) was designed to test the space quantization of angular momentum, as predicted by the old quantum theory (Sommerfeld's model). It provided the first direct experimental evidence that electron spin is quantized in space.

Theory

Classical prediction: If a beam of atoms with magnetic moment $\vec{\mu}$ passes through a non-uniform magnetic field, each atom experiences a force

$$\vec{F} = (\vec{\mu}\cdot\vec{\nabla})\,\vec{B}$$

For a field varying along the $z$-axis,

$$F_z = \mu_z \frac{\partial B_z}{\partial z}$$

Classically, $\mu_z$ can take any value between $-\mu$ and $+\mu$. Hence the beam should spread into a continuous smear on the screen.

Quantum prediction: According to space quantization, the component of angular momentum along $z$ is quantized:

$$L_z = m_l \hbar, \qquad m_l = -l, \ldots, +l$$

So $\mu_z$ takes only discrete values, and the beam should split into an odd number $(2l+1)$ of discrete beams.

Experimental Setup

1. Source (Oven): Silver (Ag) is heated in an oven to about 1000 °C. Silver atoms escape through a small hole, forming a beam.
2. Collimators: Two narrow slits produce a well-defined narrow beam.
3. Non-uniform magnetic field: The beam passes between specially shaped pole pieces — one sharp knife-edge and one flat — producing a strong field gradient $\partial B_z/\partial z$ along the vertical direction.
4. Detector (Glass plate): The beam strikes a glass plate where silver atoms deposit. The plate is developed (like a photographic plate) to reveal the pattern.

Observation

• A beam of silver atoms (which have zero net orbital angular momentum, $l=0$) was used.
• Classically, a continuous smear was expected.
• Experimentally, the beam split into exactly two distinct traces, symmetric about the centre.

Conclusion

1. Space quantization is real: The beam splits into discrete components, confirming that angular momentum is quantized in space.
2. Electron spin: Since silver atoms have $l=0$ (no orbital contribution), the observed splitting must arise from intrinsic angular momentum — electron spin, with $s = \tfrac{1}{2}$.
3. Two spin states: The number of beams is $2s+1 = 2$, giving $m_s = +\tfrac{1}{2}$ and $-\tfrac{1}{2}$.
4. Failure of classical picture: A classical continuous distribution is ruled out; the result is purely quantum mechanical.
5. Stern–Gerlach experiment thus provides direct evidence for the existence of electron spin and its spatial quantization.

Oscillator: An oscillator is an electronic circuit that generates a continuous, periodic (usually sinusoidal) output waveform of a required frequency without any external input signal, by converting DC energy from a DC power supply into AC energy. It works on the principle of positive feedback, where a part of the output is fed back to the input in phase with it so that the circuit sustains its own oscillations.

Need (necessity) of an oscillator:

1. To generate carrier waves for radio, TV and communication transmitters.
2. To produce clock pulses and timing signals in digital computers, microprocessors and digital circuits.
3. To generate audio and radio frequency signals for testing, measurement and signal generators (CRO, function generators).
4. To produce high-frequency AC for induction and dielectric heating in industry.
5. To generate oscillations in receivers (local oscillator in superheterodyne receivers) and in radar, sonar and medical equipment.

Colpitt's oscillator: It is a tuned LC oscillator in which the tank circuit consists of two capacitors $C_1$ and $C_2$ in series with an inductor $L$ in parallel. The feedback is taken from the capacitor divider, so the feedback is capacitive. It is the dual of the Hartley oscillator.

Circuit diagram (common-emitter Colpitts oscillator):

                          +Vcc
                            |
                  R1        RFC
                  |         |
                  |         +---------+--------------+
                  |         |         |              |
            Cb    |       Collector     |              |
   +--------||----+---B  Q (NPN)       C2             |
   |              |        Emitter      |              |
   |              R2         |          +----- L ------+
   |              |          |          |
   |              |         RE || CE     C1
   |              |          |          |
  GND            GND        GND        GND

Redrawn in words: the tank circuit is the inductor $L$ in parallel with the series pair $C_1$–$C_2$. One end of the tank goes to the collector (through $C_2$ side) and the other end to the base (through $C_1$ side); the junction of $C_1$ and $C_2$ is at ac ground, which is the emitter. $R_1$, $R_2$, $R_E$ and the bypass capacitor $C_E$ provide stable biasing, $C_b$ is a coupling/blocking capacitor, and the RFC feeds $V_{cc}$ to the collector while blocking the ac signal.

Working:

1. When the supply $V_{cc}$ is switched on, the transistor amplifies the small noise (transient) signal present in the circuit. This causes the capacitors $C_1$ and $C_2$ to charge.
2. The charged capacitors discharge through the inductor $L$, and the energy oscillates between the electric field of the capacitors and the magnetic field of the inductor. Thus the tank circuit produces damped oscillations at its resonant frequency.
3. The transistor provides the necessary amplification to overcome the losses in the tank circuit (due to resistance of the coil and load), so the oscillations are sustained.
4. A part of the output is fed back from the capacitor divider ($C_1$, $C_2$) to the input in proper phase (positive feedback) and with sufficient magnitude, so the oscillations are continuous and undamped.
5. The frequency of oscillation is the resonant frequency of the tank circuit:

$f = \frac{1}{2\pi\sqrt{L C_{eq}}}, \qquad C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$

Component providing positive feedback: The capacitor divider formed by $C_1$ and $C_2$ (the two capacitors of the tank circuit) provides the positive feedback — the voltage developed across $C_2$ is fed back to the input (emitter–base) in phase with the output. The RFC (radio frequency choke) only isolates the AC signal from the DC supply and does not provide feedback.

TU BSc — Bachelor Level (4 Yrs.) 2nd Year

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(a) Theory of Laser

LASER stands for Light Amplification by Stimulated Emission of Radiation. The laser action is explained by three fundamental processes involving two energy levels $E_1$ (lower) and $E_2$ (upper) of an atom.

1. Absorption: An atom in the lower state $E_1$ absorbs a photon of energy $h\nu = E_2 - E_1$ and jumps to the upper state $E_2$.

2. Spontaneous emission: An atom in the excited state $E_2$ spontaneously returns to $E_1$, emitting a photon of energy $h\nu$ in a random direction with random phase. This gives ordinary (incoherent) light.

3. Stimulated emission: A photon of energy $h\nu$ incident on an atom already in state $E_2$ stimulates it to drop to $E_1$, emitting a second photon in phase, of the same frequency, direction and polarization as the incident photon. Thus one photon produces two identical photons — this is the key to amplification.

Population inversion: Under thermal equilibrium, $N_1 > N_2$ (Boltzmann distribution), so absorption dominates and no amplification occurs. For laser action, the population must be inverted, i.e. $N_2 > N_1$. This is achieved by pumping (optical, electrical or chemical excitation) with the help of a metastable state where atoms live long enough to accumulate.

Optical resonator: The active medium is placed between two mirrors — one fully reflecting and one partially reflecting. Photons travel back and forth, repeatedly causing stimulated emissions and building up an intense, highly monochromatic, coherent, unidirectional beam, which emerges through the partial mirror as the laser output.

Conditions for laser action: (i) population inversion, (ii) metastable state, (iii) resonant cavity for feedback.

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(b) Raman Scattering

When a beam of monochromatic light (usually laser) is passed through a transparent medium (gas, liquid or solid), a small fraction of the scattered light is found to have frequencies different from that of the incident light. This phenomenon is called Raman scattering or the Raman effect, discovered by C. V. Raman in 1928.

Explanation: The incident photon of energy $h\nu_0$ interacts with a molecule of the medium. The molecule may gain or lose energy in a vibrational or rotational transition.

• If the molecule gains energy $\Delta E$ from the photon, the scattered photon has lower energy: $h\nu_s = h\nu_0 - \Delta E$ This shifted component is called the Stokes line.

• If the molecule is already excited and gives energy $\Delta E$ to the photon, the scattered photon has higher energy: $h\nu_a = h\nu_0 + \Delta E$ This is called the anti-Stokes line.

The frequency shift $\Delta\nu = \nu_0 - \nu_s$ is called the Raman shift and is independent of the frequency of the incident light; it depends only on the nature of the scattering molecule. The anti-Stokes lines are weaker than the Stokes lines because fewer molecules are initially in the excited state.

Key features:

• Scattered light contains the original frequency $\nu_0$ (Rayleigh line) plus shifted frequencies.
• The effect is very weak; intense sources such as lasers are needed.
• It is used to study molecular structure and vibrational energy levels (Raman spectroscopy).

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(c) Meson Theory of Nuclear Forces

Proposed by Yukawa (1935), this theory explains the strong, short-range attractive force that binds nucleons (protons and neutrons) inside the nucleus.

Basic idea: Just as the electromagnetic force between two charges is due to the exchange of photons, the nuclear force between two nucleons is due to the continuous exchange of mesons (particles of intermediate mass, ~200–300 times the electron mass). The exchange is described by the uncertainty principle: $\Delta E \,\Delta t \approx \hbar$

Taking the rest energy of the exchanged particle as $\Delta E = m c^2$ and the time of exchange as $\Delta t = r/c$ (where $r$ is the range of the force): $(m c^2)\left(\frac{r}{c}\right) \approx \hbar \quad\Rightarrow\quad r \approx \frac{\hbar}{m c}$

This gives the range of nuclear force in terms of the mass of the exchanged meson. For a pion of mass $m_\pi \approx 270\,m_e$, $r \approx \frac{\hbar}{m_\pi c} \approx 1.4 \times 10^{-15}\ \text{m} \approx 1.4\ \text{fm}$ which agrees well with the observed range of nuclear forces (~1–2 fm).

Types of mesons exchanged:

• π-mesons (pions) — $\pi^+,\ \pi^-,\ \pi^0$ — responsible for the long-range part of the nuclear force.
• ρ, ω, σ mesons — responsible for the short-range repulsive core.

Exchange processes (examples): $p \leftrightarrow n + \pi^+,\qquad n \leftrightarrow p + \pi^-,\qquad p \leftrightarrow p + \pi^0$

Characteristics explained:

• Short range — because the exchanged meson has finite mass.
• Charge independence — the $p$–$p$, $n$–$n$ and $p$–$n$ forces are nearly equal.
• Saturation — each nucleon interacts only with nearest neighbours.
• Spin dependence and existence of a repulsive core at very short distances.

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(d) Zener Diode

A Zener diode is a heavily doped p–n junction diode designed to operate safely in the reverse breakdown region and to maintain a constant voltage across its terminals.

Construction: It is made by doping both p- and n-regions heavily so that the depletion layer is very thin. It is always connected in reverse bias in circuits.

Symbol: A normal diode symbol with bent ends on the bar (cathode) — the bar is drawn with two bends like the letter "Z".

V–I characteristics:

• In forward bias it behaves like an ordinary diode (low resistance, ~0.7 V knee for Si).
• In reverse bias, a very small reverse saturation current flows until the Zener voltage $V_Z$ is reached.
• At $V_Z$ the current increases sharply while the voltage across the diode remains almost constant at $V_Z$.

Breakdown mechanisms:

1. Zener breakdown — occurs in diodes with $V_Z < 6\ \text{V}$. The strong electric field in the thin depletion layer directly pulls electrons from the valence band to the conduction band (field emission / tunneling).
2. Avalanche breakdown — occurs in diodes with $V_Z > 6\ \text{V}$. Carriers accelerated by the field collide with atoms and create electron–hole pairs, which multiply in an avalanche.

Uses:

• Voltage regulator — the Zener is connected in reverse bias across the load; the output voltage remains constant at $V_Z$ even if the input voltage or load current varies.
• Voltage reference in power supplies and meters.
• Wave clipping and clamping circuits.
• Protection of sensitive devices against over-voltage.

Key point: The Zener diode is never used in forward bias for regulation; its usefulness lies in the sharp, stable reverse breakdown characteristic.

(a) Huygen's Wave Theory of Light

Huygen's wave theory (proposed by Christian Huygens in 1678) explains the propagation of light as a wave phenomenon.

Postulates:

1. Luminous source as centre of disturbance: Every point on a luminous body acts as a source of secondary disturbances. The medium through which light travels is called ether.

2. Wavefront: The locus of all particles of the medium vibrating in the same phase is called a wavefront. Light travels perpendicular to the wavefront.

3. Secondary wavelets: Every point on a given wavefront acts as a fresh source of new disturbance, called secondary wavelets, which travel in all directions with the velocity of light in that medium.

4. New wavefront (Huygens' principle): The forward envelope (common tangent) of all secondary wavelets at any instant gives the new position of the wavefront at that instant.

Explanation of reflection and refraction:

• Reflection: When a wavefront AB strikes a plane mirror, each point on AB sends secondary wavelets. The envelope of these wavelets after time $t$ gives the reflected wavefront $A'B'$. Since the secondary wavelets travel with the same speed in the same medium, the angle of incidence equals the angle of reflection — hence $\angle i = \angle r$.

• Refraction: When light passes from a rarer to a denser medium, the speed of secondary wavelets in the denser medium is less ($v_2 < v_1$). The envelope of secondary wavelets gives the refracted wavefront, leading to Snell's law: $\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \mu$

Limitations: It could not explain the photoelectric effect, the rectilinear propagation of light in a perfectly homogeneous medium, and polarization satisfactorily.

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(b) Moseley's Law

Statement: Moseley's law states that the square root of the frequency of a characteristic X-ray spectral line of an element is directly proportional to the atomic number corrected for screening, $(Z - b)$. Equivalently, the frequency varies as the square of $(Z - b)$.

$\sqrt{\nu} \propto (Z - b)$

or

$\nu = a(Z - b)^2$

where $\nu$ is the frequency of the characteristic X-ray, $Z$ is the atomic number, $a$ is a proportionality constant, and $b$ is the screening constant (approximately 1 for $K_\alpha$ lines).

Explanation:

Moseley (1913) measured the frequencies of characteristic X-rays emitted by various elements used as anticathodes in an X-ray tube. He found that the frequency increases regularly with atomic number, unlike the irregular behaviour predicted by Mendeleev's periodic law.

For the $K_\alpha$ line, using Bohr's theory for a hydrogen-like atom with effective nuclear charge $(Z - b)$:

$\nu = Rc(Z - b)^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

where $R$ is the Rydberg constant and $c$ is the velocity of light. For the $K_\alpha$ line, $n_1 = 1$, $n_2 = 2$, and $b = 1$:

$\nu = Rc(Z - 1)^2\left(1 - \frac{1}{4}\right) = \frac{3}{4}Rc(Z - 1)^2$

Significance:

1. It established the atomic number as the fundamental property of an element (rather than atomic weight).
2. It corrected the positions of several elements in Mendeleev's periodic table.
3. It led to the discovery of new elements (e.g., hafnium, rhenium) by predicting their X-ray frequencies.

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(c) Thevenin's Theorem

Statement: Thevenin's theorem states that any linear active network with two terminals can be replaced by a single equivalent voltage source $V_{th}$ in series with a single equivalent resistance $R_{th}$.

• $V_{th}$ (Thevenin voltage) is the open-circuit voltage across the terminals.
• $R_{th}$ (Thevenin resistance) is the resistance measured across the terminals with all independent sources replaced by their internal resistances (voltage sources short-circuited, current sources open-circuited).

Explanation:

Consider a linear network containing resistances and sources, with two terminals A and B connected to a load $R_L$. According to the theorem, the entire network to the left of AB can be replaced by a single voltage source $V_{th}$ in series with resistance $R_{th}$, and the same current flows through $R_L$ as in the original network.

Steps to find the Thevenin equivalent:

1. Remove the load resistance $R_L$ from terminals A and B.
2. Find the open-circuit voltage $V_{th}$ across A and B.
3. Replace all independent voltage sources by short circuits and current sources by open circuits.
4. Find the equivalent resistance $R_{th}$ looking into terminals A and B.
5. Reconnect $R_L$ to the Thevenin equivalent circuit and calculate: $I_L = \frac{V_{th}}{R_{th} + R_L}$

Advantages: It simplifies the analysis of complex networks, especially when the load is varied, since the Thevenin equivalent remains unchanged for different load values.

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(d) UJT as a Relaxation Oscillator

UJT (Uni-Junction Transistor): A UJT is a three-terminal semiconductor device having one p-n junction. It consists of a lightly doped n-type silicon bar with two ohmic contacts at the ends (base B₁ and B₂) and a heavily doped p-type emitter E near B₂. It exhibits negative resistance characteristics.

Circuit as relaxation oscillator:

The circuit consists of a UJT, a capacitor $C$ connected between emitter E and base B₁, a resistor $R$ connected from supply $V_{BB}$ to emitter, and resistors $R_1$ and $R_2$ in the base circuit. The supply $V_{BB}$ is applied across B₂ and B₁.

Working:

1. When the supply is switched on, the capacitor $C$ charges through resistor $R$ with time constant $\tau = RC$. The emitter voltage $V_E$ rises exponentially.

2. As long as $V_E < V_P$ (peak point voltage, $V_P = \eta V_{BB} + V_D$), the UJT remains OFF and the capacitor continues to charge.

3. When $V_E$ reaches $V_P$, the p-n junction becomes forward biased and the UJT turns ON. The emitter resistance drops sharply (negative resistance region), and the capacitor discharges rapidly through E–B₁ and $R_1$.

4. The discharge current produces a sharp voltage pulse across $R_1$, which is the output.

5. When $V_E$ falls to the valley point $V_V$, the UJT turns OFF again and the capacitor starts charging once more. The cycle repeats, producing a continuous train of pulses.

Frequency of oscillation: $f = \frac{1}{T} \approx \frac{1}{RC \ln\left(\frac{1}{1-\eta}\right)}$

where $\eta$ is the intrinsic stand-off ratio.

Applications: Used in pulse generators, timing circuits, sawtooth generators, and thyristor triggering circuits.

Given:

• Wavelength, $\lambda = 5900\ \text{Å} = 5900 \times 10^{-10}\ \text{m} = 5.9 \times 10^{-7}\ \text{m}$
• Diameter of $10^{\text{th}}$ dark ring, $D_{10} = 50\ \text{mm} = 0.05\ \text{m}$
• Order, $n = 10$

Formula: For the $n^{\text{th}}$ dark ring in reflected light, $D_n^2 = 4n\lambda R$ $\therefore R = \frac{D_n^2}{4n\lambda}$

Radius of curvature: $R = \frac{(0.05)^2}{4 \times 10 \times 5.9 \times 10^{-7}}$ $R = \frac{0.0025}{2.36 \times 10^{-5}}$ $R \approx 105.93\ \text{m}$

Thickness of air film: At the $n^{\text{th}}$ dark ring, $t = \frac{n\lambda}{2}$ $t = \frac{10 \times 5.9 \times 10^{-7}}{2}$ $t = 2.95 \times 10^{-6}\ \text{m}$ $t = 2.95\ \mu\text{m}$

Answer:

• Radius of curvature of the lens, $R \approx 105.93\ \text{m}$
• Thickness of air film at the $10^{\text{th}}$ dark ring, $t = 2.95 \times 10^{-6}\ \text{m} = 2.95\ \mu\text{m}$

Given: [ N=6000\ \text{lines/cm} ] Grating element: [ d=\frac{1}{N}=\frac{1}{6000}\ \text{cm} =1.6667\times 10^{-4}\ \text{cm} =1.6667\times 10^{-6}\ \text{m} ] Wavelength: [ \lambda=600\ \text{nm}=6\times 10^{-7}\ \text{m} ]

For normal incidence, the grating equation is [ d\sin\theta_m=m\lambda ] so [ \sin\theta_m=\frac{m\lambda}{d} ]

For the 2nd order, (m=2): [ \sin\theta_2=\frac{2\times 6\times 10^{-7}}{1.6667\times 10^{-6}} =0.72 ] [ \theta_2=\sin^{-1}(0.72)=46.05^\circ ]

For the 3rd order, (m=3): [ \sin\theta_3=\frac{3\times 6\times 10^{-7}}{1.6667\times 10^{-6}} =1.08 ]

But (\sin\theta_3=1.08>1), which is impossible. Hence the 3rd order spectrum does not exist.

Maximum possible order: [ m_{\max}=\frac{d}{\lambda} =\frac{1.6667\times 10^{-6}}{6\times 10^{-7}} =2.77 ] So only orders (m=0,1,2) are possible.

Therefore, the difference in angles of deviation between the 2nd and 3rd order spectra cannot be found because the 3rd order spectrum is absent.

Given:

• For platinum: (Z_1=78), (\lambda_1=1.321\ \text{Å})
• For unknown substance: (\lambda_2=4.174\ \text{Å})
• Screening constant for (L_\alpha) line: (b=7.4)

For the (L_\alpha) line, the transition is from (n=3) to (n=2). By Moseley’s law,

[ \frac{1}{\lambda}=R(Z-b)^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right) ]

Since the same transition occurs in both substances,

[ \frac{1}{\lambda}\propto (Z-b)^2 ]

So,

[ \frac{Z_2-b}{Z_1-b}=\sqrt{\frac{\lambda_1}{\lambda_2}} ]

[ Z_2-b=(Z_1-b)\sqrt{\frac{\lambda_1}{\lambda_2}} ]

Substituting values:

[ Z_2-7.4=(78-7.4)\sqrt{\frac{1.321}{4.174}} ]

[ Z_2-7.4=70.6\sqrt{0.3165} ]

[ Z_2-7.4=70.6\times 0.5626 ]

[ Z_2-7.4=39.72 ]

[ Z_2=39.72+7.4=47.12 ]

Hence, the atomic number of the unknown substance is

[ \boxed{Z \approx 47} ]

So the unknown substance is silver ((Ag)).

Given:

• Power, (P = 30{,}000\ \text{kW} = 3 \times 10^7\ \text{W})
• Time, (t = 1000\ \text{h} = 1000 \times 3600 = 3.6 \times 10^6\ \text{s})
• Energy released per fission of (U^{235}), (E_f \approx 200\ \text{MeV})

Total energy produced:

[ E = P t = (3 \times 10^7)(3.6 \times 10^6) ]

[ E = 1.08 \times 10^{14}\ \text{J} ]

Convert fission energy into joule:

[ E_f = 200\ \text{MeV} = 200 \times 1.6 \times 10^{-13}\ \text{J} ]

[ E_f = 3.2 \times 10^{-11}\ \text{J} ]

Number of (U^{235}) atoms used up:

[ N = \frac{E}{E_f} ]

[ N = \frac{1.08 \times 10^{14}}{3.2 \times 10^{-11}} ]

[ N = 3.375 \times 10^{24} ]

[ \boxed{N \approx 3.38 \times 10^{24}\ \text{atoms of } U^{235}} ]

Mass of (U^{235}) used:

[ m = \frac{N}{N_A} = \frac{3.375 \times 10^{24}}{6 \times 10^{26}} ]

[ m = 5.625 \times 10^{-3}\ \text{kg} = 5.625\ \text{g} ]

So, about (\boxed{3.38 \times 10^{24}}) atoms of (U^{235}) would be used up in 1000 hours.

For a differential amplifier,

[ A_d=\frac{V_{o1}-V_{o2}}{V_{i1}-V_{i2}} ]

[ A_c=\frac{V_{o1}+V_{o2}}{V_{i1}+V_{i2}} ]

1. Differential gain (A_d)

Given:

[ V_{i1}=0.1\sin\omega t,\qquad V_{i2}=-0.1\sin\omega t ]

[ V_{o1}=-5\sin\omega t,\qquad V_{o2}=5\sin\omega t ]

[ A_d=\frac{-5\sin\omega t-5\sin\omega t}{0.1\sin\omega t-(-0.1\sin\omega t)} ]

[ A_d=\frac{-10\sin\omega t}{0.2\sin\omega t}=-50 ]

So,

[ |A_d|=50 ]

2. Common-mode gain (A_c)

When both inputs are equal,

[ V_{i1}=V_{i2}=2\sin\omega t ]

and

[ V_{o1}=V_{o2}=0.05\sin\omega t ]

[ A_c=\frac{0.05\sin\omega t+0.05\sin\omega t}{2\sin\omega t+2\sin\omega t} ]

[ A_c=\frac{0.1\sin\omega t}{4\sin\omega t}=0.025 ]

3. CMRR

[ \text{CMRR}=\frac{|A_d|}{A_c} ]

[ \text{CMRR}=\frac{50}{0.025}=2000 ]

In decibels,

[ \text{CMRR(dB)}=20\log_{10}(2000) ]

[ =20\log_{10}(2\times 10^3) ]

[ =20(0.3010+3) ]

[ =66.02\ \text{dB} ]

[ \boxed{\text{CMRR} \approx 66\ \text{dB}} ]

For a Colpitt’s oscillator, the two capacitors are effectively in series, so

$C_{eq}=\frac{C_1C_2}{C_1+C_2}$

Taking the given capacitor values as $0.2\ \mu\text{F}$ and $0.02\ \mu\text{F}$ (the unit $\mu\text{H}$ appears to be a typo):

$C_{eq}=\frac{0.2\times 0.02}{0.2+0.02}\ \mu\text{F}$ $C_{eq}=\frac{0.004}{0.22}\ \mu\text{F}=0.01818\ \mu\text{F}$ $C_{eq}=0.01818\times 10^{-6}\ \text{F}=1.818\times 10^{-8}\ \text{F}$

The oscillator frequency is

$f=\frac{1}{2\pi\sqrt{LC_{eq}}}$

Rearranging for $L$,

$L=\frac{1}{4\pi^2 f^2 C_{eq}}$

Given $f=10\ \text{kHz}=10^4\ \text{Hz}$,

$L=\frac{1}{4\pi^2(10^4)^2(1.818\times 10^{-8})}$

$L=\frac{1}{71.78}$

$L=0.01393\ \text{H}$

$\boxed{L \approx 13.9\ \text{mH}}$

So, the required inductor is approximately 13.9 mH.

Diffraction Through a Single Slit and a Double Slit

Diffraction is the bending or spreading of light around the edges of an obstacle or through an aperture, producing a pattern of alternate bright and dark fringes on a screen. It occurs when the size of the aperture or obstacle is comparable to the wavelength of light.

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1. Diffraction at a Single Slit

Let a plane wavefront of monochromatic light of wavelength $\lambda$ be incident normally on a single slit of width $a$. The slit is divided into a large number of narrow strips (secondary wavelets) which interfere on a screen placed at distance $D$ ($D \gg a$).

The path difference between waves from the two edges of the slit at an angle $\theta$ is

$\Delta = a \sin\theta$

Conditions:

• Minima (dark fringes): $a \sin\theta = n\lambda, \qquad n = 1, 2, 3, \dots$

• Maxima (bright fringes): $a \sin\theta = (2n+1)\frac{\lambda}{2}, \qquad n = 1, 2, 3, \dots$

The central maximum at $\theta = 0$ is the brightest and widest, with width

$W = \frac{2\lambda D}{a}$

The intensity distribution follows

$I = I_0 \left(\frac{\sin\alpha}{\alpha}\right)^2, \qquad \alpha = \frac{\pi a \sin\theta}{\lambda}$

The intensity of successive maxima falls off rapidly on either side of the central maximum.

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2. Diffraction at a Double Slit

When light passes through two parallel narrow slits separated by a distance $d$, each slit produces its own diffraction pattern. These two patterns overlap and interfere, giving a combined pattern of equally spaced interference fringes whose brightness is modulated by the single-slit diffraction envelope.

Conditions:

• Constructive interference (bright fringes): $d \sin\theta = n\lambda, \qquad n = 0, 1, 2, \dots$

• Destructive interference (dark fringes): $d \sin\theta = (2n+1)\frac{\lambda}{2}$

The fringe width is

$\beta = \frac{\lambda D}{d}$

The resultant intensity is

$I = 4I_0 \cos^2\!\left(\frac{\pi d \sin\theta}{\lambda}\right) \left(\frac{\sin\alpha}{\alpha}\right)^2$

where the first factor gives the interference term and the second gives the diffraction envelope. The interference fringes are visible only within the diffraction envelope; where the envelope has a minimum (missing order), the fringes disappear.

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3. Differences Between Single-Slit and Double-Slit Diffraction

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Conclusion

A single slit gives a diffraction pattern with a broad central maximum and rapidly fading secondary maxima. A double slit gives a pattern in which the two diffracted beams interfere, producing many equally spaced bright and dark fringes whose intensity is governed by the single-slit diffraction envelope. Thus double-slit diffraction is essentially interference modified by diffraction.

Vector Atom Model

1. Introduction

The Vector atom model was developed (by Sommerfeld, Debye, and others) to remove the shortcomings of Bohr's atomic model. In Bohr's model the electron was treated simply as a negatively charged particle revolving in a circular orbit. The vector atom model treats the electron as having two intrinsic vector properties — the spin angular momentum $\vec{S}$ and the magnetic moment $\vec{\mu}_s$ — in addition to its orbital angular momentum $\vec{L}$ and orbital magnetic moment $\vec{\mu}_l$. These quantities are treated as vectors that obey the rules of vector addition, and the model successfully explains the fine structure of spectral lines, the Zeeman effect, the Stern–Gerlach experiment, etc.

2. Basic Concepts of the Vector Atom Model

(i) Quantization of direction (Space quantization).
According to Bohr, the magnitude of the orbital angular momentum is quantized: $L = \frac{nh}{2\pi}, \qquad n = 1, 2, 3, \dots$ The vector atom model further assumes that the direction of this angular momentum vector in space is also quantized. The component of $\vec{L}$ along a fixed reference direction (say the $Z$-axis, taken along an applied magnetic field) can take only discrete values: $L_z = m_l \hbar, \qquad m_l = 0, \pm 1, \pm 2, \dots, \pm l$ where $\hbar = h/2\pi$ and $l$ is the orbital (azimuthal) quantum number. Thus the vector $\vec{L}$ can point only along certain allowed directions in space — this is called space quantization.

(ii) Quantization of magnitude.
The magnitude of the orbital angular momentum is $|\vec{L}| = \sqrt{l(l+1)}\,\hbar$ and the magnitude of the spin angular momentum is $|\vec{S}| = \sqrt{s(s+1)}\,\hbar, \qquad s = \tfrac{1}{2}$ The spin magnetic moment is $\mu_s = \frac{e}{m}S = 2\sqrt{s(s+1)}\,\mu_B = \sqrt{3}\,\mu_B \approx 1.73\,\mu_B$ where $\mu_B = \dfrac{eh}{4\pi m}$ is the Bohr magneton.

(iii) Quantization of the resultant (total) angular momentum.
The orbital and spin angular momenta couple to give the total angular momentum $\vec{J} = \vec{L} + \vec{S}, \qquad |\vec{J}| = \sqrt{j(j+1)}\,\hbar$ with $j = l \pm \tfrac{1}{2}$. The component along the field direction is $J_z = m_j \hbar, \qquad m_j = -j, -(j-1), \dots, +j \quad (2j+1 \text{ values})$

(iv) Magnetic moment of the atom.
The resultant magnetic moment is $\vec{\mu} = \vec{\mu}_l + \vec{\mu}_s = -\frac{e}{2m}\left(\vec{L} + 2\vec{S}\right) = -g\,\frac{e}{2m}\vec{J}$ where $g$ is the Lande $g$-factor: $g = 1 + \frac{j(j+1) + s(s+1) - l(l+1)}{2j(j+1)}$

3. Quantum Numbers

The model requires four quantum numbers to specify completely the state of an electron in an atom:

4. Verification of Space Quantization — Stern–Gerlach Experiment

Space quantization is directly verified by the Stern–Gerlach experiment (1922).

Apparatus. A beam of neutral silver (or sodium) atoms is produced by heating silver in an electric oven. The beam is collimated by narrow slits and passed through a strong, non-uniform magnetic field produced by specially shaped pole pieces (one sharply pointed, the other grooved). The atoms are finally collected on a glass plate (detector).

Principle. A magnetic atom in a non-uniform field experiences a force $F_z = \mu_z \frac{\partial B}{\partial z}$ If the direction of $\vec{\mu}$ were arbitrary (classical picture), $\mu_z$ would vary continuously and the beam would spread into a continuous band on the plate. But if space quantization is real, $\mu_z$ takes only discrete values, so the beam must split into a finite number of discrete spots.

Observation.

• With silver atoms, the beam splits into two distinct spots, symmetric about the undeflected position.
• No spot appears at the centre.
• The separation of the spots agrees with the value $\mu_z = \pm \mu_B$ (one Bohr magneton).

Conclusion.
The splitting into exactly two beams proves that the magnetic moment (and hence the angular momentum) of the atom can have only two allowed orientations in space, i.e. space quantization is real. For silver, the outer electron has $l = 0$, so the whole magnetic moment is due to spin with $m_s = \pm \tfrac12$, giving exactly two orientations — in complete agreement with the vector atom model. (For atoms with $l \neq 0$, the beam splits into $2j+1$ components, as observed in the Zeeman effect.)

5. Other Supporting Evidence

1. Anomalous Zeeman effect — the splitting of spectral lines in a magnetic field into $2j+1$ components is explained only by space quantization of $\vec{J}$.
2. Fine structure of spectral lines — arises from the different orientations of $\vec{S}$ relative to $\vec{L}$ (spin–orbit coupling).
3. Lande $g$-factor — measured values agree with the vector model prediction.

6. Merits and Demerits

Merits: Explains fine structure, normal and anomalous Zeeman effect, Stern–Gerlach result, and the four quantum numbers.

Demerits: It is still a semi-classical model — it uses classical pictures of spinning electrons and vector addition, and cannot explain intensities of spectral lines or the spectra of many-electron atoms. These required the later quantum mechanical (wave mechanical) model.

Zener Diode

A Zener diode is a heavily doped p-n junction diode designed to operate safely in the reverse breakdown region. Because of heavy doping, the depletion layer is very thin and the breakdown voltage (Zener voltage, $V_Z$) is low and sharply defined.

Working

Forward bias: It behaves like an ordinary diode — conducts when $V > 0.7\ \text{V}$ (silicon).

Reverse bias: Up to $V_Z$, only a small leakage current flows. When the reverse voltage reaches $V_Z$, a large reverse current suddenly flows and the voltage across the diode stays almost constant at $V_Z$, even if the current changes over a wide range. This is the operating region of the Zener diode.

Two mechanisms cause breakdown:

• Zener breakdown — in diodes with $V_Z < 6\ \text{V}$. The strong electric field in the thin depletion layer directly tears electrons from covalent bonds (field emission). It has a negative temperature coefficient.
• Avalanche breakdown — in diodes with $V_Z > 6\ \text{V}$. Thermally generated carriers gain enough energy from the field to knock out more carriers, causing a carrier multiplication (avalanche). It has a positive temperature coefficient.

Characteristic Features

1. Sharp breakdown voltage: $V_Z$ is fixed and nearly independent of current, so it gives a stable reference voltage.
2. Constant voltage in breakdown: A large change in reverse current produces a very small change in voltage — this is the basis of voltage regulation. The dynamic resistance is $r_Z = \frac{\Delta V_Z}{\Delta I_Z}$ which is very small (a few ohms to a few tens of ohms).
3. Wide current range: It can be operated continuously in breakdown without damage, provided the current is limited by an external series resistor and the power rating $P_Z = V_Z I_Z$ is not exceeded.
4. Temperature dependence: $V_Z$ drifts slightly with temperature — negative for Zener breakdown, positive for avalanche breakdown. Near $V_Z \approx 5\text{–}6\ \text{V}$ the two effects cancel, giving the best temperature stability.
5. Low leakage and fast response: Leakage below $V_Z$ is small, and the diode responds quickly to voltage changes.
6. Available in a wide range: Zener voltages from about $2\ \text{V}$ to $200\ \text{V}$, with power ratings from $0.25\ \text{W}$ to $50\ \text{W}$ or more.

Applications

1. Voltage regulator (shunt regulator): The Zener is connected in reverse across the load with a series resistor $R_S$. The output stays at $V_Z$ even when the input voltage or load current varies. The series resistor absorbs the extra voltage: $R_S = \frac{V_{in} - V_Z}{I_Z + I_L}$
2. Voltage reference: Provides a stable reference voltage in power supplies, comparators and measuring instruments.
3. Voltage clipper / limiter: Two back-to-back Zeners clip or limit a signal to $\pm V_Z$, protecting circuits from overvoltage.
4. Wave shaping: Converts a sine wave into a square (trapezoidal) wave by clipping the peaks.
5. Overvoltage protection: A Zener across a sensitive device (IC, transistor) clamps surges to a safe level.
6. Meter protection: Limits the voltage applied to a voltmeter or galvanometer.

Conclusion

The Zener diode is the simplest and cheapest solid-state voltage regulator. Its sharply defined, current-independent breakdown voltage makes it indispensable as a voltage reference, clipper and protective device in electronic circuits.

b) Why Ramsden eyepiece is more preferable than Huygens' eyepiece

A Ramsden eyepiece consists of two plano-convex lenses of equal focal length placed a distance apart equal to about two-thirds of the focal length, with their convex surfaces facing each other. The field lens (nearer the objective) and the eye lens are separated by $d = \frac{2f}{3}$.

Reasons for preference:

1. Cross-wire / micrometer scale can be used: In a Ramsden eyepiece the field lens is placed below the focal plane of the eye lens, so the real image formed by the objective falls outside the eyepiece. A cross-wire or graduated scale can therefore be mounted at the focal plane of the eye lens without disturbing the lenses. In a Huygens' eyepiece the field lens lies between the objective and the image, so the image forms inside the eyepiece and no cross-wire can be used.

2. Greater flexibility for measurement: Because the cross-wire is accessible, the Ramsden eyepiece is used in telescopes and microscopes meant for accurate measurement (e.g., spectrometer, travelling microscope).

3. Simple construction and easy adjustment: The separation $\frac{2f}{3}$ makes the eyepiece nearly achromatic for the axial rays and the focal plane is external, so the eyepiece can be focused independently.

Drawback (why Huygens' is still used elsewhere): In a Ramsden eyepiece the rays between the two lenses are not parallel but convergent, so spherical and chromatic aberration are only partly corrected; hence it is inferior to Huygens' eyepiece for mere visual observation where no scale is needed.

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c) Importance and conclusions of Moseley's law

Statement (Moseley's law): The frequency $\nu$ of the characteristic $K_\alpha$ X-ray line of an element of atomic number $Z$ is given by

$\sqrt{\nu} = a(Z - b)$

where $a$ is a constant for a given series and $b$ is the screening constant ($b \approx 1$ for $K$ series).

Importance:

1. Established atomic number as the fundamental property of an element: Moseley showed that it is $Z$, not atomic weight, that decides the position of an element in the periodic table. This corrected several wrong placements (e.g., Co–Ni, Te–I) based on atomic weight.

2. Predicted new elements and gaps: Using the law, the number of missing elements between known ones could be fixed exactly, leading to the discovery of elements such as hafnium ($Z = 72$).

3. Determination of atomic number of unknown elements: By measuring the X-ray frequency of an unknown sample, its $Z$ can be found — a standard method in X-ray spectroscopy.

4. Verification of Bohr's theory: The law agrees with Bohr's model when the screening constant is introduced, confirming the shell structure of the atom.

Conclusions:

1. $\sqrt{\nu}$ varies linearly with $Z$ for a given series — a straight line when plotted.
2. The frequency of characteristic X-rays depends only on the target material, not on the accelerating voltage (once the voltage exceeds the excitation potential).
3. Characteristic X-rays arise from transitions in the inner shells of the atom, so they are unaffected by chemical combination or physical state.
4. The screening constant $b$ accounts for the partial shielding of the nuclear charge by the remaining electrons.

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d) Phase inversion in CE-amplifier

Meaning: In a common-emitter (CE) amplifier, the output signal is 180° out of phase with the input signal. This is called phase inversion (or phase reversal).

Explanation:

1. The input signal $v_i$ is applied between base and emitter, and the output $v_o$ is taken between collector and emitter. The emitter is common to both.

2. During the positive half-cycle of input, the base becomes more positive (forward bias increases), so base current $i_b$ and hence collector current $i_c = \beta i_b$ increase.

3. The collector current flows through the load resistor $R_C$, so the voltage drop across $R_C$ ($i_c R_C$) increases. Since $v_o = V_{CC} - i_c R_C,$ an increase in $i_c R_C$ makes $v_o$ decrease. Thus the output goes negative while the input is positive.

4. During the negative half-cycle of input, the base becomes less positive, $i_b$ and $i_c$ decrease, the drop $i_c R_C$ decreases, and hence $v_o$ increases — the output goes positive while the input is negative.

5. Therefore the output completes its positive half-cycle when the input is in its negative half-cycle, i.e. the output is shifted by 180° with respect to the input.

Conclusion: The CE amplifier gives a voltage gain and a current gain, but introduces a phase reversal of 180° between input and output. This is because the collector voltage is obtained as $V_{CC} - i_c R_C$, so an increase in collector current produces a decrease in output voltage.

a) Applications of Ruby laser

The ruby laser (first laser, built by Maiman in 1960) is a pulsed three-level solid-state laser emitting red light at 694.3 nm. Its main applications are:

1. Holography — its long coherence length and pulsed high energy make it suitable for recording holograms of large objects.
2. Distance measurement (range finding) — short, intense pulses are reflected from a target; the time of flight gives distance. Used in military range finders and satellite ranging.
3. Medical applications — dermatology and ophthalmology: removal of tattoos, birthmarks and pigmented skin lesions, and retinal photocoagulation.
4. Material processing — drilling fine holes in diamonds and hard metals, welding, and cutting.
5. Scientific research — study of nonlinear optics, plasma generation, and as a pump source for other lasers.
6. Holographic interferometry and metrology — measurement of small displacements and vibrations.

b) Pair production

Pair production is the conversion of a high-energy photon (γ-ray) into an electron–positron pair in the strong Coulomb field of a heavy nucleus (or an electron).

• Conservation laws: Charge is conserved (0 → +e and −e); momentum is conserved by recoil of the heavy nucleus; energy is conserved.
• Threshold energy: The minimum photon energy required is $E_{min} = 2m_0c^2 = 2 \times 0.511\ \text{MeV} = 1.022\ \text{MeV}$ where (m_0c^2 = 0.511) MeV is the rest energy of an electron. In practice the threshold is slightly higher because the nucleus must recoil.
• Energy balance: If the photon has energy (h\nu), then $h\nu = 2m_0c^2 + K_{e^-} + K_{e^+}$ so the excess energy appears as kinetic energy of the pair.
• Reverse process: The inverse of pair production is pair annihilation, in which an electron and a positron combine to give two photons, each of energy (0.511) MeV.
• Significance: It is a dominant mode of absorption of high-energy γ-rays in matter and is direct experimental evidence of the mass–energy relation (E = mc^2).

c) Steps to solve a typical electric circuit by Thevenin's Theorem

1. Identify the load — remove the load resistance (R_L) across whose terminals the equivalent circuit is required, and mark the terminals as (A) and (B).
2. Find (V_{th}) (open-circuit voltage) — with the load removed, calculate the voltage across terminals (A) and (B) using any network method (series–parallel, mesh, nodal, superposition).
3. Find (R_{th}) (Thevenin resistance) — replace all independent voltage sources by their internal resistances (short circuit ideal voltage sources) and all independent current sources by open circuits. Then find the resistance looking into terminals (A) and (B).
   • Alternatively, (R_{th} = V_{oc}/I_{sc}), where (I_{sc}) is the short-circuit current through (A)–(B).
4. Draw the Thevenin equivalent circuit — a single voltage source (V_{th}) in series with resistance (R_{th}), with the load (R_L) reconnected across (A)–(B).
5. Calculate the load current and voltage: $I_L = \frac{V_{th}}{R_{th} + R_L}, \qquad V_L = I_L R_L$

d) Why electrons cannot be present inside the nucleus

1. Energy (uncertainty) argument: If an electron were confined inside a nucleus of radius $r \approx 10^{-15}$ m, the uncertainty principle gives $\Delta p \approx \frac{\hbar}{r} \approx \frac{1.05 \times 10^{-34}}{10^{-15}} \approx 1.05 \times 10^{-19}\ \text{kg m/s}.$ The non-relativistic estimate $\dfrac{(\Delta p)^2}{2m} \approx 6\times10^{-9}$ J (about $4\times10^{4}$ MeV) is not valid, because such a momentum makes the electron highly relativistic. Using $E \approx pc$ instead: $E \approx (1.05\times10^{-19})(3\times10^{8}) \approx 3\times10^{-11}\ \text{J} \approx 200\ \text{MeV}.$ Either way the energy is far larger than the energy of β-particles emitted by nuclei (a few keV to a few MeV) and than the nuclear binding energy per nucleon (about 8 MeV). An electron with such energy could not be bound in the nucleus, so electrons cannot be nuclear constituents.

2. Spin and statistics: Protons and neutrons are fermions with spin (1/2). If electrons (also spin (1/2)) were present, the nuclear spin and magnetic moment values would not agree with observation. Nuclear magnetic moments are of the order of the nuclear magneton, not the much larger Bohr magneton expected from electrons.

3. β-decay evidence: In β-decay the emitted electron is created at the instant of decay from the transformation of a neutron into a proton: $n \rightarrow p + e^- + \bar{\nu}_e$ The electron is thus a product of the decay, not a pre-existing particle inside the nucleus. This also explains the continuous energy spectrum of β-particles.

Given:

• Distance of screen from slit, (D = 2\ \text{m})
• Slit width, (a = 0.3\ \text{mm} = 0.3 \times 10^{-3}\ \text{m})
• Position of first minimum from central maximum, (y = 6\ \text{mm} = 6 \times 10^{-3}\ \text{m})
• Order of minimum, (n = 1)

Formula: For Fraunhofer diffraction at a single slit, the condition for minima is [ a\sin\theta = n\lambda ] For small angle, [ \sin\theta \approx \tan\theta = \frac{y}{D} ] So, [ a\frac{y}{D} = n\lambda ] [ \lambda = \frac{a y}{nD} ]

Substitution: [ \lambda = \frac{(0.3 \times 10^{-3})(6 \times 10^{-3})}{1 \times 2} ] [ \lambda = \frac{1.8 \times 10^{-6}}{2} ] [ \lambda = 0.9 \times 10^{-6}\ \text{m} ] [ \lambda = 900\ \text{nm} ]

Result: [ \boxed{\lambda = 900\ \text{nm} = 9000\ \text{Å}} ]

So, the wavelength of light used is 900 nm.

Using the two-constant Cauchy equation,

$n=A+\frac{B}{\lambda^2}$

For blue line, $\lambda_1=4358\ \text{\AA}$, $n_1=1.6525$:

$1.6525=A+\frac{B}{(4358)^2}$

For green line, $\lambda_2=5461\ \text{\AA}$, $n_2=1.6245$:

$1.6245=A+\frac{B}{(5461)^2}$

Subtracting,

$1.6525-1.6245=B\left(\frac{1}{4358^2}-\frac{1}{5461^2}\right)$

$0.0280=B\left(5.2653\times10^{-8}-3.3532\times10^{-8}\right)$

$0.0280=B(1.9121\times10^{-8})$

$B=\frac{0.0280}{1.9121\times10^{-8}}=1.4643\times10^6\ \text{\AA}^2$

Now,

$A=n_1-\frac{B}{\lambda_1^2}$

$A=1.6525-\frac{1.4643\times10^6}{(4358)^2}$

$A=1.6525-0.0771=1.5754$

Hence,

$\boxed{A=1.5754,\qquad B=1.4643\times10^6\ \text{\AA}^2}$

If $\lambda$ is taken in metres, then

$B=1.4643\times10^{-14}\ \text{m}^2$

Given:

• Radius of circular path, (r = 0.32\ \text{m})
• Frequency of applied emf, (f = 10\ \text{MHz} = 10^7\ \text{Hz})
• Mass of deuteron, (m = 3.32 \times 10^{-27}\ \text{kg})
• Charge of deuteron, (q = 1.6 \times 10^{-19}\ \text{C})

Formula: [ f=\frac{qB}{2\pi m} ] [ B=\frac{2\pi m f}{q} ]

For magnetic flux density: [ B=\frac{2\pi (3.32\times10^{-27})(10^7)}{1.6\times10^{-19}} ] [ B=\frac{2.086\times10^{-19}}{1.6\times10^{-19}} ] [ B \approx 1.30\ \text{T} ]

Velocity of emerging deuteron: [ v=\omega r=2\pi f r ] [ v=2\pi (10^7)(0.32) ] [ v \approx 2.01\times10^7\ \text{m/s} ]

Answer: [ \boxed{B \approx 1.30\ \text{T}} ] [ \boxed{v \approx 2.01\times10^7\ \text{m/s}} ]

So, the magnetic flux density is about 1.30 T, and the deuteron emerges from the cyclotron with a speed of about (2.01\times10^7\ \text{m/s}).

Given:

• Power of reactor, (P = 30\ \text{MW} = 30\times10^6\ \text{W} = 3.0\times10^7\ \text{J/s})
• Time, (t = 1000\ \text{hr} = 1000\times3600\ \text{s} = 3.6\times10^6\ \text{s})
• Energy released per fission, (E = 200\ \text{MeV})
• Avogadro’s number, (N_A = 6.02\times10^{23}\ \text{atoms/g})

Formula: [ \text{Total energy produced, } E_{\text{total}} = P t ] [ \text{Number of U}^{235}\text{ atoms fissioned, } N = \frac{E_{\text{total}}}{E} ] [ \text{Mass used, } m = \frac{N}{N_A} ]

Calculation: [ E_{\text{total}} = 3.0\times10^7 \times 3.6\times10^6 ] [ E_{\text{total}} = 1.08\times10^{14}\ \text{J} ]

Now, [ 1\ \text{MeV} = 1.602\times10^{-13}\ \text{J} ] [ E = 200\ \text{MeV} = 200\times1.602\times10^{-13} ] [ E = 3.204\times10^{-11}\ \text{J} ]

Number of atoms used: [ N = \frac{1.08\times10^{14}}{3.204\times10^{-11}} ] [ N = 3.37\times10^{24}\ \text{atoms} ]

Mass in grams: [ m = \frac{3.37\times10^{24}}{6.02\times10^{23}} ] [ m = 5.60\ \text{g} ]

Mass in kilograms: [ m = 5.60\times10^{-3}\ \text{kg} ]

Result: [ \boxed{m \approx 5.6\times10^{-3}\ \text{kg}} ]

So, about (5.6\ \text{g}) or (5.6\times10^{-3}\ \text{kg}) of (\text{U}^{235}) would be used up in 1000 hours of operation.

The circuit diagram is not included in the question as recorded, so the resistor values, supply voltage and biasing arrangement are unknown, and a numerical output voltage and input impedance cannot be computed from the text alone. The method for a transistor circuit with $\beta = 50$ is as follows; substitute the values from the diagram.

Step 1: d.c. (bias) analysis. Take $V_{BE} \approx 0.7$ V (silicon). For a fixed-bias circuit, $I_B = \dfrac{V_{CC} - V_{BE}}{R_B}$, then $I_C = \beta I_B = 50\,I_B$ and $V_{CE} = V_{CC} - I_C R_C$ (including $R_E$ if present).

Step 2: small-signal parameters. The dynamic emitter resistance is $r_e = \frac{25\ \text{mV}}{I_E} \approx \frac{25\ \text{mV}}{I_C}, \qquad r_\pi = \beta r_e.$

Step 3: a.c. results (common-emitter, bypassed emitter).

• Input impedance of the transistor: $Z_{in(base)} = \beta r_e$; the circuit's input impedance is $Z_{in} = R_B \parallel \beta r_e$ (for a divider bias, $R_1 \parallel R_2 \parallel \beta r_e$).
• Voltage gain: $A_v = -\dfrac{R_C \parallel R_L}{r_e}$.
• Output voltage: $v_o = A_v\, v_i$.

If an unbypassed emitter resistor $R_E$ is present, use $Z_{in(base)} = \beta(r_e + R_E)$ and $A_v \approx -\dfrac{R_C \parallel R_L}{r_e + R_E}$.

The question should be corrected by adding the circuit values before a final numerical answer is given.