Single slit. Monochromatic light of wavelength $\lambda$ falls on a narrow slit of width $a$, and a screen is placed at distance $D$. Every point of the slit acts as a secondary source (Huygens' principle), and the wavelets from different parts of the same slit interfere. For an angle $\theta$, the waves from the two edges differ in path by $a\sin\theta$.

• Minima (dark): $a\sin\theta = n\lambda,\; n=\pm1,\pm2,\dots$
• Secondary maxima (approximately): $a\sin\theta = (2n+1)\dfrac{\lambda}{2},\; n=\pm1,\pm2,\dots$
• The central maximum at $\theta=0$ is the brightest and widest, with width $2\lambda D/a$. The secondary maxima are much weaker and are half as wide.

Double slit. Light falls on two parallel narrow slits, each of width $a$ and separated by $d$ (centre to centre). Each slit diffracts the light, and the two diffracted beams then overlap and interfere. The pattern is a set of interference fringes under the single-slit diffraction envelope.

• Bright fringes: $d\sin\theta = n\lambda,\; n=0,\pm1,\pm2,\dots$
• Dark fringes: $d\sin\theta = (2n+1)\dfrac{\lambda}{2}$
• Fringe width: $\beta = \dfrac{\lambda D}{d}$, the same for all fringes.
• If an interference maximum falls at a diffraction minimum, that order is missing. This happens when $d/a = n/m$ for integers $n$ and $m$.

Difference between single-slit and double-slit patterns

Binary 111001 to decimal

$111001_2 = 1\times32 + 1\times16 + 1\times8 + 0\times4 + 0\times2 + 1\times1 = 57$

Binary 110101 to decimal

$110101_2 = 1\times32 + 1\times16 + 0\times8 + 1\times4 + 0\times2 + 1\times1 = 53$

Sum

$57 + 53 = 110$

Check by adding in binary: $111001_2 + 110101_2 = 1101110_2 = 64+32+8+4+2 = 110$.

Interference by division of wavefront and division of amplitude

Interference needs two coherent sources. Since independent sources are incoherent, coherent beams are obtained from a single source in two ways.

1. Division of wavefront: The wavefront from a narrow source is divided into two parts by slits, mirrors or prisms, and the parts are superposed. Both parts come from the same wavefront, so they stay coherent. The source must be small (a point or narrow slit).

• Examples: Young's double slit, Fresnel's biprism, Lloyd's mirror.
• Fringes are usually equally spaced, e.g. $\beta = \dfrac{\lambda D}{d}$ in Young's experiment.

2. Division of amplitude: A beam is divided into a reflected and a transmitted part by partial reflection at a surface; the amplitude (intensity) is shared and the parts are later recombined. An extended source can be used.

• Examples: thin films, wedge fringes, Newton's rings, Michelson interferometer.
• Path difference in a thin film (reflected light): $2\mu t\cos r = (2n+1)\dfrac{\lambda}{2}$ for maxima.

Coherence

Temporal coherence is the correlation of phase of the wave at the same point at different times. It depends on the spectral purity (bandwidth $\Delta\nu$) of the source. A wave train lasts for a coherence time $\tau_c \approx 1/\Delta\nu$, giving the coherence length $L_c = c\,\tau_c = \frac{\lambda^2}{\Delta\lambda}$ Fringes are visible only if the path difference is less than $L_c$.

Spatial coherence is the correlation of phase at different points of the wavefront at the same instant. It depends on the size of the source. For a source of angular size $\theta$, the transverse coherence length is $l_c \approx \frac{\lambda}{\theta}$ Fringes are sharp only when the two slits lie within this width.

A laser has both high temporal and high spatial coherence; a sodium lamp has good spatial coherence only when a narrow slit is used.

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OR

Plane transmission grating

A plane transmission grating is a large number $N$ of equal, parallel, equidistant slits ruled on a glass plate, each of width $a$ and separated by opaque width $b$. $(a+b)$ is the grating element (grating constant), $a+b = 1/(\text{lines per metre})$.

Theory: Monochromatic parallel light falls normally on the grating. Light diffracted at angle $\theta$ from corresponding points of adjacent slits has path difference $(a+b)\sin\theta$. The diffracted waves from all slits are in phase (principal maxima) when $(a+b)\sin\theta = n\lambda, \quad n = 0, 1, 2, \dots$ This is the grating equation; $n$ is the order. The intensity distribution is $I = I_0\left(\frac{\sin\alpha}{\alpha}\right)^2\left(\frac{\sin N\beta}{\sin\beta}\right)^2,\quad \alpha=\frac{\pi a\sin\theta}{\lambda},\ \beta=\frac{\pi(a+b)\sin\theta}{\lambda}$

• Principal maxima at $\beta = n\pi$, i.e. the grating equation.
• Minima at $N\beta = m\pi$ ($m\ne 0, N, 2N,\dots$), i.e. $(a+b)\sin\theta = \dfrac{m}{N}\lambda$.
• Between two principal maxima there are $(N-1)$ minima and $(N-2)$ weak secondary maxima.
• With white light, the central maximum is white and higher orders are spectra (violet nearest the centre).

Maximum number of orders: Since $\sin\theta \le 1$, $n_{max} \le \frac{a+b}{\lambda}$ $n_{max}$ is the largest integer not exceeding $(a+b)/\lambda$. Maxima occur on both sides of the centre, so the total number of maxima is $2n_{max}+1$. (Example: 6000 lines/cm and $\lambda = 5890$ Å: $a+b = 1.667\times10^{-6}$ m, $(a+b)/\lambda = 2.83$, so $n_{max}=2$ and 5 maxima are seen.)

Resolving power: The resolving power of a grating is its ability to separate two close spectral lines of wavelengths $\lambda$ and $\lambda + d\lambda$: $R = \frac{\lambda}{d\lambda}$ By Rayleigh's criterion, the $n$th maximum of $\lambda+d\lambda$ falls on the first minimum next to the $n$th maximum of $\lambda$: $n(\lambda + d\lambda) = \left(n + \frac1N\right)\lambda \;\Rightarrow\; n\,d\lambda = \frac{\lambda}{N}$ $\boxed{R = \frac{\lambda}{d\lambda} = nN}$ Resolving power is proportional to the order $n$ and the total number of rulings $N$ illuminated.

Frank-Hertz Experiment, Bohr's Model and Schrödinger's Modification

1. Frank-Hertz Experiment

The Frank-Hertz experiment (1914) gave direct experimental proof of the quantisation of atomic energy levels, confirming Bohr's postulates.

Apparatus: A glass tube containing mercury vapour at low pressure, a heated filament (cathode) F, a grid G (accelerating electrode) and a plate P (collector). A small retarding potential $V_r$ (about 0.5 V) is applied between G and P.

Working:

• Electrons emitted thermionically from F are accelerated by the potential $V$ between F and G.
• Electrons passing through the grid reach P only if they still have enough energy to overcome $V_r$. The plate current $I_p$ is measured as a function of $V$.

Observation: The current rises, then falls sharply at $V = 4.9$ V, rises again, falls at $9.8$ V, and so on: peaks repeat at intervals of 4.9 V.

Explanation: At $V = 4.9$ V an electron has kinetic energy

$E = eV = 1.6\times10^{-19}\times 4.9 = 7.84\times10^{-19}\ \text{J} \approx 4.9\ \text{eV}$

This equals the energy needed to raise a mercury atom from the ground state to its first excited state. The electron loses almost all its energy in an inelastic collision and cannot overcome the retarding potential, so the current drops. The excited atom returns to the ground state emitting a photon of wavelength

$\lambda = \frac{hc}{E} = \frac{12400}{4.9} \approx 2530\ \text{Å}$

which agrees with the mercury ultraviolet line at 2537 Å (energy 4.886 eV). The same process repeats at $2\times4.9 = 9.8$ V, etc.

Conclusion: Atoms absorb energy only in discrete quanta, confirming stationary energy levels.

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2. Advantages of Bohr's Model

1. Explains the hydrogen spectrum: gives $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)$ and predicts the Lyman, Balmer, Paschen, Brackett and Pfund series.
2. Explains the stability of the atom: electrons in stationary orbits do not radiate.
3. Introduces quantisation: $mvr = n\hbar$ and $E_n = -13.6/n^2$ eV.
4. Explains ionisation and excitation potentials of hydrogen and hydrogen-like ions (He⁺, Li²⁺).
5. Gives the correct Rydberg constant and Bohr radius $a_0 = 0.529$ Å.
6. Explains the Frank-Hertz results and Moseley's X-ray law qualitatively.

3. Limitations of Bohr's Model

1. Works only for one-electron atoms; fails for helium and heavier atoms.
2. Cannot explain fine structure (needs spin and relativistic corrections).
3. Cannot explain the Zeeman effect or the Stark effect.
4. Cannot explain molecular spectra or chemical bonding.
5. Violates the uncertainty principle by assigning a definite orbit (exact $r$ and $p$).
6. Mixes classical and quantum ideas; gives no reason why only certain orbits are allowed.
7. Cannot explain relative intensities of lines or selection rules.
8. Does not include electron spin or the Pauli exclusion principle.

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4. How Schrödinger Modified Bohr's Model

Schrödinger (1926) replaced Bohr's definite orbits with probability distributions (wave mechanics). The electron is described by a wave function $\psi(x,y,z,t)$ satisfying

$-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi = i\hbar\frac{\partial\psi}{\partial t}$

For a stationary state, $\psi = \psi_0 e^{-iEt/\hbar}$, giving the time-independent equation

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

Results:

1. Energy of hydrogen: $E_n = -\dfrac{13.6}{n^2}$ eV, the same as Bohr's, so the hydrogen spectrum is reproduced.
2. Angular momentum: $L = \sqrt{l(l+1)}\,\hbar$, $l = 0,1,\dots,(n-1)$, different from Bohr's $n\hbar$.
3. Magnetic quantum number: $m_l = -l,\dots,+l$ explains the Zeeman effect and space quantisation.
4. Electron spin: $m_s = \pm\tfrac12$ (Uhlenbeck and Goudsmit) explains fine structure and the Stern-Gerlach experiment.
5. Probability density: $|\psi|^2 dV$ is the probability of finding the electron in $dV$; the radial probability of the ground state peaks at $r = a_0$, recovering Bohr's radius as the most probable distance.
6. Selection rules $\Delta l = \pm1$, $\Delta m_l = 0,\pm1$ follow from the mathematics.
7. The Pauli exclusion principle and electron configurations of multi-electron atoms follow naturally.

Conclusion: Bohr's model is a limiting case of Schrödinger's theory for large $n$ (correspondence principle).

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OR

Fine Structure of Hydrogen and Helium: Comparison

Fine structure is the splitting of a spectral line into closely spaced components due to spin-orbit coupling and the relativistic correction to the electron mass.

Hydrogen

• System: one electron, one proton. In simple Bohr theory the energy depends only on $n$.
• Corrections: relativistic mass correction + spin-orbit interaction + Darwin term.
• Result: levels of the same $n$ but different $j$ split:

$E_{n,j} = -\frac{13.6}{n^2}\left[1 + \frac{\alpha^2}{n^2}\left(\frac{n}{j+\tfrac12} - \frac{3}{4}\right)\right]\text{ eV}$

where $\alpha = \dfrac{e^2}{4\pi\varepsilon_0\hbar c} \approx \dfrac{1}{137}$ is the fine structure constant.

• States with the same $n$ and $j$ but different $l$ stay degenerate (e.g. $2s_{1/2}$ and $2p_{1/2}$); only the Lamb shift lifts this slightly.
• Scale: splitting $\sim \alpha^2 E_n$, of the order of $10^{-4}$ eV.

Helium

• System: two electrons, one nucleus. Spin-orbit coupling and electron-electron (exchange) interaction both contribute.
• Two systems: parahelium (spins antiparallel, $S=0$, singlet) and orthohelium (spins parallel, $S=1$, triplet). Transitions between them (intercombination lines) are very weak.
• Result: separate singlet and triplet series; triplet levels show three-component fine structure.
• Scale: exchange splittings are much larger, of the order of $10^{-2}$ to $10^{-1}$ eV, while the true spin-orbit fine structure is small.

Comparison

Conclusion: Both show fine structure due to spin-orbit coupling, but in hydrogen it is a small correction to a one-electron problem, while in helium electron-electron interaction dominates and produces two distinct term systems.

Current Gain in CE, CB and CC Configurations

Common Emitter (CE) Configuration

In CE configuration, the emitter is common to input and output, input is applied between base and emitter, and output is taken between collector and emitter. The current gain is the ratio of collector current to base current at constant collector-emitter voltage:

$\beta = \frac{I_C}{I_B}\Big|_{V_{CE}}$

Since $I_C$ is much larger than $I_B$ (typically $\beta \approx 20$–$500$), CE gives high current gain and also high voltage and power gain. It is the most widely used configuration.

Common Base (CB) Configuration

Here the base is common, input is applied between emitter and base, and output is taken between collector and base. The current gain is the ratio of collector current to emitter current at constant collector-base voltage:

$\alpha = \frac{I_C}{I_E}\Big|_{V_{CB}}$

Since $I_C$ is slightly less than $I_E$ (some current is lost as base current), $\alpha$ is less than unity (typically $0.95$–$0.99$). Hence CB gives no current gain but high voltage gain.

Common Collector (CC) Configuration

Here the collector is common, input is applied between base and collector, and output is taken between emitter and collector. The current gain is the ratio of emitter current to base current at constant collector-emitter voltage:

$\gamma = \frac{I_E}{I_B}\Big|_{V_{CE}}$

Since $I_E \approx I_C$ and $I_B$ is small, $\gamma$ is very high (slightly greater than $\beta$), so CC gives high current gain but voltage gain less than unity. It is used as a buffer (emitter follower).

Relation between $\alpha$ and $\beta$

By Kirchhoff's current law in the transistor,

$I_E = I_C + I_B \quad \text{...(1)}$

Step 1. Divide (1) by $I_C$:

$\frac{I_E}{I_C} = 1 + \frac{I_B}{I_C}$

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

Step 2. Inverting,

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

Step 3. From (1), divide by $I_B$:

$\frac{I_E}{I_B} = \frac{I_C}{I_B} + 1 \;\Rightarrow\; \gamma = \beta + 1$

Also, from $\alpha = \dfrac{\beta}{\beta+1}$,

$\alpha(\beta + 1) = \beta \;\Rightarrow\; \alpha\beta + \alpha = \beta \;\Rightarrow\; \beta(1-\alpha) = \alpha$

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

Since $\alpha < 1$, the denominator $(1-\alpha)$ is small, so $\beta$ is large. For example, if $\alpha = 0.98$, then $\beta = 0.98/0.02 = 49$.

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OR

Inverting and Non-inverting Modes of an Operational Amplifier

An operational amplifier (op-amp) is a high-gain, direct-coupled differential amplifier. Its output is given by

$V_o = A_{OL}(V_2 - V_1)$

where $V_1$ is the voltage at the inverting ($-$) input, $V_2$ at the non-inverting ($+$) input, and $A_{OL}$ is the open-loop gain. By connecting external resistors, two useful modes are obtained.

(a) Inverting Mode

The input signal is applied to the inverting input through $R_1$, while the non-inverting input is grounded. A feedback resistor $R_f$ connects the output back to the inverting input. The inverting input is a virtual ground ($V_1 \approx V_2 = 0$).

Current through $R_1$: $I_1 = \dfrac{V_{in}}{R_1}$

Since no current enters the ideal op-amp input, the same current flows through $R_f$:

$I_f = I_1 \;\Rightarrow\; \frac{0 - V_o}{R_f} = \frac{V_{in}}{R_1}$

$\boxed{V_o = -\frac{R_f}{R_1}\,V_{in}}$

The output is 180° out of phase with the input, and the voltage gain $A_v = -R_f/R_1$ depends only on external resistors.

(b) Non-inverting Mode

The input signal is applied to the non-inverting input, and $R_1$ is connected from the inverting input to ground with $R_f$ from output to inverting input. Since $V_1 = V_2 = V_{in}$ and the same current flows through $R_1$ and $R_f$:

$\frac{V_o - V_{in}}{R_f} = \frac{V_{in}}{R_1}$

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

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

The output is in phase with the input, and the gain is always greater than unity. If $R_f = 0$ and $R_1 = \infty$, the gain becomes 1, giving the voltage follower.

Differential Mode Gain

When equal and opposite signals are applied to the two inputs ($V_1 = -V_2 = V_d/2$), the output is

$V_o = A_d\,V_d, \qquad A_d = \frac{V_o}{V_d}$

$A_d$ is the differential gain — the amplification of the difference signal. For an ideal op-amp, $A_d \to \infty$; practically it is very large ($\sim 10^5$).

Common Mode Gain

When the same signal is applied to both inputs ($V_1 = V_2 = V_c$), the output should ideally be zero. In practice a small output appears:

$V_o = A_c\,V_c$

$A_c$ is the common mode gain, which is very small. The quality of an op-amp is measured by the Common Mode Rejection Ratio:

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

For an ideal op-amp, $A_c = 0$ and CMRR $= \infty$; practically CMRR is very high (about $10^5$), meaning the op-amp amplifies difference signals but rejects signals common to both inputs.

Double Refraction

When a ray of light enters certain crystals (calcite, quartz, tourmaline, etc.), 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 or birefringence.

Outline sketch:

                 Incident ray
                      |
                      |  (unpolarised)
        ──────────────┼──────────────  ← crystal surface
             O ray  ↙      ↘  E ray
          (obeys      (does not obey
           Snell's)    Snell's law)

The two rays are:

The principal section is any plane containing the optic axis and the direction of the incident ray. The O-ray vibrates perpendicular to this section, while the E-ray vibrates in it. The two rays are therefore polarised in mutually perpendicular planes.

Example: A calcite crystal (Iceland spar) has refractive indices $\mu_o = 1.658$ and $\mu_e = 1.486$. A ray incident normally on its surface still splits into two — the O-ray goes straight through, while the E-ray is deviated sideways. This is why a dot viewed through calcite appears as two dots.

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Nicol Prism

A Nicol prism is a polarising device made from a calcite crystal, used to produce and analyse plane-polarised light.

Construction:

        A ─────────────── B
         \               /
          \   Canada    /     ← E-ray absorbed
           \  balsam   /        at cemented face
            \         /
             \       /
        A' ─────────────── B'
              ↑
          O-ray transmitted
          (plane-polarised)

• A calcite crystal is cut with its length about 3 times its breadth, with end faces inclined at $68°$ and side faces at $71°$.
• The crystal is cut diagonally along a plane perpendicular to the principal section and the two halves are cemented together with Canada balsam (refractive index $\mu = 1.55$).

Working:

1. For the O-ray, $\mu_o = 1.658 > \mu_{\text{balsam}} = 1.55$, so it passes from a denser to a rarer medium at the balsam layer. Since the angle of incidence exceeds the critical angle, the O-ray suffers total internal reflection and is absorbed by the blackened side of the prism.
2. For the E-ray, $\mu_e = 1.486 < \mu_{\text{balsam}} = 1.55$, so it passes from a rarer to a denser medium and is transmitted straight through without deviation.
3. The emergent light is therefore plane-polarised, with vibrations in the principal section.

Uses: Producing plane-polarised light, analysing polarised light, in polarimeters and saccharimeters.

Stern–Gerlach Experiment

Theory

The Stern–Gerlach experiment (1922), performed by Otto Stern and Walther Gerlach, was designed to test the spatial quantization of angular momentum predicted by the old quantum theory (Sommerfeld–Bohr model).

Principle: A beam of neutral atoms with a permanent magnetic moment $\vec{\mu}$ is passed through a non-uniform magnetic field. The field exerts a net force on the magnetic dipole:

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

If the field is non-uniform along the $z$-axis only, the force is

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

Classical prediction: The magnetic moment $\vec{\mu}$ can take any orientation in space, so $\mu_z = \mu\cos\theta$ varies continuously from $-\mu$ to $+\mu$. The beam should therefore spread into a continuous smear on the screen.

Quantum prediction: The component $\mu_z$ is quantized. For orbital angular momentum quantum number $l$, the magnetic quantum number takes $2l+1$ discrete values:

$\mu_z = -m_l\,\mu_B,\quad m_l = -l,\ -l+1,\ \dots,\ +l$

So 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 evaporate and escape through a small hole, forming a narrow beam. Silver is chosen because it has a single unpaired outer electron and behaves as a neutral atom (so it is unaffected by electric fields).

2. Collimating slits: Two narrow slits produce a well-defined, pencil-thin beam.

3. Non-uniform magnetic field: The beam passes between specially shaped pole pieces — one flat, one knife-edged — so that the field is strong near the knife edge and weak near the flat pole. This gives a large, constant gradient $\partial B_z/\partial z$ along the vertical direction.

4. Detector (glass plate): The atoms strike a glass plate and are detected by the silver deposit they leave (developed photographically).

Observation

• Classically, a continuous vertical smear was expected.
• Actually observed: The beam split into two distinct, sharp traces on the plate — one deflected upward, one downward — with nothing in between.

Conclusion

1. Space quantization is real: The magnetic moment (and hence angular momentum) of an atom can take only certain discrete orientations in space — it is quantized, not continuous. This confirms the quantum picture and rules out the classical one.

2. Two beams, not three: For silver, the outer electron is in an $s$-state, so the orbital angular momentum is zero ($l = 0$), which would give $2l+1 = 1$ beam. The observed two beams therefore cannot come from orbital angular momentum.

3. Existence of electron spin: The splitting into exactly two beams shows the electron possesses an intrinsic angular momentum — spin — with $s = \tfrac{1}{2}$, giving $2s+1 = 2$ orientations ($m_s = +\tfrac{1}{2}, -\tfrac{1}{2}$). This was the first direct evidence of electron spin (later explained by Uhlenbeck and Goudsmit, 1925).

4. Magnetic moment of the electron: The measured deflection gives $\mu_z = \pm \mu_B$ (one Bohr magneton), confirming the spin magnetic moment of the electron.

5. Foundation for quantum mechanics: The experiment provides direct experimental proof of the quantization of angular momentum and the projection of a quantum state, and it is the basis of modern spin and Stern–Gerlach spin-filter experiments in quantum measurement theory.

Oscillator: An oscillator is an electronic circuit that generates a continuous, periodic (usually sinusoidal) output of a desired frequency without any external input signal. It converts DC energy from the supply into AC energy. It works on the principle of positive feedback: a part of the output is fed back to the input in phase with it.

Need of an oscillator:

1. To generate carrier waves in radio and TV transmitters and communication systems.
2. To produce clock pulses and timing signals in digital circuits, computers and microprocessors.
3. To generate RF signals in radar, induction heating and dielectric heating.
4. To produce audio-frequency signals in signal generators, function generators and test equipment.
5. To supply the local oscillator signal in superheterodyne receivers.
6. For electronic watches, calculators, medical equipment and ultrasonic generators.

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Circuit diagram of Colpitts oscillator

               +Vcc
                 |
               [RFC]
                 |
        +--------+--------+  
        |        | C      |
       [C1]   +--+--+     |
        |     |  Q  |    [L]
        +-----+ E   |     |
        |     |   B +--+  |
       [C2]   +-----+  |  |
        |               |  |
        +---------------+--+

Tank circuit: inductor $L$ in parallel with the series combination of $C_1$ and $C_2$. The junction of $C_1$ and $C_2$ is connected to the emitter (ground for AC); the collector goes to the top end of the tank and the base to the bottom end. Biasing resistors and coupling/bypass capacitors are omitted for clarity.

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Working

1. Tank circuit: When the supply is switched on, noise or the transient surge of collector current makes the LC circuit oscillate at its natural frequency.
2. Amplifier action: The voltage across $C_2$ is applied between base and emitter. The transistor amplifies it and gives a larger output (phase-inverted, $180^\circ$) at the collector, across $C_1$.
3. Positive feedback: The grounded junction of $C_1$ and $C_2$ makes the voltages across them $180^\circ$ out of phase. This provides another $180^\circ$ shift, so the total loop phase shift is $360^\circ$ (0°) and the feedback is positive (regenerative).
4. Sustained oscillations: Losses in the tank are made up from the DC supply through the transistor. When the loop gain satisfies $A\beta = 1$, oscillations are continuous and of constant amplitude. The feedback fraction is $\beta = C_1/C_2$, so the condition is $h_{fe} \ge C_2/C_1$.
5. Frequency of oscillation: $f = \frac{1}{2\pi\sqrt{L\,C_{eq}}}, \qquad C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$

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Component providing positive feedback

The capacitive voltage divider formed by $C_1$ and $C_2$ provides the positive feedback. The voltage across $C_2$ is fed back to the base-emitter input, and with the $180^\circ$ shift of the transistor plus the $180^\circ$ shift of the divider, the net phase shift is zero.

a) Theory of Laser

LASER stands for Light Amplification by Stimulated Emission of Radiation. It is based on the interaction of radiation with matter, which involves three fundamental processes:

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

2. Spontaneous Emission: An atom in the excited state $E_2$ spontaneously returns to the ground state $E_1$, emitting a photon of energy $h\nu$ in a random direction with random phase.

3. Stimulated Emission: A photon of energy $h\nu$ incident on an atom already in the excited state $E_2$ stimulates it to return to $E_1$, emitting a second photon that is identical in frequency, phase, direction and polarization to the stimulating photon. Thus one photon produces two coherent photons.

Condition for laser action — Population Inversion: Under normal thermal equilibrium, $N_1 > N_2$ (Boltzmann distribution), so stimulated absorption dominates and light is absorbed. For amplification, the population of the excited state must exceed that of the ground state, i.e. $N_2 > N_1$. This condition is called population inversion, achieved by pumping (optical, electrical or chemical excitation) using a three-level or four-level system.

Optical Resonator: The active medium is placed between two mirrors — one fully reflecting and one partially reflecting. Photons emitted along the axis are reflected back and forth, stimulating further emissions and building up an intense, highly monochromatic and coherent beam, which emerges through the partial mirror as the laser beam.

Properties of laser light: (i) high monochromaticity, (ii) high coherence, (iii) high directionality, (iv) very high intensity/brightness.

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b) Raman Scattering

When a beam of monochromatic light (usually from a laser) is passed through a transparent medium such as gas, liquid or solid, most of the light is scattered with the same frequency as the incident light (Rayleigh scattering). However, a small fraction of the scattered light is found to have frequencies different from that of the incident light. This phenomenon was discovered by C. V. Raman in 1928 and is called Raman scattering (Raman effect); he received the Nobel Prize in 1930.

Explanation: Let the incident photon of frequency $\nu_0$ interact with a molecule of the medium. The molecule may be in its ground state or in a vibrational/rotational excited state.

• If the molecule gains energy $\Delta E$ from the photon, the scattered photon has lower energy: $h\nu' = h\nu_0 - \Delta E \quad \Rightarrow \quad \nu' = \nu_0 - \frac{\Delta E}{h}$ These lower-frequency lines form the Stokes lines.

• If the molecule was already excited and gives energy $\Delta E$ to the photon, the scattered photon has higher energy: $\nu' = \nu_0 + \frac{\Delta E}{h}$ These higher-frequency lines form the anti-Stokes lines.

The frequency shift $\Delta\nu = |\nu' - \nu_0|$ 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. Anti-Stokes lines are weaker than Stokes lines because fewer molecules are initially in the excited state.

Significance: Raman spectroscopy is used to study molecular structure, vibrational and rotational energy levels, and is applied in chemistry, biology and materials science.

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 surrounding it is filled with ether, and vibrations travel through it.

2. Wavefront: The locus of all particles of the medium which are in the same phase of vibration 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 disturbances called secondary wavelets, which spread out in all directions with the same speed as the original wave.

4. New wavefront (Huygens' principle): The forward-going envelope (common tangent surface) of all the secondary wavelets at any later instant gives the new position of the wavefront. The backward envelope is ineffective.

Explanation of reflection and refraction:

• Reflection: When a wavefront strikes a reflecting surface, secondary wavelets from different points travel back into the same medium. The envelope of these wavelets gives the reflected wavefront. Since the speed is unchanged, the angle of incidence equals the angle of reflection.

• Refraction: When light passes from a rarer to a denser medium, the speed of secondary wavelets decreases. The envelope of these wavelets gives the refracted wavefront, which bends towards the normal. Snell's law $\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2} = \mu$ follows directly.

Merits: Explains reflection, refraction, interference, diffraction and polarisation. Demerits: Could not explain rectilinear propagation convincingly, the photoelectric effect, or the existence of ether.

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b) Moseley's law

Statement: The frequency of the characteristic X-ray spectral line emitted by an element is directly proportional to the square of the atomic number of the element.

$\nu \propto (Z-b)^2 \quad \text{or} \quad \sqrt{\nu} = a(Z-b)$

where $Z$ is the atomic number, $b$ is the screening constant (about 1 for the $K_\alpha$ line) and $a$ is a constant.

Explanation: Moseley (1913) bombarded various elements with high-speed electrons and measured the frequency of the emitted characteristic X-rays. He found that the frequency depends only on the atomic number of the target element and not on the nature of the bombarding particle. Plotting $\sqrt{\nu}$ against $Z$ gives a straight line.

Physical reason: The characteristic X-ray is emitted when an electron jumps from an outer shell to a vacancy in an inner shell. The energy difference between these shells depends on the nuclear charge, i.e. on $Z$. Hence the frequency of the emitted X-ray is governed by $Z$.

Importance:

1. It established atomic number as the fundamental property of an element (more basic than atomic weight).
2. It was used to predict new elements and to fix their positions in the periodic table.
3. It confirmed Bohr's theory of the atom and the shell structure of electrons.

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c) Thevenin's theorem

Statement: 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 $R_{th}$, where:

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

Steps to apply the theorem:

1. Remove the load resistance $R_L$ from the terminals A and B.
2. Find the open-circuit voltage $V_{th}$ across A and B.
3. Replace all voltage sources by short circuits and all current sources by open circuits; find the resistance $R_{th}$ looking into A and B.
4. Reconnect $R_L$ in series with $V_{th}$ and $R_{th}$.
5. Load current: $I_L = \dfrac{V_{th}}{R_{th} + R_L}$.

Explanation: The theorem works because the terminal behaviour of any linear network is completely described by its open-circuit voltage and its internal resistance. It greatly simplifies the analysis of complex networks, especially when only the current through one particular branch (the load) is required. It is the dual of Norton's theorem.

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d) UJT as a relaxation oscillator

Construction: A Uni-Junction Transistor (UJT) consists of a lightly doped n-type silicon bar with two ohmic contacts $B_1$ (base 1) and $B_2$ (base 2) at the two ends, and a heavily doped p-type emitter E alloyed near $B_2$. The p-n junction between E and the bar gives it the name "uni-junction". The resistance between $B_1$ and $B_2$ with E open is $R_{BB}$ (typically 5–10 kΩ).

Circuit: A capacitor $C$ is connected from emitter to ground, a resistor $R$ from supply $V_{BB}$ to emitter, and $R_1$, $R_2$ are base resistors. The UJT is connected across $V_{BB}$.

Working:

1. When $V_{BB}$ is switched on, the capacitor $C$ charges through $R$ with time constant $\tau = RC$. The emitter voltage $V_E$ rises exponentially.
2. As long as $V_E < V_P$ (peak voltage, $V_P = \eta V_{BB} + V_D$ where $\eta$ is the intrinsic stand-off ratio), the emitter junction is reverse biased and the UJT is OFF. The capacitor keeps charging.
3. When $V_E$ reaches $V_P$, the emitter junction becomes forward biased. The emitter region injects holes into the n-bar, which lowers the resistance between $B_1$ and $E$. This is negative resistance region: as $I_E$ increases, $V_E$ falls.
4. The capacitor now discharges rapidly through $E$–$B_1$ and $R_1$, producing a sharp pulse of current through $R_1$.
5. When $V_E$ falls to the valley voltage $V_V$, the junction again becomes reverse biased, the UJT turns OFF, and the capacitor starts charging again.

Result: A sawtooth (relaxation) oscillation is obtained across the capacitor and sharp pulses across $R_1$.

Frequency of oscillation:

$f = \frac{1}{T} \approx \frac{1}{RC \ln\left(\dfrac{1}{1-\eta}\right)}$

Uses: Triggering SCRs and triacs, timing circuits, pulse generators and sawtooth wave generators.

Given:

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

Formula: For Newton’s rings in reflected light, the diameter of the (n^{\text{th}}) dark ring is

[ D_n^2 = 4n\lambda R ]

So,

[ R = \frac{D_n^2}{4n\lambda} ]

The thickness of the air film at the (n^{\text{th}}) dark ring is

[ t = \frac{n\lambda}{2} ]

Calculation of radius of curvature (R):

[ R = \frac{(0.05)^2}{4 \times 10 \times 5.9 \times 10^{-7}} ]

[ R = \frac{0.0025}{2.36 \times 10^{-5}} ]

[ R = 105.93\ \text{m} ]

Calculation of thickness (t):

[ t = \frac{10 \times 5.9 \times 10^{-7}}{2} ]

[ t = 2.95 \times 10^{-6}\ \text{m} ]

[ t = 2.95\ \mu\text{m} ]

Result:

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

Given:

• Number of lines per cm, (N = 6000\ \text{lines/cm})
• Wavelength, (\lambda = 600\ \text{nm} = 6\times 10^{-7}\ \text{m})

Grating element:

[ d=\frac{1}{N}=\frac{1}{6000}\ \text{cm} ]

[ d=\frac{10^{-2}}{6000}\ \text{m} =1.667\times 10^{-6}\ \text{m} ]

Grating equation for normal incidence:

[ d\sin\theta_n=n\lambda ]

[ \sin\theta_n=\frac{n\lambda}{d} ]

For (n=2),

[ \sin\theta_2=\frac{2\times 600\times 10^{-9}}{1.667\times 10^{-6}} ]

[ \sin\theta_2=0.72 ]

[ \theta_2=\sin^{-1}(0.72)=46.05^\circ ]

For (n=3),

[ \sin\theta_3=\frac{3\times 600\times 10^{-9}}{1.667\times 10^{-6}} ]

[ \sin\theta_3=1.08 ]

But (\sin\theta_3=1.08>1), which is impossible. Therefore the third-order spectrum does not exist for this grating and wavelength.

The maximum possible order is

[ n_{\max}<\frac{d}{\lambda} =\frac{1.667\times 10^{-6}}{6\times 10^{-7}} =2.78 ]

So only orders (n=0,1,2) are possible.

Answer: The third-order spectrum is absent, so the difference in angles of deviation between the second and third order spectra cannot be defined.

For the (L_{\alpha}) line, Moseley’s law gives

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

Since (\nu=\dfrac{c}{\lambda}),

[ \sqrt{\frac{c}{\lambda}}=a(Z-b) ]

For two substances emitting the same (L_{\alpha}) line,

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

Given:

[ Z_1=78,\quad b=7.4,\quad \lambda_1=1.321\ \text{Å},\quad \lambda_2=4.174\ \text{Å} ]

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

[ 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,

[ \boxed{Z_2 \approx 47} ]

The unknown substance has atomic number 47, so it 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}) (standard value) (= 200\ \text{MeV})

Energy per fission in joule: [ E_f = 200\ \text{MeV} = 200 \times 1.6\times 10^{-13}\ \text{J} ] [ E_f = 3.2\times 10^{-11}\ \text{J} ]

Total energy produced: [ E = P t = 3\times 10^7 \times 3.6\times 10^6 ] [ E = 1.08\times 10^{14}\ \text{J} ]

Number of (U^{235}) atoms used up: [ N = \frac{E}{E_f} = \frac{1.08\times 10^{14}}{3.2\times 10^{-11}} ] [ N = 3.375\times 10^{24} ]

[ \boxed{N \approx 3.4\times 10^{24}\ \text{atoms of } U^{235}} ]

So, about (3.4\times 10^{24}) atoms of (U^{235}) would be used up in 1000 hours of operation.
(The given Avogadro number is not needed for the atom count itself.)

Given:

Case 1: $V_{i1}=0.1\sin\omega t,\qquad V_{i2}=-0.1\sin\omega t$ $V_{o1}=-5\sin\omega t,\qquad V_{o2}=5\sin\omega t$

Differential input: $V_d=V_{i1}-V_{i2}=0.1\sin\omega t-(-0.1\sin\omega t)=0.2\sin\omega t$

Differential output: $V_{od}=V_{o1}-V_{o2}=-5\sin\omega t-5\sin\omega t=-10\sin\omega t$

So differential gain: $A_d=\left|\frac{V_{od}}{V_d}\right|=\frac{10}{0.2}=50$

Case 2: both inputs are equal, so $V_{i1}=V_{i2}=2\sin\omega t$ $V_{o1}=0.05\sin\omega t,\qquad V_{o2}=-0.05\sin\omega t$

Common-mode input: $V_c=2\sin\omega t$

Differential output: $V_{od}=V_{o1}-V_{o2}=0.05\sin\omega t-(-0.05\sin\omega t)=0.1\sin\omega t$

So common-mode gain: $A_c=\left|\frac{V_{od}}{V_c}\right|=\frac{0.1}{2}=0.05$

Now, $\text{CMRR}=\frac{A_d}{A_c}=\frac{50}{0.05}=1000$

In decibels: $\text{CMRR(dB)}=20\log_{10}(1000)$ $=20\times 3=60\ \text{dB}$

$\boxed{\text{CMRR}=60\ \text{dB}}$

For a Colpitt’s oscillator, the two capacitors are in series, so their equivalent capacitance is

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

Taking the capacitor values as
$C_1=0.2\ \mu\text{F},\qquad C_2=0.02\ \mu\text{F}$

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

The frequency of a Colpitt’s oscillator is

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

So,

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

Given,

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

Substituting,

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

$L=\frac{1}{3.9478\times 10^9\times 1.818\times 10^{-8}}$

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

$L=0.01393\ \text{H}$

$\boxed{L=13.93\ \text{mH}}$

So, the inductor used in the Colpitt’s oscillator is approximately 13.93 mH.

Vector Atom Model and Experimental Verification of Space Quantization

1. Vector Atom Model

The vector atom model was developed (by Sommerfeld, Debye, and others) to remove the shortcomings of Bohr's atomic model. In this model, the various properties of an electron in an atom are represented by vectors having definite magnitude and direction.

The two fundamental vectors are:

1. Orbital angular momentum vector ($\vec{L}$) — due to the orbital motion of the electron around the nucleus. Its magnitude is $\left|\vec{L}\right| = \sqrt{l(l+1)}\,\frac{h}{2\pi}$ where $l = 0,1,2,\dots,(n-1)$ is the orbital quantum number.

2. Spin angular momentum vector ($\vec{S}$) — due to the spin (intrinsic rotation) of the electron about its own axis. Its magnitude is $\left|\vec{S}\right| = \sqrt{s(s+1)}\,\frac{h}{2\pi},\qquad s = \tfrac{1}{2}$

The total angular momentum is the vector sum $\vec{J} = \vec{L} + \vec{S},\qquad \left|\vec{J}\right| = \sqrt{j(j+1)}\,\frac{h}{2\pi}$

Associated with these vectors are the magnetic moment vectors $\vec{\mu}_l$ and $\vec{\mu}_s$, which are oppositely directed to $\vec{L}$ and $\vec{S}$ respectively.

Basic postulates of the vector atom model

• Quantization of direction (space quantization): The angular momentum vectors can take only certain discrete orientations in space, so that their components along a fixed reference direction (say the $Z$-axis) are integral or half-integral multiples of $h/2\pi$.
• Quantization of magnitude: The magnitude of each vector is fixed by its quantum number, as given above.
• Spinning electron: The electron possesses an intrinsic spin and hence a spin magnetic moment.
• Coupling of vectors: $\vec{L}$ and $\vec{S}$ couple to give the resultant $\vec{J}$ (spin–orbit coupling).

Space quantization

If a magnetic field is applied along the $Z$-direction, the orbital angular momentum vector $\vec{L}$ can orient itself only in those directions for which the component along $Z$ is $L_z = m_l\,\frac{h}{2\pi},\qquad m_l = -l,\,-l+1,\dots,0,\dots,+l$ Thus there are $(2l+1)$ possible orientations. This restriction of the orientation of an angular momentum vector to discrete directions is called space quantization.

Similarly, for the spin vector, $S_z = m_s\,\frac{h}{2\pi},\qquad m_s = \pm\tfrac{1}{2}$

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2. Experimental Verification — Stern–Gerlach Experiment

Space quantization was directly verified by Stern and Gerlach (1922) using a beam of silver atoms.

Experimental arrangement

• Silver is heated in an electric furnace and the vapour is allowed to escape through a fine hole, producing a narrow beam of silver atoms.
• The beam is collimated by slits and passed through a highly non-uniform magnetic field produced by specially shaped pole pieces (one sharp knife-edge, the other flat).
• The atoms are finally collected on a glass plate (detector), where the deposit is examined.

Working

A silver atom has one unpaired electron in the outermost shell, so its magnetic moment is due to that electron. In a non-uniform field the atom experiences a force $F_z = \mu_z\frac{\partial B}{\partial z}$

• Classical prediction: If the orbital angular momentum could have any orientation, $\mu_z$ would vary continuously and the beam should produce a continuous smear on the plate.
• Quantum prediction: Since $L_z$ (and hence $\mu_z$) is space-quantized, the beam should split into a discrete number of beams.

Result

The beam was found to split into two distinct traces (a symmetrical splitting), not a continuous band. This confirms space quantization directly.

Interpretation

For silver, the orbital angular momentum is zero ($l=0$), so the splitting cannot be explained by orbital quantization. It is explained only by the spin of the electron with $s = 1/2$, giving $m_s = \pm 1/2$, i.e. exactly two orientations. Hence the experiment verifies:

1. Space quantization of angular momentum, and
2. The existence of electron spin (spin quantum number $s = 1/2$).

Other supporting evidence

• Zeeman effect: Splitting of spectral lines in a magnetic field into discrete components is explained only by space quantization of $\vec{L}$ and $\vec{S}$.
• Anomalous Zeeman effect confirms the spin magnetic moment and the coupling $\vec{J} = \vec{L} + \vec{S}$.

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Conclusion

The vector atom model represents orbital and spin angular momenta as vectors with quantized magnitude and direction. Space quantization — the restriction of these vectors to $(2l+1)$ or $(2s+1)$ discrete orientations — is experimentally verified by the Stern–Gerlach experiment (two-fold splitting of a silver atomic beam) and by the Zeeman effect.

Zener Diode

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

Working

1. Forward bias: It behaves like an ordinary diode, conducting at about 0.7 V.
2. Reverse bias (below $V_Z$): Only a small reverse saturation current flows.
3. At $V_Z$: Breakdown occurs and a large reverse current flows. In a Zener diode this breakdown is non-destructive (if power rating is not exceeded), so the diode can be operated continuously in this region.
4. Voltage regulation: In the breakdown region the voltage across the diode stays almost constant at $V_Z$ even when the current changes greatly. The series resistor $R_S$ absorbs the excess voltage:

$V_S = V_{in} - V_Z, \qquad I_Z = I_L + I_S$

If $V_{in}$ rises, $I_Z$ rises and the extra voltage drops across $R_S$, keeping $V_Z$ (and hence the load voltage) constant. If $V_{in}$ falls, $I_Z$ falls and the voltage is again maintained. Thus the Zener acts as a voltage reference/regulator.

Characteristic Features

• Sharp breakdown voltage: $V_Z$ is fixed and precise (from about 2.4 V to 200 V).
• Constant voltage in breakdown: Voltage remains nearly constant over a wide range of reverse current.
• Two breakdown mechanisms:
  • Zener breakdown — dominant below about 6 V; due to direct rupture of covalent bonds by a strong electric field (negative temperature coefficient).
  • Avalanche breakdown — dominant above about 6 V; due to collision multiplication of carriers (positive temperature coefficient).
• Dynamic resistance $r_Z = \Delta V_Z / \Delta I_Z$ is very small in the breakdown region.
• Power rating $P_Z = V_Z I_Z$ must not be exceeded, otherwise the diode is destroyed.
• Heavily doped, so the depletion region is narrow and breakdown is sharp.
• Available in standard voltages (e.g. 5.1 V, 6.2 V, 9.1 V, 12 V) and power ratings (0.5 W, 1 W, 5 W).

Applications

1. Voltage regulator — maintains a constant DC output against changes in input voltage or load current.
2. Voltage reference — provides a stable reference voltage in power supplies, voltmeters and comparator circuits.
3. Wave clipping / limiting — clamps or limits signal amplitude in wave-shaping circuits.
4. Overvoltage protection — protects sensitive circuits and meters from excessive voltage surges.
5. Meter protection — shunts excess current away from a galvanometer or meter movement.
6. Voltage shifter — shifts the DC level of a signal by a fixed amount.
7. Square-wave generation — used with an amplifier to produce square waves from sine waves.

Conclusion: The Zener diode is the simplest and most widely used solid-state device for voltage regulation and reference, because its reverse breakdown voltage is sharp, stable and repeatable.

Answer any TWO of the following.

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a) What is holography? Explain.

Holography is the technique of recording and reproducing the complete three-dimensional image of an object by means of the interference pattern of light waves. The recorded photograph is called a hologram.

Explanation:

1. A beam of coherent light from a laser is split into two parts: the object beam and the reference beam.
2. The object beam is made to fall on the object, and the light scattered from the object falls on a photographic plate.
3. The reference beam is made to fall directly on the same photographic plate.
4. The two beams interfere and produce a complex interference pattern on the plate. This plate, after development, is the hologram.
5. When the hologram is illuminated by the reference beam (or a similar coherent beam), the light diffracts from the recorded pattern and reconstructs the original wavefront, giving a three-dimensional image of the object.

Key features:

• The image is truly three-dimensional — it shows parallax and depth.
• No lens is needed for recording or reconstruction.
• Any small part of the hologram can reproduce the whole image.

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

1. Cross-wire arrangement: In Ramsden's eyepiece the field lens is placed below the cross-wires (between the objective and the cross-wires), so the cross-wires can be adjusted at the focus of the eye lens and are clearly visible. In Huygens's eyepiece the cross-wires cannot be used, as they would be blurred.
2. Real image formation: Ramsden's eyepiece forms a real image outside the eyepiece, so a micrometer or scale can be mounted at the image plane for measurement. This is not possible in Huygens's eyepiece.
3. Wider field of view and less spherical aberration: The two lenses in Ramsden's eyepiece are placed farther apart, giving a larger field of view and reduced spherical aberration compared with Huygens's eyepiece.

For these reasons Ramsden's eyepiece is preferred in optical instruments where measurement and cross-wires are required.

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c) What are the importance and conclusions of Moseley's law?

Moseley's law: The frequency of the characteristic X-ray emitted when an electron jumps from the L-shell to the K-shell is given by

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

where $Z$ is the atomic number, and $a$, $b$ are constants for a given series.

Conclusions:

1. The frequency of characteristic X-rays depends on the atomic number of the target element, not on its atomic weight.
2. The quantity $\sqrt{\nu}$ varies linearly with $Z$, so a plot of $\sqrt{\nu}$ against $Z$ is a straight line.
3. The atomic number is the fundamental property that decides the position of an element in the periodic table.

Importance:

1. It established atomic number as the basic criterion for arranging elements in the periodic table, correcting the earlier order based on atomic weight.
2. It helped in the discovery of new elements and in fixing the position of doubtful elements (e.g. correcting the positions of Co–Ni, Te–I).
3. It gave direct experimental evidence for the Bohr model of the atom and for the shell structure of electrons.
4. It is used in X-ray spectroscopy for the identification of elements and in chemical analysis.

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d) Explain the phase inversion in CE-amplifier.

In a common-emitter (CE) amplifier, the input signal is applied between base and emitter, and the output is taken between collector and emitter. The output voltage is 180° out of phase with the input voltage — this is called phase inversion.

Explanation:

1. During the positive half-cycle of the input signal, the base voltage increases, so the base current increases. This increases the collector current $I_C$.
2. The collector voltage is $V_{CE} = V_{CC} - I_C R_C$. As $I_C$ increases, the drop $I_C R_C$ increases, so $V_{CE}$ (the output) decreases. Thus a positive input gives a negative output.
3. During the negative half-cycle of the input, the base current decreases, $I_C$ decreases, the drop $I_C R_C$ decreases, and hence $V_{CE}$ increases. Thus a negative input gives a positive output.

Since the output changes in the opposite direction to the input, the output is shifted by 180° in phase with respect to the input. Hence a CE amplifier introduces a phase reversal of 180° between input and output.

(a) Applications of the ruby laser

The ruby laser (Cr³⁺ ions in an Al₂O₃ crystal) is a pulsed three-level solid-state laser that emits red light of wavelength 694.3 nm. Its main applications are:

1. Holography: its coherence and high pulse energy make it suitable for recording holograms of large objects.
2. Pulsed laser deposition: it ablates a target to deposit thin films of oxides and superconductors.
3. Range finding and LIDAR: short, high-energy pulses allow accurate measurement of long distances.
4. Dentistry and dermatology: drilling teeth, removing tattoos and treating pigmented skin lesions.
5. Industry and defence: target designation, drilling holes in diamonds and hard metals, and welding.
6. Research: nonlinear optics, plasma generation and studies of stimulated emission.

(b) Pair production

Pair production is the conversion of a high-energy photon ($\gamma$-ray) into an electron-positron pair in the strong Coulomb field of a nucleus. It is a direct consequence of $E = mc^2$.

• The photon energy must be at least the rest energy of the two particles: $E_{\min} = 2m_0c^2 = 2 \times 0.511\ \text{MeV} = 1.022\ \text{MeV}$
• The nucleus takes up some recoil momentum, so energy and momentum are both conserved. A photon cannot produce a pair in empty space.
• The excess photon energy appears as kinetic energy of the pair: $h\nu = 2m_0c^2 + K_e + K_p$
• The positron soon meets an electron and annihilates, giving two photons of 0.511 MeV each.

(c) Steps to solve a circuit by Thevenin's theorem

1. Remove the load resistance $R_L$ and leave the two terminals open.
2. Calculate the open-circuit voltage $V_{th}$ across these terminals (by mesh, nodal analysis or superposition).
3. Replace all independent voltage sources by short circuits and all independent current sources by open circuits.
4. Find the equivalent resistance $R_{th}$ looking into the open terminals.
5. Draw the Thevenin equivalent, which is $V_{th}$ in series with $R_{th}$, and reconnect $R_L$ across it.
6. 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. Uncertainty principle. For an electron confined in a nucleus of size $\Delta x \approx 10^{-14}$ m, $\Delta p \approx \frac{h}{2\pi \Delta x} \approx \frac{6.63\times10^{-34}}{2\pi \times 10^{-14}} \approx 1.06\times10^{-20}\ \text{kg m/s}$ The corresponding kinetic energy (relativistically $E \approx pc$) is about $E \approx pc \approx 1.06\times10^{-20}\times 3\times10^{8} \approx 3.2\times10^{-12}\ \text{J} \approx 20\ \text{MeV}$ This is much more than the binding energy per nucleon (about 8 MeV), so an electron could not be held in the nucleus.
2. Beta decay. $\beta$-particles are emitted with a continuous energy spectrum. They are created at the moment of decay ($n \rightarrow p + e^- + \bar{\nu}_e$) and do not pre-exist in the nucleus.
3. Nuclear spin. Counting protons and electrons as constituents gives wrong spin values. For example, nitrogen-14 would have 21 particles (odd), which would give half-integer spin, but the observed spin is 1.
4. Magnetic moment. Measured nuclear magnetic moments are of the order of the nuclear magneton. If electrons were present, they would be of the order of the much larger Bohr magneton.

Hence electrons are not constituents of the nucleus; they appear only in $\beta$-decay.

Given:

• Distance of screen from lens, (D = 2\ \text{m})
• Slit width, (a = 0.3\ \text{mm} = 0.3 \times 10^{-3}\ \text{m})
• Distance 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, [ \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 = 9 \times 10^{-7}\ \text{m} ]

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

Thus, the wavelength of light used is 900 nm (or 9000 Å).

The two-constant Cauchy equation is

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

Given:

• For blue line: $\lambda_1=4358\ \text{Å}$, $n_1=1.6525$
• For green line: $\lambda_2=5461\ \text{Å}$, $n_2=1.6245$

So,

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

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

Subtracting,

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

$0.0280=B\left(\frac{1}{18{,}992{,}164}-\frac{1}{29{,}822{,}521}\right)$

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

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

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

$B=1.464\times10^6\ \text{Å}^2$

Now,

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

$A=1.6525-\frac{1.464\times10^6}{18{,}992{,}164}$

$A=1.6525-0.0771$

$A=1.5754$

Hence,

$\boxed{A=1.5754}$

$\boxed{B=1.464\times10^6\ \text{Å}^2}$

If $\lambda$ is taken in nm, then

$B=1.464\times10^4\ \text{nm}^2$