HINTS & SOLUTIONS

Answer: (B) repel with a force $\frac{F}{2}$ Explanation: Initial force $F \propto \frac{(q)(-2q)}{r^2} = -\frac{2q^2}{r^2}$ (attractive). When brought in contact, total charge $= q + (-2q) = -q$. They share it equally, so each gets $-q/2$. New force $F' \propto \frac{(-q/2)(-q/2)}{(r/2)^2} = \frac{q^2/4}{r^2/4} = +\frac{q^2}{r^2}$ (repulsive). Ratio magnitude $|F'| / |F| = (q^2/r^2) / (2q^2/r^2) = 1/2$. Thus, $F' = F/2$ and it is repulsive.

Answer: (C) R Explanation: Electric field $E = -\frac{dV}{dx}$. The magnitude of force $F = qE$ is maximum where the magnitude of the slope of the V-x graph is maximum (steepest). Point R lies on the steepest segment of the generic potential-distance curve provided.

Answer: (B) $\frac{\mu_{0}I}{\pi a\sqrt{2}}$ and directed along OD Explanation: Distance from each corner to center is $r = \frac{a}{\sqrt{2}}$. The magnetic field from each wire at the center has magnitude $B_0 = \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{\sqrt{2}\pi a}$. Applying Right-Hand Grip Rule: Fields from the three downward currents and one upward current create pairs of vectors. A and C vectors add up, B and D vectors act, resulting in a net field directly diagonal along the symmetry axis OD.

Answer: (D) any one or more of the factors given in (A), (B) and (C) Explanation: Magnetic flux $\Phi = \vec{B} \cdot \vec{A} = BA \cos\theta$. It can be changed by changing the magnetic field strength (B), the area of the loop (A), or the angle/orientation ($\theta$).

Answer: (D) ac is less dangerous. Explanation: This statement is FALSE. For the same specified RMS voltage, AC is actually *more* dangerous than DC because its peak voltage ($V_{rms} \times \sqrt{2}$) is significantly higher, and the alternating nature interferes severely with the human heart's electrical system.

Answer: (C) $7.5\times10^{2}m^{-1}$ Solution: From equation $B_{y} = B_{0} \sin(kx + \omega t)$, we have $\omega = 1.5\times 10^{11} \text{ rad/s}$ and $k = \alpha$. Speed of wave in glass $v = \frac{c}{n} = \frac{3\times 10^{8}}{1.5} = 2\times 10^{8} \text{ m/s}$. Also, $v = \frac{\omega}{k} \implies k = \frac{\omega}{v} = \frac{1.5\times 10^{11}}{2\times 10^{8}} = 0.75\times 10^{3} = 7.5\times 10^{2} \text{ m}^{-1}$.

Answer: (D) Blue light Explanation: Energy of a photon $E = \frac{hc}{\lambda}$. Energy is inversely proportional to wavelength. Among the given options (VIBGYOR), Blue has the shortest wavelength and therefore the maximum energy.

Answer: (B) Isotones Explanation: Isotones are nuclei that have the same number of neutrons but different number of protons. (Isobars = same mass number; Isotopes = same atomic number).

Answer: (A) 6.0 fm Solution: Radius of nucleus $R = R_{0} A^{1/3}$. Here, $R_{0} \approx 1.2 \text{ fm}$ and $A = 125$. $R = 1.2 \times (125)^{1/3} = 1.2 \times 5 = 6.0 \text{ fm}$.

Answer: (C) $\frac{h}{\pi}$ Solution: Energy $E_{n} = \frac{-13.6}{n^{2}} \text{ eV}$. Given $E_{n} = -3.4 \text{ eV} \implies \frac{-13.6}{n^{2}} = -3.4 \implies n^{2} = 4 \implies n = 2$. Angular momentum $L = \frac{nh}{2\pi} = \frac{2h}{2\pi} = \frac{h}{\pi}$.

Answer: (A) high resistance in reverse bias and a low resistance in forward bias Explanation: A healthy p-n junction diode acts as a closed switch (low resistance) when forward-biased and an open switch (high/infinite resistance) when reverse-biased.

Answer: (C) $\frac{V_{0}}{\sqrt{2}},0$ Explanation: For a complete sinusoidal cycle, the root-mean-square (rms) voltage is $V_{rms} = \frac{V_{0}}{\sqrt{2}}$. The average voltage over a full cycle is strictly $0$ because the positive and negative half-cycles cancel each other out.

Answer: (D) Both Assertion (A) and Reason (R) are false. Explanation: According to Faraday's Law, the induced emf depends on the rate of change of magnetic flux ($e = -d\Phi/dt$), not on the magnitude of the flux itself. Even if the flux is very large, if it is constant, the induced emf is zero.

Answer: (A) Both A and R are true and R is the correct explanation of A. Explanation: In YDSE, both bright and dark fringes are equally spaced. The formula $\beta = \frac{\lambda D}{d}$ applies to the width of both types of fringes, making the reason the correct explanation.

Answer: (C) Assertion (A) is true, but Reason (R) is false. Explanation: Energy is indeed released in heavy fission and light fusion because the products are more stable. However, the reason is false: for heavy nuclei, Binding Energy per nucleon decreases with increasing Z (or A), and for light nuclei, it increases.

Answer: (A) Both A and R are true and R is the correct explanation of A. Explanation: Photoemission is virtually instantaneous ($10^{-9}$ s). The wave theory failed to explain this because it predicted the energy would be distributed continuously, requiring significant time for a single electron to absorb enough energy to escape. This failure (the Reason) justifies why the phenomenon is instantaneous and requires the photon (particle) model.

(a) Solution: Given: Rated Power $P_{0} = 2.2 \text{ kW} = 2200 \text{ W}$, Rated Voltage $V_{0} = 220 \text{ V}$. (i) Resistance: $R = \frac{V_{0}^2}{P_{0}} = \frac{220 \times 220}{2200} = 22 \text{ }\Omega$. (ii) Heat Produced: Operated at $V' = 110 \text{ V}$ for $t = 10 \text{ mins} = 600 \text{ s}$. Power at 110V: $P' = \frac{V'^2}{R} = \frac{110 \times 110}{22} = 550 \text{ W}$. Heat $H = P' \times t = 550 \text{ W} \times 600 \text{ s} = 330,000 \text{ J}$ (or $330 \text{ kJ}$). (b) Solution: Given: $I = 4.0 \text{ A}$, $l = 1 \text{ m}$, $A = 1.0 \text{ mm}^2 = 1.0 \times 10^{-6} \text{ m}^2$, $V = 2 \text{ V}$. First, find resistance using Ohm's Law: $R = \frac{V}{I} = \frac{2}{4} = 0.5 \text{ }\Omega$. Resistivity $\rho = \frac{R A}{l} = \frac{0.5 \times (1.0 \times 10^{-6})}{1} = 5.0 \times 10^{-7} \text{ }\Omega\text{m}$.

Solution: (a) Magnetic Flux ($\Phi$): At $t=0$, the plane is parallel to the field, meaning the area vector is perpendicular ($\theta = 90^\circ$). Flux $\Phi = BA \cos(90^\circ - \omega t) = BA \sin(\omega t)$. Plot: A standard sine wave starting from origin (0,0).

(b) Induced EMF ($e$): By Faraday's Law, $e = -\frac{d\Phi}{dt} = -\frac{d}{dt}(BA \sin\omega t) = -BA\omega \cos(\omega t)$. Plot: A negative cosine wave starting from a negative peak on the y-axis at $t=0$.

Solution: Refractive index of liquid $\mu = \frac{\text{Real Depth}}{\text{Apparent Depth}} = \frac{12.5 \text{ m}}{9.0 \text{ m}} = \frac{125}{90} = 1.388$. We know, $\mu = \frac{c}{v}$, where $c$ is speed of light in vacuum ($3 \times 10^8 \text{ m/s}$) and $v$ is speed in liquid. $v = \frac{c}{\mu} = \frac{3 \times 10^8}{125/90} = \frac{3 \times 90}{125} \times 10^8 = \frac{270}{125} \times 10^8 = 2.16 \times 10^8 \text{ m/s}$.

Solution: For the first lens (focal length $f_1$), forming image at $v_1$: $\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1}$ ... (Eq 1) For the second lens (focal length $f_2$), the image $v_1$ acts as a virtual object forming final image at $v$: $\frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2}$ ... (Eq 2) Adding Eq 1 and Eq 2: $\left(\frac{1}{v_1} - \frac{1}{u}\right) + \left(\frac{1}{v} - \frac{1}{v_1}\right) = \frac{1}{f_1} + \frac{1}{f_2} \implies \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2}$. If the combination is replaced by an equivalent lens of focal length $F$, then $\frac{1}{v} - \frac{1}{u} = \frac{1}{F}$. Therefore, $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{f_2 + f_1}{f_1 f_2} \implies F = \frac{f_1 f_2}{f_1 + f_2}$.

Solution: Arsenic is a pentavalent impurity. Doping Si with it creates an n-type semiconductor. Majority carriers are electrons. At room temperature, $n_e \approx N_D$ (Donor concentration). $n_e = 5 \times 10^{22} \text{ m}^{-3}$. Using the mass action law, $n_e \times n_h = n_i^2$. Minority carrier (hole) concentration $n_h = \frac{n_i^2}{n_e} = \frac{(1.5 \times 10^{16})^2}{5 \times 10^{22}} = \frac{2.25 \times 10^{32}}{5 \times 10^{22}} = 0.45 \times 10^{10} = 4.5 \times 10^9 \text{ m}^{-3}$.

Solution: (a) Let capacitance of X (air) be $C_X = C$. Capacitance of Y (dielectric $K=4$) is $C_Y = KC = 4C$. In series, $C_{eq} = \frac{C_X C_Y}{C_X + C_Y} = \frac{C \times 4C}{5C} = \frac{4C}{5}$. Given $C_{eq} = 4 \text{ }\mu\text{F} \implies \frac{4C}{5} = 4 \implies C = 5 \text{ }\mu\text{F}$. Thus, $C_X = 5 \text{ }\mu\text{F}$ and $C_Y = 20 \text{ }\mu\text{F}$.

(b) In series, charge $Q$ is same on both. $Q = C_{eq} \times V = 4 \text{ }\mu\text{F} \times 6 \text{ V} = 24 \text{ }\mu\text{C}$. Potential across X: $V_X = \frac{Q}{C_X} = \frac{24}{5} = 4.8 \text{ V}$. Potential across Y: $V_Y = \frac{Q}{C_Y} = \frac{24}{20} = 1.2 \text{ V}$.

Solution: Vector Form of Biot-Savart Law: $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \vec{r})}{r^3}$ (or $\frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \hat{r})}{r^2}$).
Calculation: Current element $Id\vec{l} = 10 \times (10^{-2} \text{ m}) \hat{i} = 10^{-1} \hat{i} \text{ A}\cdot\text{m}$. Position vector $\vec{r} = (1\hat{i} + 1\hat{j} + 0\hat{k}) \text{ m}$. Magnitude $r = \sqrt{1^2 + 1^2} = \sqrt{2} \text{ m}$. Cross product: $d\vec{l} \times \vec{r} = (10^{-2} \hat{i}) \times (\hat{i} + \hat{j}) = 10^{-2}(\hat{i}\times\hat{i} + \hat{i}\times\hat{j}) = 10^{-2} \hat{k}$. $d\vec{B} = 10^{-7} \times \frac{10 \times (10^{-2} \hat{k})}{(\sqrt{2})^3} = \frac{10^{-8}}{2\sqrt{2}} \hat{k} = \frac{10^{-8}}{2 \times 1.414} \hat{k} \approx 3.53 \times 10^{-9} \hat{k} \text{ T}$. The field points along the positive z-axis.

Solution: Mutual Inductance: Let current $I_1$ flow in the inner long solenoid. Magnetic field inside it is $B_1 = \mu_0 n_1 I_1 = \mu_0 \left(\frac{N_1}{L}\right) I_1$. The flux linked with each turn of the outer coil (which surrounds the mid-point, so field is uniform) is confined only to the area of the inner solenoid $\pi r_1^2$ (since $B_1 = 0$ outside $r_1$). Flux through outer coil $\Phi_2 = N_2 \times (B_1 \times \text{Area}) = N_2 \left(\mu_0 \frac{N_1}{L} I_1\right) (\pi r_1^2)$. By definition, $\Phi_2 = M_{12} I_1$. Comparing, $M_{12} = \frac{\mu_0 N_1 N_2 \pi r_1^2}{L}$.
Validity of $M_{12} = M_{21}$: Yes, according to the reciprocity theorem, the mutual inductance between any two geometrically fixed circuits is symmetric. $M_{12} = M_{21}$ holds perfectly valid here.

Solution: Displacement Current ($i_d$): It is that current which comes into existence in a region wherever the electric field (and hence electric flux) is changing with time. It maintains the continuity of current in circuits containing capacitors.
Derivation: Consider a parallel plate capacitor charging with current $i_c$. Charge on plates is $q$. Electric field between plates is $E = \frac{\sigma}{\epsilon_0} = \frac{q}{A\epsilon_0}$. Electric flux $\Phi_E = E \times A = \left(\frac{q}{A\epsilon_0}\right) A = \frac{q}{\epsilon_0}$. Differentiating with respect to time $t$: $\frac{d\Phi_E}{dt} = \frac{1}{\epsilon_0} \frac{dq}{dt}$. Since rate of change of charge is conduction current ($i_c = dq/dt$), we get $i_c = \epsilon_0 \frac{d\Phi_E}{dt}$. By Maxwell's theory, this is the displacement current $i_d$ bridging the gap. Thus, $i_d = \epsilon_0 \frac{d\Phi_E}{dt}$.
For a constant voltage: The electric field is steady (not changing with time). Therefore, $\frac{d\Phi_E}{dt} = 0$, which means $i_d = 0$.

(a) Solution: (i) Features of nuclear forces: (1) They are the strongest forces in nature but have a very short range (~2-3 fm). (2) They are charge-independent (same between p-p, n-n, p-n).
(ii) Mass-Energy Conversion: Though nucleon number is conserved, the *binding state* of the nucleons changes. The rest mass of the combined nucleus is slightly less than the sum of the rest masses of individual nucleons. This "missing mass" ($\Delta m$, mass defect) is converted into kinetic energy/gamma radiation according to Einstein's mass-energy equivalence $E = \Delta m c^2$. (b) Solution: (i) Graph & Conclusions: (A standard graph showing $N(\theta)$ rapidly decaying exponentially against $\theta$, approaching $180^\circ$ asymptotically). *Conclusions:* (1) Most of the space inside the atom is empty (most pass undeflected). (2) The entire positive charge and almost all mass are concentrated in an extremely tiny, dense central core called the nucleus (large angle deflections).
(ii) Planetary Orbits: For macroscopic bodies like planets, mass and radius are huge. Applying $L = mvr = \frac{nh}{2\pi}$, the quantum number $n$ becomes incredibly large (e.g., $10^{70}$ for Earth). The energy/momentum difference between consecutive allowed orbits ($\Delta n = 1$) is so vanishingly small that the states appear continuous. Thus, quantum effects are entirely unobservable macroscopically.

Solution: Given: Work function $\phi_0 = 1.96 \text{ eV}$, Frequency $\nu = 6.4 \times 10^{14} \text{ Hz}$, $h = 6.63 \times 10^{-34} \text{ Js}$.
(a) Energy of incident photon ($E$): $E = h\nu = 6.63 \times 10^{-34} \times 6.4 \times 10^{14} = 42.432 \times 10^{-20} \text{ J}$. In eV: $E = \frac{42.432 \times 10^{-20}}{1.6 \times 10^{-19}} = \frac{4.2432}{1.6} \approx 2.65 \text{ eV}$.
(b) Maximum Kinetic Energy ($K_{max}$): From Einstein's Equation: $K_{max} = E - \phi_0 = 2.65 \text{ eV} - 1.96 \text{ eV} = 0.69 \text{ eV}$.
(c) Stopping Potential ($V_s$): Since $K_{max} = e V_s$, we have $e V_s = 0.69 \text{ eV} \implies V_s = 0.69 \text{ Volts}$.

Solution: Circuit Diagram: Features an AC input connected to the primary of a center-tapped transformer. The secondary ends are connected to the p-sides of diodes $D_1$ and $D_2$. Their n-sides are tied together and connected to one end of a load resistor $R_L$. The other end of $R_L$ is connected to the center tap.
Working: During the positive half cycle of AC, the upper end of the secondary is positive, making $D_1$ forward-biased (conducting) and $D_2$ reverse-biased (open). Current flows through $R_L$ downwards. During the negative half cycle, the lower end becomes positive. $D_2$ becomes forward-biased and $D_1$ reverse-biased. Current again flows through $R_L$ in the same downward direction.
Waveforms: Input is a continuous sine wave spanning above and below the zero axis. Output is a series of adjacent positive half-cycle pulses (pulsating DC) with no gaps.

Solution: (i) (D) $-(2\times10^{4}\frac{V}{m})\hat{k}$ Reasoning: Electric field $E = V/d = 200 / 0.01 = 20,000 \text{ V/m} = 2 \times 10^4 \text{ V/m}$. The field points from positive to negative plate (downward along z-axis, assuming standard orientation), hence $-\hat{k}$.

(ii) (B) $(3.5\times10^{15}ms^{-2})\hat{k}$ Reasoning: $a = \frac{F}{m} = \frac{eE}{m} = \frac{1.6 \times 10^{-19} \times 2 \times 10^4}{9.1 \times 10^{-31}} \approx 3.5 \times 10^{15} \text{ m/s}^2$. Since the electron is negative, force and acceleration are opposite to the E-field, pointing upward $+\hat{k}$.

(iii) (C) $1.67\times10^{-9}s$ Reasoning (a): Time $t = \frac{\text{horizontal distance}}{\text{horizontal velocity}} = \frac{0.05 \text{ m}}{3 \times 10^7 \text{ m/s}} \approx 1.67 \times 10^{-9} \text{ s}$. (B) 4.9 mm Reasoning (b): Vertical displacement $y = \frac{1}{2} a t^2 = 0.5 \times 3.5 \times 10^{15} \times (1.67 \times 10^{-9})^2 = 1.75 \times 10^{15} \times 2.78 \times 10^{-18} \approx 4.88 \times 10^{-3} \text{ m} \approx 4.9 \text{ mm}$.

(iv) (A) a Reasoning: The electron enters a uniform transverse electric field. Its horizontal velocity remains constant, while it experiences a constant vertical acceleration, generating a parabolic trajectory curving towards the positive plate.

Solution: (i) (A) Wave nature of light Reasoning: Interference (the superposition of waves resulting in bright and dark regions) is a hallmark property that can only be explained by treating light as a continuous wave.

(ii) (C) 480 nm Reasoning (a): Standard calculation for a 1.2mm fringe width with D=5m, d=2mm yields $\lambda = \frac{\beta d}{D} = \frac{1.2\times10^{-3} \times 2\times10^{-3}}{5} = 480 \text{ nm}$. (A) 1.2 mm Reasoning (b): Assuming the source is 480nm, $\beta = \frac{480\times10^{-9} \times 5}{2\times10^{-3}} = 1.2 \text{ mm}$.

(iii) (B) $7.2\times10^{-7}m$ Reasoning: For a minimum, path difference $\Delta x = (2n-1)\frac{\lambda}{2}$. If $n=2$ (second minimum), $\Delta x = \frac{3}{2} \times 480 \text{ nm} = 720 \text{ nm} = 7.2 \times 10^{-7} \text{ m}$.

(iv) (D) be compressed Reasoning: When immersed in a liquid of refractive index $\mu$, the wavelength of light decreases to $\lambda' = \lambda/\mu$. Since fringe width $\beta \propto \lambda$, the new fringe width $\beta' = \beta/\mu$, meaning the pattern shrinks or becomes compressed.

(a) Solution: (i) Wheatstone Bridge Balance: Let the bridge arms be P, Q, R, S. A galvanometer G connects the junctions. When balanced, current through G is zero ($I_g=0$). Points B and D are at the same potential. Apply KVL to loop ABDA: $I_1 P - I_2 R = 0 \implies I_1 P = I_2 R$. Apply KVL to loop BCDB: $I_1 Q - I_2 S = 0 \implies I_1 Q = I_2 S$. Dividing the two equations gives the balance condition: $\frac{P}{Q} = \frac{R}{S}$.
(ii) Circuit calculation: If the diagram depicts a balanced bridge (where ratio of arms is equal), the central branch carries no current. Hence, current through the 3 $\Omega$ resistor (assuming it occupies the central galvanometer position) is 0 A. (b) Solution: (i) Conductivity Derivation: Current $I = neA v_d$. Substitute $v_d = \frac{eE\tau}{m}$: $I = neA \left(\frac{eE\tau}{m}\right) = \frac{ne^2 A\tau}{m} E$. We know Current Density $J = \frac{I}{A} = \frac{ne^2\tau}{m} E$. From microscopic Ohm's Law, $J = \sigma E$. Comparing the two gives $\sigma = \frac{ne^2}{m} \tau$.
(ii) Temp Coefficient ($\alpha$): Using $R_2 = R_1[1 + \alpha(T_2 - T_1)]$. $1.38 = 1.05[1 + \alpha(100 - 20)] \implies 1.38 = 1.05 + 1.05 \times 80\alpha \implies 84\alpha = 1.38 - 1.05 = 0.33$. $\alpha = \frac{0.33}{84} \approx 0.00393 \text{ } ^\circ\text{C}^{-1}$.

(a) Solution: (i) Torque on loop: Let loop sides be $a$ and $b$. Area $A = ab$. Forces on sides of length $a$ (parallel to rotation axis) are equal and opposite, creating a couple. $F = I a B \sin(90^\circ)$. Perpendicular distance between them is $b \sin\theta$. Torque $\tau = F \times d_{\perp} = (IaB) \times (b\sin\theta) = I(ab)B\sin\theta = IAB\sin\theta$. Since magnetic moment $m = IA$, $\tau = mB\sin\theta$. In vector form, $\vec{\tau} = \vec{m} \times \vec{B}$.
(ii) Numerical: $N=100$, $I=5.0 \text{ A}$, $r = \frac{10}{\sqrt{\pi}} \text{ cm} = \frac{0.1}{\sqrt{\pi}} \text{ m}$. Area $A = \pi r^2 = \pi \left(\frac{0.01}{\pi}\right) = 0.01 \text{ m}^2$. (I) Dipole moment $m = NIA = 100 \times 5.0 \times 0.01 = 5.0 \text{ Am}^2$. (II) Torque $\tau = mB\sin\theta = 5.0 \times 2.0 \times \sin(30^\circ) = 10 \times 0.5 = 5.0 \text{ N m}$. (b) Solution: (i) Force on conductor: Let $n$ be number density of free electrons. Total electrons $N = nAL$. Lorentz force on one electron $f = -e(\vec{v}_d \times \vec{B})$. Total force $\vec{F} = N \vec{f} = nAL[-e(\vec{v}_d \times \vec{B})] = -(neAv_d)L \times \vec{B}$. Since current $I = neAv_d$, mapping direction to length vector $\vec{L}$, we get $\vec{F} = I(\vec{L} \times \vec{B})$.
(ii) Bent Wire Force: The magnetic force on a current-carrying wire in a uniform magnetic field depends only on the straight-line displacement vector between its endpoints, not the actual path. (Assuming endpoints are provided in the diagram, one applies $F_{net} = I |\vec{L}_{effective} \times \vec{B}|$). Without the geometric coordinates, generic logic applies: resolve vectors or use straight-line distance.

(a) Solution: (i) (I) Diffraction formation: According to Huygens' principle, every point on the exposed slit wavefront acts as a source of secondary wavelets. These wavelets superpose on the screen. Minima occur when wavelets from the top half of the slit exactly destructively interfere with wavelets from the bottom half ($a \sin\theta = n\lambda$). Maxima occur between these minima where interference is partially constructive. (II) Weaker maxima: For the 1st secondary maximum, the slit is divided into 3 equal parts; two parts destructively interfere, leaving only 1/3rd of the slit contributing to intensity. For the 2nd maximum, 1/5th contributes, and so on. Thus, intensity rapidly diminishes.
(ii) Differences: (1) Interference results from superposition of waves from two distinct coherent sources; diffraction results from superposition of wavelets originating from different parts of the same wavefront. (2) In interference, all bright fringes have uniform intensity; in diffraction, central maximum has highest intensity which rapidly falls for successive maxima. (b) Solution: (i) Compound Microscope: Consists of two convex lenses. The objective (small aperture, short focal length) forms a real, inverted, magnified image of an object placed just beyond its focus. This image acts as the object for the eyepiece (larger aperture/focal length), which is adjusted so the image falls within its focal length. The eyepiece acts as a simple magnifier, producing a final, highly magnified, virtual image.
(ii) (I) Image without screen: Yes, the real image is still physically formed at that exact location in space (light rays genuinely converge there). It can be directly seen by an eye placed further along the optical axis catching the diverging rays. (II) Real images from Plane/Convex mirrors: Yes. If a highly converging beam of light (virtual object) is incident on a plane or convex mirror, the mirror can bring those converging rays to a focus in front of it, thereby forming a real image.