CBSE Class 12 Physics: 2023 board paper, set 55/5/1

1. Torque $\tau =pE\sin \theta =q(2a)E\sin \theta$, so $q=\frac{\tau }{2aE\sin \theta }$.

2. $q=\frac{8\times 10^{-3}}{0.02\times 2\times 10^5\times 0.5}=4\times 10^{-6}$ C $=4\mu$C, option (A).

   1 mark

Answer: 4 μC, option (A)

1. $\frac{F}{l}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}=\frac{2\times 10^{-7}\times 3\times 3}{2}=9\times 10^{-7}$ N/m.

2. Currents in the same direction attract, so the force is $9\times 10^{-7}$ N/m, attractive. None of the options matches.

   1 mark

Answer: 9 × 10⁻⁷ N/m, attractive; no option is correct

1. Diamagnetic materials have ${\mu }_r<1$, i.e. $\mu <{\mu }_0$. Copper is diamagnetic (Al, CuCl₂ are paramagnetic; Ni is ferromagnetic). Option (A).

   1 mark

Answer: Copper, option (A)

1. $\epsilon =NA\frac{dB}{dt}=100\times (0.1{)}^2\times 1=1.0$ V, option (D).

   1 mark

Answer: 1.0 V, option (D)

1. Gamma rays lie at the highest-frequency end of the EM spectrum, so they have the shortest wavelength. Option (A).

   1 mark

Answer: Gamma rays, option (A)

1. Angular separation $\theta =\lambda /d$ is independent of D, so it stays constant; fringe width $\beta =\lambda D/d$ increases with D. Option (C).

   1 mark

Answer: Option (C)

1. $E=h\nu =hc/\lambda$. Option (B).

   1 mark

Answer: hc/λ, option (B)

1. $R=R_0{A}^{1/3}$, so volume ∝ A and mass ∝ A; density is independent of A. Ratio = 1, option (D).

   1 mark

Answer: 1, option (D)

1. As diffusion builds the depletion layer, the junction field grows and drift current rises until it balances diffusion current (equilibrium). Option (D).

   1 mark

Answer: Option (D)

1. $E_n=-13.6/n^2$ eV. Gaps: 3→1: 12.09 eV; 4→3: 0.66 eV; 2→1: 10.2 eV; 3→2: 1.89 eV. Largest is transition I. Option (A).

   1 mark

Answer: Transition I, option (A)

1. $p=h/\lambda$, so p ∝ 1/λ: a rectangular hyperbola decreasing with λ. Option (D).

   1 mark

Answer: Option (D)

1. Two 4 μF in series give 2 μF; this in parallel with 4 μF gives 6 μF. Option (C).

   1 mark

Answer: Option (C)

1. $Z=\sqrt{R^2+(X_L-X_C{)}^2}$; as frequency rises, Z falls to R at resonance, then rises. At resonance $V_L=V_C$, and if $X_L-X_C=R$ current lags (not leads). Option (A).

   1 mark

Answer: Option (A)

1. Huygens assumed secondary wavelets have maximum amplitude forward and zero backward, so no backwave forms. Option (B).

   1 mark

Answer: Option (B)

1. $r_n=\frac{n^2{h}^2{\epsilon }_0}{\pi m{e}^2}$, so $r_n\propto n^2$. Option (A).

   1 mark

Answer: n², option (A)

1. Heating breaks covalent bonds, creating more electron-hole pairs, so conductivity rises and resistance falls. Both true and R explains A. Option (A).

   1 mark

Answer: Option (A)

1. Per the network, the resistors are not all in simple parallel; reduction gives equivalent resistance 2R, so A is true while R (all in parallel) is false. Option (C).

   1 mark

Answer: Option (C)

1. $\vec{\tau }=\vec{m}\times \vec{B}$. If the loop plane ⟂ B, then $\vec{m}\parallel \vec{B}$ and τ = 0, so A is true. Torque is a cross product, not dot product, so R is false. Option (C).

   1 mark

Answer: Option (C)

1. Graph (described): Potential energy $U$ (MeV) is on the y-axis and separation $r$ (fm) is on the x-axis. At very small $r$ (below about 0.8 fm) $U$ is large and positive and rises steeply as $r$ decreases. The curve falls as $r$ increases, crosses zero and reaches a minimum (about $-100$ MeV) at $r=r_0\approx 0.8$ fm. Beyond $r_0$ it rises again and approaches zero for $r$ greater than about 2–3 fm, because nuclear force is short-ranged.

   1 mark

2. (a) Attractive region: for $r>r_0$. Here $U$ increases with $r$, so the force $F=-\frac{dU}{dr}$ is negative and pulls the nucleons together. Mark this on the graph to the right of the minimum.

   ½ mark

3. (b) Repulsive region: for $r<r_0$. Here $U$ decreases as $r$ increases, so $F=-\frac{dU}{dr}$ is positive and pushes the nucleons apart. Mark this on the graph to the left of the minimum.

   ½ mark

Answer: PE–separation curve with a minimum at $r_0\approx 0.8$ fm; attractive for $r>r_0$, repulsive for $r<r_0$.

1. (i) The de Broglie wavelength is $\lambda =\frac{h}{mv}$, so $\lambda \propto \frac{1}{v}$.

   ½ mark

2. Therefore, when the velocity of the electron decreases, its de Broglie wavelength increases.

   ½ mark

3. (ii) For an electron accelerated through a potential $V$, $eV=\frac{p^2}{2m}$. This gives $\lambda =\frac{h}{\sqrt{2meV}}$, so $\lambda \propto \frac{1}{\sqrt{V}}$.

   ½ mark

4. Therefore, when the accelerating potential is increased, the de Broglie wavelength decreases.

   ½ mark

Answer: (i) λ increases (λ ∝ 1/v); (ii) λ decreases (λ ∝ 1/√V)

1. (i) The stopping potential increases when the frequency is increased.

   ½ mark

2. Justification: Einstein's equation gives $e{V}_0=h\nu -{\varphi }_0$, so $V_0=\frac{h\nu }{e}-\frac{{\varphi }_0}{e}$. Hence $V_0$ increases linearly with $\nu$.

   ½ mark

3. (ii) The stopping potential remains the same when the intensity is decreased.

   ½ mark

4. Justification: The intensity only changes the number of photons, and hence the number of photoelectrons. The energy of each photon ($h\nu$) does not change. So the maximum kinetic energy of the photoelectrons, and therefore $V_0$, is independent of intensity.

   ½ mark

Answer: (i) Stopping potential increases; (ii) stopping potential remains unchanged

1. (a) The range $10^{-12}$ m to $10^{-8}$ m corresponds to X-rays.

   ½ mark

2. One use: X-rays are used as a diagnostic tool in medicine, for example to detect bone fractures. They are also used to study crystal structure.

   ½ mark

3. (b) The range $10^{-3}$ m to $10^{-1}$ m corresponds to microwaves.

   ½ mark

4. One use: microwaves are used in radar systems for aircraft navigation. They are also used in microwave ovens.

   ½ mark

Answer: (a) X-rays – medical diagnosis; (b) Microwaves – radar for aircraft navigation

1. (a) Stable equilibrium: The dipole moment $\vec{p}$ points in the same direction as $\vec{E}$, so $\theta =0^{\circ }$. In the diagram, draw the field lines pointing right and the dipole with −q on the left and +q on the right.

   ½ mark

2. Potential energy: $U=-pE\cos 0^{\circ }=-pE$, which is the minimum value.

   ½ mark

3. (b) Unstable equilibrium: The dipole moment $\vec{p}$ points opposite to $\vec{E}$, so $\theta =180^{\circ }$. In the diagram, draw the field pointing right and the dipole with +q on the left and −q on the right.

   ½ mark

4. Potential energy: $U=-pE\cos 180^{\circ }=+pE$, which is the maximum value.

   ½ mark

Answer: (a) $\vec{p}\parallel \vec{E}$, U = −pE; (b) $\vec{p}$ antiparallel to $\vec{E}$, U = +pE

1. Lorentz (magnetic) force: $\vec{F}=q(\vec{v}\times \vec{B})$, whose magnitude is $F=qvB\sin \theta$.

   ½ mark

2. The force is maximum, $F_{max}=qvB$, when $\theta =90^{\circ }$, i.e. when $\vec{v}\perp \vec{B}$.

   ½ mark

3. Work done: $\vec{F}$ is always perpendicular to $\vec{v}$, and hence to the displacement $d\vec{r}=\vec{v}dt$. So $W={\int }_{{\vec{r}}_1}^{{\vec{r}}_2}\vec{F}\cdot d\vec{r}=\int q(\vec{v}\times \vec{B})\cdot \vec{v}dt=0$, because $(\vec{v}\times \vec{B})\cdot \vec{v}=0$. Each element of work is $Fdr\cos 90^{\circ }=0$, so no work is done and the speed (kinetic energy) of the particle remains constant.

   1 mark

Answer: $\vec{F}=q(\vec{v}\times \vec{B})$; maximum when $\vec{v}\perp \vec{B}$; W = 0 since $\vec{F}\perp d\vec{r}$

1. Magnetic field due to the long straight wire at distance $d$: $B=\frac{{\mu }_{0}I}{2\pi d}$. By the right-hand thumb rule (current upward, point on the right of the wire), this field is directed into the plane of the paper.

   ½ mark

2. Force on the moving charge: $\vec{F}=q(\vec{v}\times \vec{B})$. Here $\vec{v}$ (downward) is perpendicular to $\vec{B}$ (into the page), so $F=qvB\sin 90^{\circ }$.

   ½ mark

3. Substituting for $B$: $F=\frac{{\mu }_{0}Iqv}{2\pi d}$.

   ½ mark

4. Direction: $(-\hat{j})\times (-\hat{k})=+\hat{i}$. So for a positive charge the force is to the right, i.e. away from the wire (repulsive). This is expected because the charge moves antiparallel to the current.

   ½ mark

Answer: $F=\frac{{\mu }_{0}Iqv}{2\pi d}$, directed away from the wire (towards the right)

1. (i) There is no effect on mobility.

   ½ mark

2. Mobility is $\mu =\frac{v_d}{E}=\frac{e\tau }{m}$. It depends only on the relaxation time and the mass, not on the applied voltage. Doubling $V$ doubles both $v_d$ and $E$, so their ratio stays constant.

   ½ mark

3. (ii) The current density doubles.

   ½ mark

4. Current density is $J=\frac{I}{A}=\frac{V}{RA}$. Since $R$ and $A$ are fixed, $J\propto V$, so doubling $V$ doubles $J$.

   ½ mark

Answer: (i) Mobility unchanged; (ii) current density doubles

1. (i) Flux linkage is related to current by ${\varphi }_2=M{I}_1$, so the mutual inductance $M$ is the slope of the ${\varphi }_2$–$I_1$ graph.

   ½ mark

2. From the graph, $M=\frac{15\text{ Wb}}{6\text{ A}}=2.5$ H.

   ½ mark

3. (ii) The induced emf is $|{\epsilon }_2|=M\frac{d{I}_1}{dt}$.

   ½ mark

4. So $\frac{d{I}_1}{dt}=\frac{100}{2.5}=40{\text{A s}}^{-1}$.

   ½ mark

Answer: (i) M = 2.5 H; (ii) dI₁/dt = 40 A/s

1. Construction (diagram described): Let $P{P}^{\prime }$ be the interface. A plane wavefront $AB$ is incident at angle $i$, with point $A$ on the surface. Let $\tau$ be the time the wavefront takes to travel from $B$ to $C$ on the surface, so $BC=v_1\tau$. With centre $A$ and radius $AE=v_2\tau$, draw a secondary wavelet (sphere) in medium 2. Since ${\mu }_2>{\mu }_1$, $v_2<v_1$ and so $AE<BC$. Draw the tangent plane $CE$ from $C$ to this sphere. $CE$ is the refracted wavefront and it bends towards the normal. Show the incident and refracted rays perpendicular to the wavefronts.

   1 mark

2. In right triangle $ABC$ (right-angled at $B$): $\sin i=\frac{BC}{AC}=\frac{v_1\tau }{AC}$.

   ½ mark

3. In right triangle $AEC$ (right-angled at $E$): $\sin r=\frac{AE}{AC}=\frac{v_2\tau }{AC}$, where $r$ is the angle of refraction.

   ½ mark

4. Dividing the two results: $\frac{\sin i}{\sin r}=\frac{v_1}{v_2}$. Using $v_1=c/{\mu }_1$ and $v_2=c/{\mu }_2$, this becomes $\frac{\sin i}{\sin r}=\frac{{\mu }_2}{{\mu }_1}$.

   ½ mark

5. Hence ${\mu }_1\sin i={\mu }_2\sin r$, which is Snell's law of refraction.

   ½ mark

Answer: Refracted wavefront CE (tangent from C to the wavelet of radius v₂τ); sin i/sin r = v₁/v₂ = μ₂/μ₁ ⇒ μ₁ sin i = μ₂ sin r

1. Construction (diagram described): A plane wavefront $AB$ is incident at angle $i$ on the reflecting surface $MN$, with $A$ on the surface. Let $\tau$ be the time for the wavefront to advance from $B$ to $C$, so $BC=v\tau$. With centre $A$, draw a secondary wavelet of radius $AE=v\tau$ in the same medium. Draw the tangent plane $CE$ from $C$ to this wavelet. $CE$ is the reflected wavefront. Show the incident and reflected rays normal to the wavefronts.

   1 mark

2. Compare triangles $ABC$ and $AEC$: $AE=BC=v\tau$, since both are covered in the same time in the same medium.

   ½ mark

3. $\angle ABC=\angle CEA=90^{\circ }$, because a wavefront is perpendicular to its ray and the tangent plane is perpendicular to the radius.

   ½ mark

4. $AC$ is common to both triangles, so $△ABC\cong △CEA$ by RHS congruence.

   ½ mark

5. Therefore $\angle BAC=\angle ECA$, i.e. $i=r$. This is the law of reflection.

   ½ mark

Answer: Reflected wavefront CE; congruent triangles ABC and CEA give i = r

1. (i) In X the current lags the voltage by $\pi /2$, so X is a pure inductor. In Y the current is in phase with the voltage, so Y is a pure resistor.

   1 mark

2. (ii) Inductive reactance: $X_L=\frac{V}{I}=\frac{220}{0.22}=1000\Omega$.

   ½ mark

3. Resistance: $R=\frac{220}{0.22}=1000\Omega$.

   ½ mark

4. For the series combination: $Z=\sqrt{R^2+X_L^2}=\sqrt{1000^2+1000^2}=1000\sqrt{2}\Omega \approx 1414\Omega$.

   ½ mark

5. Current: $I_{rms}=\frac{V_{rms}}{Z}=\frac{220}{1000\sqrt{2}}=\frac{0.22}{\sqrt{2}}\approx 0.16$ A.

   ½ mark

Answer: (i) X: inductor, Y: resistor; (ii) I ≈ 0.16 A (0.11√2 A)

1. Principle: An ac generator works on electromagnetic induction. Whenever the magnetic flux linked with a coil changes, an emf is induced in it.

   1 mark

2. Working: A coil (armature) of $N$ turns and area $A$ is rotated mechanically with constant angular speed $\omega$ in a uniform magnetic field $\vec{B}$. As the coil rotates, the angle between its area vector and $\vec{B}$ changes continuously, so the flux through it changes and an emf is induced. The emf is taken out through slip rings and carbon brushes to the external circuit.

   1 mark

3. Expression: At time $t$ the angle is $\theta =\omega t$, so the flux through each turn is ${\varphi }_B=BA\cos \omega t$. By Faraday's law, $\epsilon =-N\frac{d{\varphi }_B}{dt}=-NBA\frac{d}{dt}(\cos \omega t)$.

   ½ mark

4. Therefore $\epsilon =NBA\omega \sin \omega t={\epsilon }_0\sin \omega t$, where ${\epsilon }_0=NBA\omega$ is the peak emf.

   ½ mark

Answer: ε = NBAω sin ωt

1. (a) Current sensitivity is $I_s=\frac{\varphi }{I}=\frac{NBA}{k}$. It can be increased by:

   • increasing the number of turns $N$;
   • increasing the magnetic field $B$, using a strong magnet;
   • increasing the area $A$ of the coil;
   • using a suspension wire of smaller torsion constant $k$.
   1 mark

2. (b) To convert the galvanometer into a voltmeter, a high resistance is put in series with it, so $V=I_g(G+R)$. For the range $0$–$V$: $V=I_g(G+R_1)$ ...(1)

   ½ mark

3. For the range $0$–$2V$: $2V=I_g(G+R_2)$ ...(2)

   ½ mark

4. Dividing (2) by (1): $2=\frac{G+R_2}{G+R_1}$, so $2G+2R_1=G+R_2$, which gives $G=R_2-2R_1$.

   1 mark

Answer: (a) Increase N, B or A, or decrease k; (b) G = R₂ − 2R₁

1. (i) **Distance of closest approach ($r_0$):** the minimum distance from the centre of the nucleus that an alpha particle reaches in a head-on approach. At this point its entire kinetic energy has been converted into electrostatic potential energy, so it momentarily stops and reverses. It is given by $r_0=\frac{1}{4\pi {\epsilon }_0}\frac{2Z{e}^2}{K}$.

   1 mark

2. **Impact parameter ($b$):** the perpendicular distance of the initial velocity vector of the alpha particle from the centre of the nucleus. A small $b$ gives a large scattering angle; $b=0$ corresponds to a head-on collision.

   1 mark

3. (ii) Using $r_0=\frac{1}{4\pi {\epsilon }_0}\frac{2Z{e}^2}{K}$ with $K=3.95\text{ MeV}=3.95\times 10^6\times 1.6\times 10^{-19}$ J:

   $r_0=\frac{9\times 10^9\times 2\times 79\times (1.6\times 10^{-19}{)}^2}{3.95\times 10^6\times 1.6\times 10^{-19}}$

   ½ mark

4. $r_0=5.76\times 10^{-14}\text{ m}=57.6\times 10^{-15}$ m, i.e. about 57.6 fm.

   ½ mark

Answer: (ii) r₀ ≈ 5.76 × 10⁻¹⁴ m (57.6 fm)

1. (i) Postulate 1: An electron revolves round the nucleus only in certain stable orbits (stationary states). While in these orbits it does not radiate energy.

   ½ mark

2. Postulate 2: The allowed orbits are those for which the angular momentum is an integral multiple of $\frac{h}{2\pi }$, i.e. $L=mvr=\frac{nh}{2\pi }$, where $n=1,2,3,\dots$

   ½ mark

3. Postulate 3: When an electron jumps from a higher-energy orbit to a lower-energy orbit, it emits a photon whose energy equals the energy difference: $h\nu =E_i-E_f$.

   ½ mark

4. (ii) Using $L=\frac{nh}{2\pi }$ with $n=2$: $L=\frac{2h}{2\pi }=\frac{h}{\pi }$.

   ½ mark

5. Substituting: $L=\frac{6.63\times 10^{-34}}{3.14}$.

   ½ mark

6. $L\approx 2.1\times 10^{-34}{\text{kg m}}^2{\text{s}}^{-1}$

   ½ mark

Answer: (ii) L = h/π ≈ 2.1 × 10⁻³⁴ kg m² s⁻¹

1. (i) Explanation: In the absence of a field, free electrons move randomly and their average velocity is zero. When a field $\vec{E}$ is applied, each electron experiences a force $\vec{F}=-e\vec{E}$ and accelerates between collisions with the ions. Collisions keep randomising its velocity. The net result, averaged over all electrons, is a small constant velocity opposite to $\vec{E}$, called the drift velocity. It is constant at a given temperature because the relaxation time $\tau$ is fixed.

   1 mark

2. Derivation: Each electron has acceleration $\vec{a}=-\frac{e\vec{E}}{m}$. If electron $i$ had velocity ${\vec{v}}_i$ just after its last collision and time $t_i$ has passed since, its velocity now is ${\vec{V}}_i={\vec{v}}_i-\frac{e\vec{E}}{m}{t}_i$.

   ½ mark

3. The drift velocity is the average over all electrons: ${\vec{v}}_d=({\vec{V}}_i{)}_{avg}=({\vec{v}}_i{)}_{avg}-\frac{e\vec{E}}{m}(t_i{)}_{avg}$.

   ½ mark

4. Because thermal velocities are random, $({\vec{v}}_i{)}_{avg}=0$. The average time between collisions is the relaxation time, $(t_i{)}_{avg}=\tau$.

   ½ mark

5. Therefore ${\vec{v}}_d=-\frac{e\vec{E}}{m}\tau$, with magnitude $v_d=\frac{eE\tau }{m}$.

   ½ mark

6. (ii) In series, the same current $I$ flows through both wires. With $I=neA{v}_d$, this gives $v_d=\frac{I}{neA}=\frac{4I}{ne\pi D^2}$. With the same $I$ and $D$, $v_d\propto \frac{1}{n}$.

   1 mark

7. $\frac{v_{dA}}{v_{dB}}=\frac{n_B}{n_A}=\frac{n_B}{1.5n_B}=\frac{2}{3}$

   1 mark

Answer: (i) ${\vec{v}}_d=-\frac{e\vec{E}\tau }{m}$; (ii) v_A : v_B = 2 : 3

1. (i)(i) V vs R: $V=\frac{ER}{R+r}$. The graph starts at $V=0$ when $R=0$ and rises steeply at first. It then bends over and approaches $E$ asymptotically as $R\to \infty$, giving a saturating curve.

   1 mark

2. (i)(ii) V vs I: $V=E-Ir$. The graph is a straight line with negative slope $-r$. Its intercept on the V-axis is $E$ (at $I=0$) and its intercept on the I-axis is $E/r$ (short circuit).

   1 mark

3. (ii) Equivalent internal resistance: $\frac{1}{r_{eq}}=\frac{1}{2r}+\frac{1}{3r}+\frac{1}{6r}=\frac{3+2+1}{6r}=\frac{1}{r}$, so $r_{eq}=r$.

   ½ mark

4. Equivalent emf: $\frac{E_{eq}}{r_{eq}}=\frac{E}{2r}+\frac{E}{3r}+\frac{E}{6r}=\frac{E}{r}$, so $E_{eq}=E$.

   1 mark

5. (ii)(i) Current in the circuit: $I=\frac{E_{eq}}{R+r_{eq}}=\frac{E}{R+r}$.

   ½ mark

6. (ii)(ii) Terminal potential difference: $V=E_{eq}-I{r}_{eq}=E-\frac{Er}{R+r}$.

   ½ mark

7. Simplifying: $V=\frac{ER}{R+r}$.

   ½ mark

Answer: (i) V–R: rising curve approaching E; V–I: straight line from E with slope −r; (ii) I = E/(R + r), V = ER/(R + r)

1. (i) Forward-bias circuit (described): A battery is connected through a rheostat (potential divider) and a key so that its positive terminal goes to the p-side of the diode and its negative terminal to the n-side. A milliammeter is in series with the diode and a voltmeter is connected across the diode.

   1 mark

2. (ii) Reverse-bias circuit (described): The battery connections are reversed, so the positive terminal goes to the n-side and the negative terminal to the p-side. The battery is applied through a potential divider. A microammeter is in series with the diode, because the reverse current is very small, and a voltmeter is connected across the diode.

   1 mark

3. V–I characteristics of a Si diode (described): In the forward-bias quadrant, the current is negligible up to the threshold (cut-in) voltage of about 0.7 V. Beyond this, the current (in mA) rises steeply, almost exponentially. In the reverse-bias quadrant, a very small, nearly constant reverse current (in µA) flows. At the breakdown voltage $V_{br}$, the reverse current suddenly increases sharply.

   1 mark

4. (i) Minority carrier injection: In forward bias the barrier is lowered. Electrons from the n-side diffuse across the depletion region into the p-side, where they are minority carriers. Holes from the p-side cross into the n-side, where they are minority carriers. This injection of minority carriers across the junction produces the forward current.

   1 mark

5. (ii) Breakdown voltage: This is the reverse-bias voltage at which the reverse current, which is almost constant until then, suddenly increases sharply, while the voltage remains nearly constant.

   1 mark

Answer: Forward and reverse bias circuits, Si V–I curve (knee ≈ 0.7 V, sharp rise at breakdown); definitions of minority carrier injection and breakdown voltage

1. The two processes involved in forming a p-n junction are diffusion and drift.

   1 mark

2. Circuit (described): The ac input is applied to the primary of a centre-tapped transformer. The secondary ends A and B are connected to the p-sides of diodes D₁ and D₂ respectively. The n-sides of both diodes are joined together, and the load resistor $R_L$ is connected between this common point and the centre tap of the transformer. The output is taken across $R_L$.

   1 mark

3. Working: During the half cycle when A is positive with respect to the centre tap, B is negative. So D₁ is forward biased and conducts, while D₂ is reverse biased, and current flows through $R_L$. In the next half cycle, A is negative and B is positive. Now D₂ conducts and D₁ does not, and current flows through $R_L$ in the same direction. Thus output is obtained during both half cycles.

   1 mark

4. Waveforms (described): The input is a sinusoidal ac voltage. The output across $R_L$ is a series of positive half-sine pulses in both half cycles; the negative halves are inverted, giving a pulsating dc. The pulses due to D₁ and D₂ alternate.

   1 mark

5. Characteristic property: A junction diode conducts (has low resistance) only when forward biased and offers very high resistance when reverse biased. This unidirectional conduction makes it suitable for rectification.

   1 mark

Answer: Diffusion and drift; centre-tap full-wave rectifier with D₁ and D₂ conducting in alternate halves; output is unidirectional pulsating dc; diode conducts only in forward bias

1. (i) Ray diagram (described). A small object AB stands on the principal axis just beyond the focus $F_o$ of the objective (short focal length $f_o$). Rays from B pass through the objective and form a real, inverted, magnified image A′B′. A′B′ lies just inside the focus $F_e$ of the eyepiece. Rays from B′ then pass through the eyepiece and diverge. When produced backwards they meet at B″, giving a virtual, highly magnified, inverted final image A″B″ at the near point, distance $D$ from the eyepiece. Arrows on the rays show light travelling from object to eye.

   1½ marks

2. Magnification by the objective. Let $L$ be the tube length, i.e. the distance between the second focal point of the objective and the first focal point of the eyepiece. Take $h$ as the object height and $h^{\prime }$ as the height of the first image. From similar triangles, $\tan \beta =\frac{h}{f_o}=\frac{h^{\prime }}{L}$, so $m_o=\frac{h^{\prime }}{h}=\frac{L}{f_o}$.

   ½ mark

3. Magnification by the eyepiece. The eyepiece acts as a simple microscope with the final image at the near point, so $m_e=1+\frac{D}{f_e}$.

   ½ mark

4. Total magnification. $m=m_o{m}_e=\frac{L}{f_o}\left(1+\frac{D}{f_e}\right)$. Equivalently, $m=\frac{v_o}{|u_o|}\left(1+\frac{D}{f_e}\right)$.

   ½ mark

5. (ii) Image by the objective. Given $u_o=-1.5$ cm and $f_o=1.25$ cm. Using $\frac{1}{v_o}-\frac{1}{u_o}=\frac{1}{f_o}$: $\frac{1}{v_o}=\frac{1}{1.25}-\frac{1}{1.5}=\frac{2}{15}$, so $v_o=7.5$ cm.

   ½ mark

6. Magnification of the objective. $|m_o|=\frac{v_o}{|u_o|}=\frac{7.5}{1.5}=5$.

   ½ mark

7. Magnification of the eyepiece. $m_e=1+\frac{D}{f_e}=1+\frac{25}{5}=6$.

   ½ mark

8. Magnifying power. $m=m_o\times m_e=5\times 6=30$. With sign convention this is $-30$, the negative sign showing that the final image is inverted.

   ½ mark

Answer: (i) $m=\frac{L}{f_o}\left(1+\frac{D}{f_e}\right)$; (ii) magnifying power = 30

1. (i) Ray diagram (described). Parallel rays from a distant object enter the objective, which has a large focal length $f_o$, at an angle $\alpha$ to the axis. They form a real, inverted image of height $h$ at the second focal plane of the objective. This focal plane coincides with the first focal plane of the eyepiece, so the lens separation is $f_o+f_e$. Rays leave the eyepiece parallel at an angle $\beta$ to the axis, so the final image is formed at infinity (normal adjustment). Arrows on the rays show the direction of light.

   1½ marks

2. Angle subtended by the object. Magnifying power is defined as $m=\frac{\beta }{\alpha }$. For small angles, the object (or the image at the objective) subtends $\alpha \approx \tan \alpha =\frac{h}{f_o}$.

   ½ mark

3. Angle subtended by the final image. The final image subtends $\beta \approx \tan \beta =\frac{h}{f_e}$ at the eye.

   ½ mark

4. Expression. $m=\frac{\beta }{\alpha }=\frac{h/f_e}{h/f_o}=\frac{f_o}{f_e}$.

   ½ mark

5. (ii) Using the magnifying power. In normal adjustment, $m=\frac{f_o}{f_e}=2.9$, so $f_o=2.9f_e$.

   ½ mark

6. Using the separation. The tube length is $f_o+f_e=150$ cm.

   ½ mark

7. Solving. $2.9f_e+f_e=150\Rightarrow 3.9f_e=150\Rightarrow f_e\approx 38.5$ cm.

   ½ mark

8. Focal length of the objective. $f_o=150-38.5=111.5$ cm.

   ½ mark

Answer: (i) $m=f_o/f_e$; (ii) $f_o\approx 111.5$ cm, $f_e\approx 38.5$ cm

1. (i) No, the lens will not produce a single image. The upper and lower halves have different refractive indices $n_1$ and $n_2$. By the lens maker's formula, each half therefore has a different focal length, so the two halves form two separate images of AB.

   1 mark

2. (ii) Yes, the image is still formed. A real image exists wherever the refracted rays actually meet, whether or not a screen is placed there. The screen only makes the image visible.

   1 mark

3. (iii) Lens maker's formula. $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$. For a double convex lens with equal radii, $R_1=+R$ and $R_2=-R$.

   ½ mark

4. Simplifying. $\frac{1}{f}=\frac{2(n-1)}{R}$, so $R=2(n-1)f$.

   ½ mark

5. Substituting. $R=2(1.55-1)\times 20=2\times 0.55\times 20=22$ cm.

   1 mark

6. (iii) OR alternative (2 marks if this choice is attempted). For lens A, $u=-30$ cm and $f=15$ cm. From $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{15}-\frac{1}{30}=\frac{1}{30}$, we get $v_1=30$ cm behind A.

7. For the rays from B to come out parallel to the principal axis, the image formed by A must lie at the focus of B, i.e. 10 cm in front of B. Hence $d=30+10=40$ cm.

Answer: (i) No, two images form because the halves have different focal lengths; (ii) Yes, the image is still formed; (iii) R = 22 cm [OR d = 40 cm]

1. (i) Trace the connections in the figure. The bridging and cross capacitors reduce the network to two effective branches between A and B, each of capacitance $C$, connected in parallel. So $C_{AB}=C+C=2C$.

   1 mark

2. (ii) The external field $E_0$ polarises the dielectric slab, so bound charges appear on its two faces. These induced charges set up a field $E_p$ inside the slab that opposes $E_0$. The net field is therefore $E=E_0-E_p=\frac{E_0}{K}$, which is smaller than $E_0$.

   1 mark

3. (iii)(a) Potential difference. Total charge is conserved and the two capacitors are in parallel, so $C_{eq}=C+2C=3C$. Then $V=\frac{Q_{total}}{C_{eq}}=\frac{Q}{3C}$.

   1 mark

4. (iii)(b) Charge lost by A. The charge left on A is $Q_A^{\prime }=CV=C\cdot \frac{Q}{3C}=\frac{Q}{3}$. Charge lost by A $=Q-\frac{Q}{3}=\frac{2Q}{3}$, and this charge flows to B.

   1 mark

5. (iii) OR alternative (2 marks if this choice is attempted). The two slabs lie one after the other between the plates, so they act as two capacitors in series. For the $2K$ slab of thickness $d/3$: $C_1=\frac{2K{\epsilon }_{0}A}{d/3}=\frac{6K{\epsilon }_{0}A}{d}$. For the $K$ slab of thickness $2d/3$: $C_2=\frac{K{\epsilon }_{0}A}{2d/3}=\frac{3K{\epsilon }_{0}A}{2d}$.

6. Combining in series: $\frac{1}{C}=\frac{d}{6K{\epsilon }_{0}A}+\frac{2d}{3K{\epsilon }_{0}A}=\frac{5d}{6K{\epsilon }_{0}A}$. Hence $C=\frac{6K{\epsilon }_{0}A}{5d}$.

Answer: (i) 2C; (ii) the induced polarisation field opposes the applied field, so $E=E_0/K$; (iii)(a) $V=Q/3C$, (b) charge lost = $2Q/3$ [OR $C=\frac{6K{\epsilon }_{0}A}{5d}$]