CBSE Class 12 Physics: 2026 board paper, set 55/1/1

1. $E=-\frac{dV}{dx}=-\frac{d}{dx}(10-50x)=+50$ V/m.

   1 mark

2. The positive sign means $E$ is $50$ N/C along $+x$.

Answer: 50 N/C along +x, option (C)

1. The connected spheres are at the same potential: $\frac{k{q}_1}{r_1}=\frac{k{q}_2}{r_2}$.

2. $E=\frac{kq}{r^2}=\frac{V}{r}$, so $\frac{E_A}{E_B}=\frac{r_2}{r_1}$.

   1 mark

Answer: $r_2/r_1$, option (B)

1. Inside the wire, Ampere's law gives $B=\frac{{\mu }_{0}Ir}{2\pi a^2}$.

2. At $r=a/2$: $B=\frac{{\mu }_{0}I}{4\pi a}$.

   1 mark

Answer: ${\mu }_{0}I/(4\pi a)$, option (C)

1. The fringes are strictly hyperbolic, but when $D\gg d$ they become nearly straight lines.

   1 mark

Answer: Straight, option (A)

1. $n=\sqrt{{\epsilon }_r{\mu }_r}=\sqrt{\frac{3}{2}\cdot \frac{8}{3}}=2$.

2. The frequency is unchanged and ${\lambda }^{\prime }=\lambda /n=\lambda /2$, so the wavelength is halved.

   1 mark

Answer: Wavelength halved, frequency unchanged, option (C)

1. $V_L=V_C$, so the circuit is at resonance and the source voltage is $V=V_R=10$ V. Also $R=X_L$.

2. With C shorted, $Z=\sqrt{R^2+X_L^2}=\sqrt{2}R$, so $I^{\prime }=\frac{10}{\sqrt{2}R}$.

3. $V_L=I^{\prime }{X}_L=\frac{10}{\sqrt{2}}=5\sqrt{2}$ V.

   1 mark

Answer: $5\sqrt{2}$ V, option (B)

1. X-rays are used as a medical diagnostic tool, and their wavelength range is about 1 nm to $10^{-3}$ nm.

   1 mark

Answer: 1 nm to $10^{-3}$ nm, option (A)

1. At closest approach, $\frac{1}{2}m{v}^2=\frac{k(2e)(Ze)}{d}$, so $d\propto \frac{1}{v^2}$.

2. Halving $v$ makes the new distance $4d$.

   1 mark

Answer: 4d, option (D)

1. For the full lens, $\frac{1}{f}=(n-1)\frac{2}{R}$. For each half, one face is plane, so $\frac{1}{f^{\prime }}=(n-1)\frac{1}{R}$.

2. This gives $f^{\prime }=2f=20$ cm (in magnitude).

   1 mark

Answer: 20 cm, option (A)

1. Sustained interference needs coherent waves, i.e. the same frequency and a constant phase difference.

2. Waves (i), (iii) and (iv) all have frequency $\omega$ with fixed phase differences, while (ii) has $2\omega$. So the answer is (i), (iii) and (iv).

   1 mark

Answer: (i), (iii) and (iv) only, option (C)

1. $R_1=V^2/P_1$ and $R_2=V^2/P_2$, so in series $R=V^2\frac{P_1+P_2}{P_1{P}_2}$.

2. $P=\frac{(V/2{)}^2}{R}=\frac{P_1{P}_2}{4(P_1+P_2)}$.

   1 mark

Answer: $\frac{P_1{P}_2}{4(P_1+P_2)}$, option (D)

1. At equilibrium there is no net current, so the diffusion current is exactly balanced by an equal and opposite drift current.

   1 mark

Answer: Option (C)

1. The assertion is false: atoms of diamagnetic substances have zero net magnetic moment.

2. The reason is false: a current loop always behaves as a magnetic dipole with moment $m=IA$. Both are false.

   1 mark

Answer: Option (D)

1. Electrons have a de Broglie wavelength, so they diffract at a narrow slit. This wave nature (dual behaviour) is exactly what explains the assertion.

   1 mark

Answer: Option (A)

1. The assertion is true: the mass defect exists.

2. The reason is false: energy is released, not absorbed, when nucleons bind (this is the binding energy).

   1 mark

Answer: Option (C)

1. Both statements are true.

2. Quantisation of energy comes from the quantisation of angular momentum, $mvr=nh/2\pi$, not from the Coulomb force supplying the centripetal force. So R does not explain A.

   1 mark

Answer: Option (B)

1. Photon energy: $E=\frac{hc}{\lambda }=\frac{1240\text{eV nm}}{200\text{nm}}=6.2\text{eV}$

   1 mark

2. The photocurrent stops at a collector potential of $-0.80$ V, so the stopping potential is $V_0=0.80$ V. Hence $K_{max}=e{V}_0=0.80$ eV.

   ½ mark

3. Einstein's photoelectric equation gives ${\varphi }_0=\frac{hc}{\lambda }-e{V}_0=6.2-0.80=5.4\text{eV}$

   ½ mark

Answer: Work function ≈ 5.4 eV

1. In single-slit diffraction the $n$th minimum lies at $y_n=\frac{n\lambda D}{a}$.

   ½ mark

2. For the minima to coincide: $n_1{\lambda }_1=n_2{\lambda }_2\Rightarrow n_1(400)=n_2(600)\Rightarrow \frac{n_1}{n_2}=\frac{3}{2}$. The least distance corresponds to $n_1=3$, $n_2=2$.

   ½ mark

3. $y=\frac{3\times 400\times 10^{-9}\times 1.5}{1\times 10^{-3}}=1.8\times 10^{-3}$ m $=1.8$ mm

   1 mark

Answer: 1.8 mm from the central maximum

1. In YDSE the $n$th bright fringe lies at $y_n=\frac{n\lambda D}{d}$.

   ½ mark

2. For the bright fringes to coincide: $n_1(440)=n_2(660)\Rightarrow \frac{n_1}{n_2}=\frac{3}{2}$. Take the least values $n_1=3$, $n_2=2$.

   ½ mark

3. $y=\frac{3\times 440\times 10^{-9}\times 1.5}{0.6\times 10^{-3}}=3.3\times 10^{-3}$ m $=3.3$ mm

   1 mark

Answer: 3.3 mm from the central maximum

1. The torque is maximum when the field lies in the plane of the coil: ${\tau }_{max}=NIAB$. Since $N$, $I$ and $B$ are the same for both coils, ${\tau }_{max}\propto A$.

   ½ mark

2. Square coil: perimeter of one turn $=\frac{L}{N}$, so side $=\frac{L}{4N}$ and $A_s=\frac{L^2}{16N^2}$.

   ½ mark

3. Circular coil: $2\pi r=\frac{L}{N}\Rightarrow r=\frac{L}{2\pi N}$, so $A_c=\pi r^2=\frac{L^2}{4\pi N^2}$.

   ½ mark

4. $\frac{{\tau }_s}{{\tau }_c}=\frac{A_s}{A_c}=\frac{L^2/16N^2}{L^2/4\pi N^2}=\frac{\pi }{4}$

   ½ mark

Answer: ${\tau }_{square}:{\tau }_{circle}=\pi :4$

1. The drift velocity of electrons in a conductor is of the order of $10^{-4}$ m/s (about a fraction of a mm/s).

   ½ mark

2. Take a conductor of cross-sectional area $A$ with $n$ free electrons per unit volume, each drifting with speed $v_d$. In time $\Delta t$, every electron within a length $v_d\Delta t$ crosses a given section. The number crossing is $nA{v}_d\Delta t$, so the charge crossing is $\Delta Q=neA{v}_d\Delta t$.

   1 mark

3. Current: $I=\frac{\Delta Q}{\Delta t}=neA{v}_d$

   ½ mark

Answer: $v_d\sim 10^{-4}$ m/s; $I=neA{v}_d$

1. Graph: Plot potential energy $U$ (MeV) on the y-axis against separation $r$ (fm) on the x-axis.

   • For small $r$, $U$ is large and positive.
   • $U$ falls steeply and crosses zero.
   • It reaches a minimum (about $-100$ MeV) at $r_0\approx 0.8$ fm.
   • It then rises and approaches zero asymptotically for $r$ beyond about 2–3 fm.
   1 mark

2. Conclusion 1: The nuclear force is attractive for separations greater than $r_0\approx 0.8$ fm, where the slope of $U$ is positive. It is strongly repulsive for separations less than $r_0$, which keeps nucleons from collapsing into one another.

   ½ mark

3. Conclusion 2: The nuclear force is short-ranged. It becomes negligible beyond a few femtometres, since $U\to 0$. It is maximally binding (minimum potential energy) near $r_0\approx 0.8$ fm.

   ½ mark

Answer: U–r curve with a minimum at about 0.8 fm. The force is repulsive below 0.8 fm, attractive above it, and short-ranged (vanishes beyond a few fm).

1. (a) Take an infinite thin sheet with uniform surface charge density $\sigma$. By symmetry, $\vec{E}$ is normal to the sheet, points away from it on both sides (for $\sigma >0$), and has the same magnitude at equal distances. Choose as the Gaussian surface a cylinder of cross-section area $A$ whose axis is normal to the sheet and which extends equally on both sides.

   ½ mark

2. Flux through the curved surface is zero because $\vec{E}\perp d\vec{A}$ there. Flux through the two flat faces $=EA+EA=2EA$.

   ½ mark

3. Charge enclosed $=\sigma A$. By Gauss's law, $2EA=\frac{\sigma A}{{\epsilon }_0}\Rightarrow E=\frac{\sigma }{2{\epsilon }_0}$, directed normally away from the sheet, so $\vec{E}=\frac{\sigma }{2{\epsilon }_0}\hat{n}$.

   1 mark

4. (b)(i) Between the sheets the two fields, each $\frac{\sigma }{2{\epsilon }_0}$, are equal and opposite: $E_{inside}=\frac{\sigma }{2{\epsilon }_0}-\frac{\sigma }{2{\epsilon }_0}=0$.

   ½ mark

5. (b)(ii) Outside the sheets the two fields point the same way: $E_{outside}=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}$, directed away from the sheets.

   ½ mark

Answer: $E=\sigma /2{\epsilon }_0$; (i) zero inside; (ii) $\sigma /{\epsilon }_0$ outside

1. (a) Wheatstone bridge set-up:

   • Arms: $P$ (A to B), $Q$ (B to C), $R$ (A to D), $S$ (D to C).
   • Battery between A and C.
   • Galvanometer $G$ between B and D.

   At balance $I_g=0$, so the current $I_1$ in P also flows through Q, and the current $I_2$ in R also flows through S.

   ½ mark

2. Kirchhoff's loop rule:

   • Loop ABDA: $I_{1}P+I_{g}G-I_{2}R=0\Rightarrow I_{1}P=I_{2}R$
   • Loop BCDB: $I_{1}Q-I_{2}S-I_{g}G=0\Rightarrow I_{1}Q=I_{2}S$
   ½ mark

3. Dividing the two equations gives the balance condition $\frac{P}{Q}=\frac{R}{S}$.

   ½ mark

4. (b) Reading of the figure: between M and N lie the resistors

   • M–O = R, M–P (top) = R, O–N (bottom) = R, P–N = R
   • a bridge O–P = R

   This is a Wheatstone bridge with $\frac{R}{R}=\frac{R}{R}$, so it is balanced and no current flows in the O–P resistor.

   ½ mark

5. Ignoring O–P, the two paths M–O–N and M–P–N are each $2R$ and are in parallel: $R_{MN}=\frac{2R\times 2R}{4R}=R$.

   ½ mark

6. Net resistance: $R_{AB}=2R+R+3R=6R$

   ½ mark

Answer: (a) $P/Q=R/S$; (b) $R_{AB}=6R$

1. After the battery is disconnected the charge $Q=CV$ on the plates stays constant, because it has no path to flow away.

2. (a) Capacitance: with the slab $C=\frac{K{\epsilon }_{0}A}{d}$. Without it, $C^{\prime }=\frac{{\epsilon }_{0}A}{d}=\frac{C}{K}$. The capacitance decreases by a factor $K$.

   1 mark

3. (b) Energy: $U=\frac{Q^2}{2C}$ with $Q$ constant, so $U^{\prime }=\frac{Q^2}{2(C/K)}=KU$. The energy increases $K$ times. The extra energy comes from the work done in pulling the slab out against the attractive force.

   1 mark

4. (c) Potential difference: $V^{\prime }=\frac{Q}{C^{\prime }}=\frac{KQ}{C}=KV$. The p.d. increases $K$ times.

   1 mark

Answer: (a) Decreases to C/K; (b) increases to K times; (c) increases to KV

1. The electron enters midway, so to just strike $P_2^{\prime }$ it must be deflected by $y=\frac{d}{2}=1$ cm $=0.01$ m while travelling $x=3$ cm $=0.03$ m horizontally.

   ½ mark

2. Time between the plates: $t=\frac{x}{v}=\frac{0.03}{3\times 10^7}=1\times 10^{-9}$ s

   ½ mark

3. From $y=\frac{1}{2}a{t}^2$: $a=\frac{2y}{t^2}=\frac{2\times 0.01}{(10^{-9}{)}^2}=2\times 10^{16}$ m/s²

   ½ mark

4. The acceleration comes from the field $E=\frac{V}{d}$: $a=\frac{eE}{m}=\frac{eV}{md}$

   ½ mark

5. $V=\frac{amd}{e}=\frac{2\times 10^{16}\times 9.1\times 10^{-31}\times 0.02}{1.6\times 10^{-19}}\approx 2.3\times 10^3$ V

   1 mark

Answer: V ≈ 2.3 × 10³ V (about 2275 V)

1. (a) In the positive half-cycle A is at a higher potential than B.

   • $D_2$ (anode at A, cathode towards R) is forward biased and conducts.
   • $D_1$ (anode at P, cathode at A) is reverse biased, since its cathode is at the higher potential, and does not conduct.
   1 mark

2. (b) Equivalent circuit: $D_2$ is replaced by a short wire and $D_1$ by an open switch. Current flows from A through $D_2$ to R. At R it divides:

   • one part through 3 kΩ directly to B;
   • the other through 1 kΩ (R to P) and then 2 kΩ (P to B).

   So 3 kΩ is in parallel with the series pair (1 kΩ + 2 kΩ) = 3 kΩ.

   1 mark

3. (c) At the peak, $V_i=12$ V and the ideal diode drops nothing, so $V_{RB}=12$ V.

   • Across 3 kΩ: $V=12$ V (current 4 mA).
   • Current in the 1 kΩ–2 kΩ branch: $I=\frac{12}{3\text{k}\Omega }=4$ mA.
   • $V_{1k}=4\text{ mA}\times 1\text{ k}\Omega =4$ V; $V_{2k}=4\text{ mA}\times 2\text{ k}\Omega =8$ V.
   1 mark

Answer: D₂ conducts (forward biased) and D₁ does not. At peak: 12 V across 3 kΩ, 4 V across 1 kΩ, 8 V across 2 kΩ.

1. Diffusion: When the junction forms, there is a concentration gradient across it. Holes diffuse from the p-side to the n-side, and electrons from the n-side to the p-side. This produces the diffusion current, directed from p to n.

   1 mark

2. Diffusion leaves behind uncovered immobile ions: positive donor ions on the n-side and negative acceptor ions on the p-side. This forms the depletion layer, a region with no free charge carriers, and sets up a junction electric field (barrier potential) directed from n to p.

   1 mark

3. Drift: The junction field sweeps minority carriers across the junction, electrons from p to n and holes from n to p. This produces the drift current, opposite to the diffusion current. Diffusion current initially exceeds drift current. Equilibrium is reached when the two are equal, so the net current is zero.

   1 mark

Answer: Diffusion of majority carriers, which creates the depletion layer and barrier potential; drift of minority carriers under the junction field, which balances diffusion at equilibrium.

1. (a) Ray diagram (normal adjustment):
   1. Parallel rays from a distant object fall on the objective (large focal length $f_o$), inclined at angle $\beta$ to the axis.
   2. They form a real, inverted, diminished image at the common focus of the objective and eyepiece.
   3. The eyepiece (small $f_e$) placed at distance $f_e$ from this image makes the rays emerge parallel at angle $\alpha$, giving the final image at infinity.
   4. Tube length $=f_o+f_e$.

   1½ marks

2. Angular magnification in normal adjustment: $m=\frac{\alpha }{\beta }=-\frac{f_o}{f_e}$ (magnitude $\frac{f_o}{f_e}$).

   ½ mark

3. (b) Reason 1: A mirror objective has no chromatic aberration. A paraboloidal mirror also removes spherical aberration.

   ½ mark

4. Reason 2: A mirror is lighter than a lens of equal aperture and can be supported over its whole back surface, so very large apertures are possible. This gives higher resolving power and light-gathering power at lower cost.

   ½ mark

Answer: m = −f_o/f_e. A reflector is free from chromatic (and spherical) aberration, and large mirrors are easier and cheaper to make and support, giving higher resolving power.

1. (a) Total internal reflection occurs when:
   (i) light travels from an optically denser medium to an optically rarer medium; and
   (ii) the angle of incidence in the denser medium is greater than the critical angle for the pair.

   1 mark

2. (b) At the A–B interface, Snell's law gives $n\sin \theta =\frac{3}{4}n\sin r\Rightarrow \sin r=\frac{4}{3}\sin \theta$. The interfaces are parallel, so $r$ is the angle of incidence at the B–C interface.

   ½ mark

3. Critical angle at the B–C interface (B denser): $\sin C=\frac{n_C}{n_B}=\frac{(2/3)n}{(3/4)n}=\frac{8}{9}$

   ½ mark

4. The beam does not enter C if $\sin r\ge \sin C$: $\frac{4}{3}\sin \theta \ge \frac{8}{9}\Rightarrow \sin \theta \ge \frac{2}{3}$. The beam is then totally reflected at B–C, so it never enters C. Equivalently, $n\sin \theta =\frac{2}{3}n\sin {\theta }_C$ would require $\sin {\theta }_C=\frac{3}{2}\sin \theta \ge 1$, which is impossible.

   1 mark

Answer: TIR conditions: denser to rarer, and i > critical angle. The beam is totally reflected at B–C when sinθ ≥ 2/3.

1. (I) A radial magnetic field keeps the plane of the coil always parallel to $\vec{B}$, so $\tau =NIAB$ at every orientation. Option (B).

   1 mark

2. (II) Current sensitivity is $\frac{\varphi }{I}=\frac{NAB}{k}$. Coil area and magnet strength are limited in practice, so the convenient (best) way to raise it is to increase the number of turns $N$. Option (A).

   1 mark

3. (III) $\tau =NIAB=50\times 5\times 4.0\times 10^{-3}\times 0.25=0.25$ N m. Option (D).

   OR: Series resistance $R=\frac{V}{I_g}-G=\frac{12}{3\times 10^{-3}}-15=3985\Omega$. Option (B).

   1 mark

4. (IV) Shunt $S=\frac{I_{g}G}{I-I_g}=\frac{5\times 10^{-3}\times 20}{10-0.005}\approx 0.01\Omega$, connected in parallel. Option (C).

   1 mark

Answer: (I) B; (II) A; (III) D [or alternative: B]; (IV) C

1. (I) Threshold frequencies are ${\nu }_1$ (A) and ${\nu }_2$ (B), and ${\varphi }_0=h{\nu }_0$. The work functions are therefore $h{\nu }_1$ and $h{\nu }_2$. Option (C).

   1 mark

2. (II) $K_{max}=h\nu -{\varphi }_0$. Since ${\varphi }_A<{\varphi }_B$, $K_{max}$ is greater for A. Option (A).

   1 mark

3. (III) Intensity affects only the number of photons. The slope $h/e$ and the threshold frequencies stay the same, but more electrons are emitted per second. Option (D).

   1 mark

4. (IV)(a) $E_1=h(3{\nu }_0)-h{\nu }_0=2h{\nu }_0$ and $E_2=6h{\nu }_0-h{\nu }_0=5h{\nu }_0$, so $\frac{E_1}{E_2}=\frac{2}{5}$. Option (C).

   (IV)(b) $V_s=\frac{h}{e}\nu -\frac{{\varphi }_0}{e}$, so the slope is $m=\frac{h}{e}$ and $h=me$. Option (A).

   1 mark

Answer: (I) C; (II) A; (III) D; (IV) C [or alternative: A]

1. (a) Place $-q$ at B and $+q$ at A, separated by $2a$, with centre O. Let P lie on the equatorial line with OP $=r$. Each charge is at distance $\sqrt{r^2+a^2}$ from P, so
   $|{\vec{E}}_{+q}|=|{\vec{E}}_{-q}|=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2+a^2}$

   1 mark

2. Resolve the two fields. Their components normal to the dipole axis cancel. The components along the axis add, both pointing from $+q$ towards $-q$, i.e. opposite to $\vec{p}$. If $\theta$ is the angle each field makes with the axis, $\cos \theta =\frac{a}{\sqrt{r^2+a^2}}$.

   ½ mark

3. $E=2E_{+q}\cos \theta =\frac{2qa}{4\pi {\epsilon }_0(r^2+a^2{)}^{3/2}}$. With $\vec{p}=q(2a)\hat{p}$: $\vec{E}=-\frac{\vec{p}}{4\pi {\epsilon }_0(r^2+a^2{)}^{3/2}}$

   1 mark

4. For $r\gg a$: $\vec{E}=-\frac{\vec{p}}{4\pi {\epsilon }_0{r}^3}$

   ½ mark

5. (b) The field $\vec{E}=2\hat{i}$ N/C is uniform. Net force: $\vec{F}=q\vec{E}+(-q)\vec{E}=0$.

   1 mark

6. Dipole moment (from $-q$ to $+q$): $\vec{p}=q(a-b)\hat{i}$. Torque: $\vec{\tau }=\vec{p}\times \vec{E}=q(a-b)\cdot 2(\hat{i}\times \hat{i})=0$, since $\vec{p}\parallel \vec{E}$.

   1 mark

Answer: $\vec{E}=-\frac{\vec{p}}{4\pi {\epsilon }_0(r^2+a^2{)}^{3/2}}$, which becomes $-\frac{\vec{p}}{4\pi {\epsilon }_0{r}^3}$ for $r\gg a$; (b) $\vec{F}=0$, $\vec{\tau }=0$

1. (a) Let currents $I_1$ and $I_2$ leave the positive terminals of cells 1 and 2, with total current $I=I_1+I_2$. Let $V$ be the common terminal p.d. Then
   $V=E_1-I_1{r}_1\Rightarrow I_1=\frac{E_1-V}{r_1}$, and $V=E_2-I_2{r}_2\Rightarrow I_2=\frac{E_2-V}{r_2}$

   1 mark

2. $I=\frac{E_1}{r_1}+\frac{E_2}{r_2}-V\left(\frac{1}{r_1}+\frac{1}{r_2}\right)$, which rearranges to
   $V=\frac{E_1{r}_2+E_2{r}_1}{r_1+r_2}-I\frac{r_1{r}_2}{r_1+r_2}$

   1 mark

3. Comparing with $V=E_{eq}-I{r}_{eq}$:
   $E_{eq}=\frac{E_1{r}_2+E_2{r}_1}{r_1+r_2}$, $r_{eq}=\frac{r_1{r}_2}{r_1+r_2}$

   1 mark

4. (b) $E_{eq}=\frac{E\cdot R+3E\cdot R}{R+R}=2E$ and $r_{eq}=\frac{R\cdot R}{2R}=\frac{R}{2}$

   1 mark

5. $I=\frac{E_{eq}}{2R+r_{eq}}=\frac{2E}{2R+R/2}=\frac{4E}{5R}$

   1 mark

Answer: $E_{eq}=\frac{E_1{r}_2+E_2{r}_1}{r_1+r_2}$, $r_{eq}=\frac{r_1{r}_2}{r_1+r_2}$; current through 2R $=\frac{4E}{5R}$

1. (a) Take a thin lens of refractive index $n_2$ in a medium of index $n_1$, with surface radii $R_1$ and $R_2$. An object is at distance $u$. Refraction at the first surface would form an image $I_1$ at $v_1$:
   $\frac{n_2}{v_1}-\frac{n_1}{u}=\frac{n_2-n_1}{R_1}$ ...(1)

   1 mark

2. $I_1$ acts as a virtual object for the second surface, since the lens is thin. The final image at $v$ satisfies
   $\frac{n_1}{v}-\frac{n_2}{v_1}=\frac{n_1-n_2}{R_2}$ ...(2)

   1 mark

3. Adding (1) and (2): $n_1\left(\frac{1}{v}-\frac{1}{u}\right)=(n_2-n_1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$. For $u=\infty$, $v=f$, so
   $\frac{1}{f}=(n_{21}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$, where $n_{21}=\frac{n_2}{n_1}$.

   1 mark

4. (b) Lens $L_1$: $u=-80$ cm, $f=+40$ cm. $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{40}-\frac{1}{80}=\frac{1}{80}$, so $v=80$ cm to the right of $L_1$.

   ½ mark

5. Lens $L_2$ is 120 cm from $L_1$, so the image lies 40 cm to its left: $u=-40$ cm $=-f$. The image is at infinity, i.e. the rays leaving $L_2$ are parallel.

   ½ mark

6. Lens $L_3$ focuses these parallel rays at its focus, 40 cm to its right. $L_3$ is $120+20=140$ cm from $L_1$, so the final image is $180$ cm right of $L_1$.

   ½ mark

7. Distance from the object: $80+180=260$ cm.

   ½ mark

Answer: $\frac{1}{f}=(n_{21}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$; final image is 260 cm from the object (40 cm to the right of L₃)

1. (a) Ray diagram (described in words). Draw a concave mirror with pole $P$, focus $F$ and centre of curvature $C$ on the principal axis. Place a small object $AB$ with $B$ on the axis between $F$ and $C$ and $A$ above the axis.

   • Ray 1: from $A$ parallel to the axis, striking the mirror at $M$ and reflecting through $F$.
   • Ray 2: from $A$ to the pole $P$, reflecting so that it makes an equal angle with the axis on the other side.

   The reflected rays meet at $A^{\prime }$ below the axis, beyond $C$. Draw the perpendicular $A^{\prime }{B}^{\prime }$ to the axis. The image $A^{\prime }{B}^{\prime }$ is real, inverted and magnified.

   1 mark

2. Similar triangles. Consider paraxial rays, so $M$ is close to $P$ and $MP$ can be taken as a straight line perpendicular to the axis, with $MP=AB$.

   • $△A^{\prime }{B}^{\prime }F\sim △MPF$ (equal vertically opposite angles at $F$, and right angles at $B^{\prime }$ and $P$). Hence $\frac{B^{\prime }{A}^{\prime }}{PM}=\frac{B^{\prime }F}{FP}$, which gives $\frac{B^{\prime }{A}^{\prime }}{BA}=\frac{B^{\prime }F}{FP}$ ...(1)
   • $△A^{\prime }{B}^{\prime }P\sim △ABP$ (angle of incidence equals angle of reflection at $P$, and right angles at $B$ and $B^{\prime }$). Hence $\frac{B^{\prime }{A}^{\prime }}{BA}=\frac{B^{\prime }P}{BP}$ ...(2)

   From (1) and (2):
   $$\frac{B^{\prime }F}{FP}=\frac{B^{\prime }P}{BP},\text{where }{B}^{\prime }F=B^{\prime }P-FP.$$

   1 mark

3. Apply the sign convention. All distances are measured from $P$, and distances against the incident light are negative. So $BP=-u$, $B^{\prime }P=-v$ and $FP=-f$.

   Substituting:
   $$\frac{-v+f}{-f}=\frac{-v}{-u}$$
   $$\Rightarrow \frac{v-f}{f}=\frac{v}{u}$$
   $$\Rightarrow \frac{v}{f}-1=\frac{v}{u}$$

   Dividing throughout by $v$:
   $$\frac{1}{f}-\frac{1}{v}=\frac{1}{u}$$
   $$\Rightarrow \boxed{\frac{1}{v}+\frac{1}{u}=\frac{1}{f}}$$

   This is the mirror formula.

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4. (b) Find the image distance. The image is virtual and erect, so $m=+2$. The object is in front of the mirror, so $u=-10$ cm.

   $$m=-\frac{v}{u}=2$$
   $$\Rightarrow v=-2u=-2(-10)=+20\text{ cm}$$

   The image is 20 cm behind the mirror.

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5. Find the focal length.
   $$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{20}+\frac{1}{-10}=\frac{1-2}{20}=-\frac{1}{20}$$
   $$\Rightarrow f=-20\text{ cm}$$

   The focal length of the concave mirror is 20 cm.

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Answer: (a) Mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$; (b) $f=-20$ cm (focal length 20 cm)

1. (a) Faraday's law. The magnitude of the emf induced in a circuit equals the time rate of change of magnetic flux linked with it. Its direction is such that it opposes that change (Lenz's law).
   $$\epsilon =-N\frac{d{\Phi }_B}{dt}$$

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2. (b) Field and flux linkage. Let a current $I$ flow in the solenoid. The number of turns per unit length is $n=N/l$.

   For a long solenoid, the field inside is uniform:
   $$B={\mu }_{0}nI=\frac{{\mu }_{0}NI}{l}$$

   The flux through each turn is $\Phi =BA$. So the total flux linkage is:
   $$N\Phi =N\cdot \frac{{\mu }_{0}NI}{l}\cdot A=\frac{{\mu }_0{N}^{2}A}{l}I$$

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3. Self-inductance. Since $N\Phi =LI$:
   $$\boxed{L=\frac{{\mu }_0{N}^{2}A}{l}}(\text{or }{\mu }_0{n}^{2}Al)$$

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4. (c) Convert the data to SI units.

   • $l=50$ cm $=0.5$ m
   • $B=4.0$ mT $=4.0\times 10^{-3}$ T
   • $\omega =60$ rpm $=\frac{60\times 2\pi }{60}=2\pi$ rad s${}^{-1}$

   For a rod rotating about one end, the emf is:
   $$\epsilon =\frac{1}{2}B\omega l^2$$

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5. Substitute.
   $$\epsilon =\frac{1}{2}\times 4.0\times 10^{-3}\times 2\pi \times (0.5{)}^2=\pi \times 10^{-3}\text{ V}$$
   $$\approx 3.14\times 10^{-3}\text{ V}=3.14\text{ mV}$$

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Answer: (a) $\epsilon =-Nd\Phi /dt$; (b) $L={\mu }_0{N}^{2}A/l$; (c) $\epsilon \approx 3.14\times 10^{-3}$ V

1. (a) Diagram (described in words). Draw a closed rectangular laminated soft-iron core.

   • On one arm, wind the primary coil $P$ with few turns ($N_p$) and connect it to the ac input voltage $V_p$.
   • On the other arm, wind the secondary coil $S$ with many more turns ($N_s>N_p$) and connect it to the load across the output $V_s$.
   • Label the core, the primary, the secondary, the input and the output.
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2. Principle. A transformer works on mutual induction. An alternating current in the primary produces a changing magnetic flux in the core. This changing flux links the secondary and induces an emf in it.

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3. Ratio in terms of turns. Assume an ideal transformer with no flux leakage. The same flux $\Phi$ links each turn of both coils.

   $${\epsilon }_s=-N_s\frac{d\Phi }{dt},{\epsilon }_p=-N_p\frac{d\Phi }{dt}$$

   With negligible coil resistance, $V_p={\epsilon }_p$ and $V_s={\epsilon }_s$. Therefore:
   $$\frac{V_s}{V_p}=\frac{N_s}{N_p}$$

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4. Ratio in terms of currents. For an ideal transformer with no energy losses, input power equals output power:
   $$V_p{I}_p=V_s{I}_s$$

   Therefore:
   $$\boxed{\frac{V_s}{V_p}=\frac{N_s}{N_p}=\frac{I_p}{I_s}}$$

5. (b)(i) Current in the primary.
   $$I_p=\frac{P}{V_p}=\frac{5000\text{ W}}{200\text{ V}}=25\text{ A}$$

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6. (b)(ii) Output voltage.
   $$V_s=V_p\frac{N_s}{N_p}=200\times \frac{5}{1}=1000\text{ V}$$

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