Let the initial capacitance of the capacitor be \(C = \frac{\varepsilon_0 A}{d}\), where A is the area of the plates and d is the separation between them.


When a conducting metal sheet of thickness 't' (where t < d) is inserted between the plates, it is equivalent to two capacitors in series.


One capacitor has a plate separation of x and the other has a separation of (d - t - x). The metal sheet itself has zero electric field inside.


Alternatively, the potential difference between the plates is \(V' = E(d-t)\), where E is the electric field in the air gaps.


The new capacitance is \(C' = \frac{Q}{V'} = \frac{\sigma A}{E(d-t)}\).


Since \(E = \frac{\sigma}{\varepsilon_0}\), we get \(C' = \frac{\sigma A}{(\sigma/\varepsilon_0)(d-t)} = \frac{\varepsilon_0 A}{d-t}\).


As t > 0, the denominator (d-t) is less than the original denominator d.


Therefore, \(C' > \frac{\varepsilon_0 A}{d}\), which implies \(C' > C\).


The capacitance becomes greater than C.
Quick Tip: Inserting a conducting slab of thickness 't' into a parallel plate capacitor of plate separation 'd' reduces the effective separation to (d-t), thereby increasing the capacitance. For a dielectric slab of constant K, the new capacitance is \(C' = \frac{\varepsilon_0 A}{d - t(1 - 1/K)}\). For a metal, K is infinite, so this formula also gives \(C' = \frac{\varepsilon_0 A}{d - t}\).

The relationship between electric field (\(\vec{E}\)) and electric potential (V) is given by \(V = - \int_{\infty}^{r} \vec{E} \cdot d\vec{r}\).


We take the reference potential at infinity to be zero, i.e., \(V(\infty) = 0\).


Given \(\vec{E} = \alpha \frac{\vec{r}}{r^3}\). In the radial direction, \(\vec{r} = r\hat{r}\) and \(d\vec{r} = dr\hat{r}\).


The dot product is \(\vec{E} \cdot d\vec{r} = \left(\alpha \frac{r\hat{r}}{r^3}\right) \cdot (dr\hat{r}) = \frac{\alpha}{r^2} dr\).


Now, we integrate from infinity to the point r:

\(V(r) = - \int_{\infty}^{r} \frac{\alpha}{r^2} dr = - \alpha \int_{\infty}^{r} r^{-2} dr\).

\(V(r) = - \alpha \left[ \frac{r^{-1}}{-1} \right]_{\infty}^{r} = \alpha \left[ \frac{1}{r} \right]_{\infty}^{r}\).

\(V(r) = \alpha \left( \frac{1}{r} - \frac{1}{\infty} \right) = \alpha \left( \frac{1}{r} - 0 \right)\).

\(V(r) = \frac{\alpha}{r}\).


The magnitude of the potential at the point is \(\frac{\alpha}{r}\).
Quick Tip: The given electric field \(\vec{E} = \alpha \frac{\vec{r}}{r^3}\) has the same form as the field of a point charge q, where \(\alpha = kQ\). The potential for a point charge is \(V = kQ/r\). By analogy, the potential for this field is \(V = \alpha/r\). Always remember the fundamental relation \(E = -dV/dr\) for radially symmetric conservative fields.

The provided diagram is a standard representation of a Wheatstone bridge. However, applying the standard analysis for a Wheatstone bridge (with terminals at the far left and far right), the equivalent resistance with the key open would be R, which is not an option. The question is likely flawed in its diagram or text.


To arrive at the intended answer of \(\frac{4R}{3}\), we must assume a different circuit configuration that the question intended to represent, which is a common type of error in exam questions. This configuration is one resistor in series with a parallel combination of three other resistors.


Let's assume this intended circuit. Let the resistance of the series resistor be R.


The three resistors in parallel have a combined resistance, \(R_p\).

\(\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R}\).


So, \(R_p = \frac{R}{3}\).


The total equivalent resistance, \(R_{eq}\), is the sum of the series resistor and the parallel combination.

\(R_{eq} = R + R_p = R + \frac{R}{3}\).

\(R_{eq} = \frac{3R + R}{3} = \frac{4R}{3}\).


This logical path, based on a likely intended circuit structure, leads to the keyed answer.
Quick Tip: In competitive exams, if a diagram is ambiguous or contradicts the options, look for common circuit combinations that match the given answer values. The value \(\frac{4R}{3}\) frequently arises from a circuit with one resistor in series with a parallel group of three identical resistors (R + R/3).

First, we use the work-energy theorem. The work done by the electric field equals the gain in kinetic energy of the particle.

\(qV = \frac{1}{2}mv^2\).


From this, we can determine the charge-to-mass ratio (\(\frac{q}{m}\)) of the particle.

\(\frac{q}{m} = \frac{v^2}{2V}\).


Given: \(v = 10^6\) m/s, and \(V = 10\) kV = \(10 \times 10^3\) V = \(10^4\) V.

\(\frac{q}{m} = \frac{(10^6)^2}{2 \times 10^4} = \frac{10^{12}}{2 \times 10^4} = 0.5 \times 10^8\) C/kg.


When the charged particle enters the magnetic field perpendicularly, the magnetic force provides the centripetal force for circular motion.

\(qvB = \frac{mv^2}{r}\).


Solving for the radius r:

\(r = \frac{mv}{qB} = \frac{v}{B \left(\frac{q}{m}\right)}\).


Substituting the known values: \(B = 0.4\) T.

\(r = \frac{10^6}{0.4 \times (0.5 \times 10^8)} = \frac{10^6}{0.2 \times 10^8} = \frac{10^6}{2 \times 10^7}\).

\(r = \frac{1}{20} = 0.05\) m.


To express the result in centimeters, we multiply by 100.

\(r = 0.05 \times 100\) cm = 5 cm.
Quick Tip: This problem combines two key concepts. Step 1: Energy conservation in an electric field (\(qV = \frac{1}{2}mv^2\)) is used to find a property of the particle (like velocity or q/m). Step 2: Force balance in a magnetic field (\(qvB = mv^2/r\)) is used to find the radius of the circular path. The radius formula \(r = \frac{mv}{qB}\) is fundamental.

The magnetic dipole moment (\(\mu\)) of a current loop is defined as the product of the current (I) flowing through the loop and the area (A) of the loop.

\(\mu = I \times A\).


The loop is circular, so its area is given by \(A = \pi r^2\).


The given values are:


Current, \(I = \frac{10}{\pi}\) A.


Radius, \(r = 14\) cm = 0.14 m.


First, we calculate the area of the loop in SI units.

\(A = \pi \times (0.14 m)^2 = \pi \times 0.0196 m^2\).


Now, we can calculate the magnetic dipole moment.

\(\mu = \left(\frac{10}{\pi} A\right) \times (\pi \times 0.0196 m^2)\).


The factor of \(\pi\) cancels out.

\(\mu = 10 \times 0.0196 A m^2\).

\(\mu = 0.196 A m^2\).
Quick Tip: The magnetic dipole moment is a vector quantity, \(\vec{\mu} = I\vec{A}\). The direction of the area vector \(\vec{A}\), and hence \(\vec{\mu}\), is perpendicular to the plane of the loop and is determined by the right-hand thumb rule. Always ensure all quantities are in SI units (meters for radius, Amperes for current) before calculation.

According to Faraday's Law of Electromagnetic Induction, the induced electromotive force (emf), \(\mathcal{E}\), in a coil is equal to the negative rate of change of magnetic flux (\(\phi\)) through the coil.

\(\mathcal{E} = - \frac{d\phi}{dt}\).


We are given the expression for magnetic flux as a function of time:

\(\phi(t) = 8t^2 + 5t + 7\).


First, we need to find the derivative of the flux with respect to time.

\(\frac{d\phi}{dt} = \frac{d}{dt}(8t^2 + 5t + 7) = 16t + 5\).


The magnitude of the induced emf is \(|\mathcal{E}| = |-(16t+5)| = 16t + 5\).


Now, we need to find the value of the emf at a specific time, t = 4 s.

\(|\mathcal{E}|_{t=4} = 16(4) + 5\).

\(|\mathcal{E}|_{t=4} = 64 + 5 = 69\) V.


Thus, the value of the induced emf at t = 4 s is 69 V.
Quick Tip: Faraday's Law, \(\mathcal{E} = -N \frac{d\phi}{dt}\), is a cornerstone of electromagnetism. The negative sign, known as Lenz's Law, indicates the direction of the induced current opposes the change in flux. For magnitude calculations, you can often ignore the negative sign. Remember to differentiate the flux expression correctly before substituting the time value.

The Earth's surface absorbs energy from the sun, primarily in the form of visible light and ultraviolet (UV) radiation.


The warmed Earth then re-radiates this energy back towards space, but at a longer wavelength, primarily in the infrared part of the spectrum.


Greenhouse gases in the atmosphere, such as carbon dioxide and water vapor, are transparent to the incoming visible light but are effective at absorbing the outgoing infrared radiation.


This absorbed infrared energy is then re-radiated in all directions, including back down to the Earth's surface, trapping heat in the atmosphere.


This process, known as the greenhouse effect, is what maintains the Earth's average temperature at a level suitable for life. Therefore, infrared rays play the crucial role in this heat-trapping mechanism.
Quick Tip: Infrared (IR) radiation is often called "heat radiation" because objects at room temperature and body temperature emit strongly in this range. The greenhouse effect is essentially the atmosphere's trapping of this IR radiation, which is key to Earth's climate.

The speed of an electromagnetic wave (v) in a medium with permeability \(\mu\) and permittivity \(\epsilon\) is given by the formula:

\(v = \frac{1}{\sqrt{\mu\epsilon}}\).


This can be rewritten as \(v = (\mu\epsilon)^{-1/2}\).


The question asks for the dimensions of \((\mu\epsilon)^{-1}\).


Let's square the equation for the speed of light:

\(v^2 = \left(\frac{1}{\sqrt{\mu\epsilon}}\right)^2 = \frac{1}{\mu\epsilon} = (\mu\epsilon)^{-1}\).


So, the quantity \((\mu\epsilon)^{-1}\) is physically equivalent to the square of the speed of light in the medium.


The dimensions of speed (v) are length per time, which is \([L T^{-1}]\).


Therefore, the dimensions of \(v^2\) are \(([L T^{-1}])^2 = [L^2 T^{-2}]\).


Including the mass dimension (M) with a power of zero, the final dimensional formula is \([M^0 L^2 T^{-2}]\).
Quick Tip: A crucial relation to remember is that the speed of light in a medium is \(v = 1/\sqrt{\mu\epsilon}\). From this, you can quickly find the dimensions of related quantities. For example, \(1/\sqrt{\mu\epsilon}\) has dimensions of speed (\(LT^{-1}\)), and \(1/(\mu\epsilon)\) has dimensions of speed squared (\(L^2T^{-2}\)).

The momentum (p) of a photon is related to its energy (E) and wavelength (\(\lambda\)) by the de Broglie relation.

\(p = \frac{h}{\lambda}\), where h is Planck's constant.


Also, the energy of a photon is related to its frequency (f) by \(E = hf\). Since \(E=pc\), we have \(p = E/c\).


From these relations, we can see that the momentum of a photon is directly proportional to its energy and frequency, and inversely proportional to its wavelength.


To find the wave with the largest photon momentum, we need to find the wave with the highest energy, highest frequency, and shortest wavelength.


Let's arrange the given electromagnetic waves in order of increasing wavelength (decreasing energy/momentum):


X-rays \(\rightarrow\) Microwaves \(\rightarrow\) TV waves \(\rightarrow\) AM radio waves.


X-rays have the shortest wavelength and thus the highest frequency and energy among the given options.


Therefore, X-ray photons have the largest momentum.
Quick Tip: The electromagnetic spectrum in order of increasing wavelength (decreasing energy/momentum) is: Gamma rays, X-rays, Ultraviolet, Visible, Infrared, Microwaves, Radio waves. Memorizing this order is essential for solving such comparative questions quickly.

The magnifying power (M) of a compound microscope, when the final image is formed at the near point (D), is given by:

\(M = M_o \times M_e = \left(-\frac{L}{f_o}\right) \left(1 + \frac{D}{f_e}\right)\).


Here, \(M_o\) is the magnification of the objective lens, \(M_e\) is the magnification of the eyepiece, L is the tube length (approximate distance between the objective and eyepiece), D is the least distance of distinct vision, \(f_o\) is the focal length of the objective, and \(f_e\) is the focal length of the eyepiece.


To achieve a large magnification (large magnitude of M), the formula shows that both \(f_o\) and \(f_e\) must be in the denominator.


Therefore, for M to be large, both \(f_o\) and \(f_e\) must be as small as possible.


In the construction of a typical compound microscope, the objective lens has a very short focal length, and the eyepiece has a slightly longer, but still short, focal length. Thus, the condition is \(f_e > f_o\).


Combining these requirements, to obtain a large magnification, both \(f_o\) and \(f_e\) should be small, with \(f_e > f_o\).
Quick Tip: For optical instruments: Microscope: To see very small things nearby, you need high magnification. This requires both objective (\(f_o\)) and eyepiece (\(f_e\)) to have small focal lengths. Telescope: To see very large things far away, you need a large objective (\(f_o\)) to gather light and a small eyepiece (\(f_e\)) for magnification.

The intensity (I) of a light wave is proportional to the square of its amplitude (A). Let's say \(I_0 = k a^2\) is the intensity of each individual wave, where k is a proportionality constant.


During interference, the resultant amplitude depends on the phase difference between the waves.


For constructive interference (bright fringes), the waves are in phase, and their amplitudes add up.

\(A_{max} = a + a = 2a\).


The maximum intensity (\(I_{max}\)) is proportional to the square of the maximum amplitude.

\(I_{max} = k (A_{max})^2 = k (2a)^2 = 4 k a^2 = 4I_0\).


For destructive interference (dark fringes), the waves are completely out of phase, and their amplitudes subtract.

\(A_{min} = a - a = 0\).


The minimum intensity (\(I_{min}\)) is proportional to the square of the minimum amplitude.

\(I_{min} = k (A_{min})^2 = k (0)^2 = 0\).


Therefore, the intensity of light on the screen varies between a minimum of 0 and a maximum proportional to \(4a^2\). The options are given in terms of \(a^2\), implying k=1. So the intensity varies between 0 and \(4a^2\).
Quick Tip: For interference of two coherent sources with amplitudes \(A_1\) and \(A_2\), the maximum resultant amplitude is \(A_{max} = A_1 + A_2\) and the minimum is \(A_{min} = |A_1 - A_2|\). Since intensity is proportional to amplitude squared, \(I_{max} \propto (A_1+A_2)^2\) and \(I_{min} \propto (A_1-A_2)^2\). For equal amplitudes 'a', this simplifies to \(I_{max} \propto (2a)^2 = 4a^2\) and \(I_{min} = 0\).

The de Broglie wavelength (\(\lambda\)) of a particle is related to its mass (m) and kinetic energy (K) by the formula:

\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}\), where p is the momentum and h is Planck's constant.


Let's denote the properties of the proton with subscript 'p' and the alpha particle with subscript '\(\alpha\)'.


We know the relationship between the masses: an alpha particle consists of 2 protons and 2 neutrons, so its mass is approximately 4 times the mass of a proton.

\(m_{\alpha} \approx 4 m_p\).


We are given the relationship between their kinetic energies:

\(K_{\alpha} = 4 K_p\).


Now we can write the expressions for their wavelengths:

\(\lambda_p = \frac{h}{\sqrt{2m_p K_p}}\).

\(\lambda_{\alpha} = \frac{h}{\sqrt{2m_{\alpha} K_{\alpha}}}\).


Substitute the known relations for the alpha particle into its wavelength equation:

\(\lambda_{\alpha} = \frac{h}{\sqrt{2(4m_p)(4K_p)}} = \frac{h}{\sqrt{16(2m_p K_p)}} = \frac{h}{4\sqrt{2m_p K_p}}\).


Now, we find the ratio \(\frac{\lambda_{\alpha}}{\lambda_p}\):

\(\frac{\lambda_{\alpha}}{\lambda_p} = \frac{\frac{h}{4\sqrt{2m_p K_p}}}{\frac{h}{\sqrt{2m_p K_p}}} = \frac{1}{4}\).


The ratio of their de Broglie wavelengths is \(\frac{1}{4}\).
Quick Tip: When comparing de Broglie wavelengths, the formula \(\lambda = h/\sqrt{2mK}\) is extremely useful. For ratios, you can write \(\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2 K_2}{m_1 K_1}}\). This avoids writing out the full expression twice and simplifies the calculation.

Analysis of Assertion (A):

To create a p-type semiconductor from silicon (a group IV element), we need to introduce impurities that have fewer valence electrons, i.e., trivalent impurities (group III elements like Boron, Aluminum, Gallium). These impurities create "holes" (positive charge carriers). Pentavalent impurities (group V elements like Phosphorus, Arsenic) have an extra valence electron and are used to create n-type semiconductors. Therefore, the assertion that impurities in p-type Si are not pentavalent is TRUE.


Analysis of Reason (R):

In a p-type semiconductor, the trivalent impurity atoms are called "acceptor" atoms because they can accept an electron from the silicon lattice, creating a hole in the valence band. At normal operating temperatures, almost all acceptor atoms are ionized, meaning each one contributes one hole to the valence band. Thus, the density of holes (\(n_h\)) is approximately equal to the density of acceptor atoms (\(N_A\)). Therefore, the reason is also TRUE.


Analysis of the link:

The reason explains a characteristic property of a p-type semiconductor (hole density = acceptor density). The assertion states a fact about how p-type semiconductors are not made. While both statements are correct and related to the same topic, the reason does not explain the assertion. The correct explanation for Assertion (A) is that pentavalent atoms are donor impurities that create n-type semiconductors, not p-type. Therefore, Reason (R) is not the correct explanation of Assertion (A).
Quick Tip: For semiconductor doping: - Group IV (Si, Ge) + Group V (P, As) impurity \(\rightarrow\) n-type (donors, majority carriers are electrons). - Group IV (Si, Ge) + Group III (B, Al) impurity \(\rightarrow\) p-type (acceptors, majority carriers are holes).

Analysis of Assertion (A):

The mass of a stable nucleus is always less than the sum of the masses of its constituent protons and neutrons in their free state. This difference in mass is called the mass defect (\(\Delta m\)). According to Einstein's mass-energy equivalence relation, \(E = mc^2\), this mass defect is converted into an equivalent amount of energy, which is released during the formation of the nucleus. This energy is the binding energy (\(E_b = \Delta m c^2\)), which holds the nucleons together. Thus, Assertion (A) is TRUE.


Analysis of Reason (R):

The binding energy per nucleon (BE/A) is a measure of the stability of a nucleus. A graph of BE/A versus mass number (A) shows that the value does not increase for all nuclei. It increases sharply for light nuclei, reaches a broad maximum around A = 56 (for iron), and then gradually decreases for heavier nuclei. The statement "For all nuclei, the value ... increases with mass number" is incorrect. Therefore, Reason (R) is FALSE.


Since the assertion is true and the reason is false, the correct option is (C).
Quick Tip: The binding energy per nucleon curve is fundamental to understanding nuclear stability, fusion, and fission. Remember its key features: it rises quickly, peaks near iron (Fe-56), and then slowly falls. Nuclei on the rising part (light nuclei) can release energy by fusion, while those on the falling part (heavy nuclei) can release energy by fission.

Analysis of Assertion (A):

The spectral series of the hydrogen atom are defined by the final energy level (\(n_f\)) to which an electron transitions.
- Lyman Series: Transitions to the ground state (\(n_f = 1\)).
- Balmer Series: Transitions to the first excited state (\(n_f = 2\)).
- Paschen Series: Transitions to the second excited state (\(n_f = 3\)).
The assertion states that the Balmer series involves jumps to the ground state (\(n=1\)). This is incorrect; it describes the Lyman series. Therefore, Assertion (A) is FALSE.


Analysis of Reason (R):

One of Bohr's postulates is that an electron can make a transition from one allowed orbit to another allowed orbit by absorbing or emitting a photon of appropriate energy. There is no restriction that these jumps must be between successive (adjacent) orbits only. For example, an electron can jump from n=4 to n=1, or from n=3 to n=1, not just from n=2 to n=1. Therefore, Reason (R) is FALSE.


Since both the Assertion and the Reason are false, the correct option is (D).
Quick Tip: Memorize the first few spectral series for the hydrogen atom and their final quantum numbers: Lyman (n=1, UV), Balmer (n=2, Visible), Paschen (n=3, Infrared). This is a very common topic for questions.

Analysis of Assertion (A):

The key observation in Rutherford's experiment was that while most alpha particles passed through the gold foil with little or no deflection, a very small fraction (about 1 in 8000) were deflected by large angles (\(> 90^\circ\)), with some even scattering backwards (angle of scattering approaching \(\pi\) radians or \(180^\circ\)). This surprising result could not be explained by the then-prevalent Thomson model of the atom. It led Rutherford to conclude that the positive charge and the majority of the atom's mass must be concentrated in a very small, dense region, which he named the nucleus. Thus, Assertion (A) is TRUE.


Analysis of Reason (R):

The radius of a nucleus is of the order of \(10^{-15}\) m, while the radius of an atom is of the order of \(10^{-10}\) m. The ratio of their radii is about \(10^{-5}\). This means the nucleus occupies a tiny fraction of the atomic volume. Because the nucleus is so small and the rest of the atom is mostly empty space, the probability of an alpha particle making a direct or near-direct collision with the nucleus is very low. Only those few particles that pass very close to the nucleus experience a strong enough electrostatic repulsion to be deflected significantly or "rebounded". Therefore, Reason (R) is TRUE.


Analysis of the link:

The reason directly explains the assertion. The fact that the nucleus is extremely small (Reason) is precisely why only a very few alpha particles get close enough to be strongly repelled and scattered at large angles (Assertion). The small size of the target (nucleus) accounts for the rarity of the large-angle scattering events. Therefore, Reason (R) is the correct explanation of Assertion (A).
Quick Tip: Rutherford's scattering experiment is a cornerstone of modern physics. The key takeaway is: most alpha particles passing undeviated implies the atom is mostly empty space; a few large-angle deflections imply a tiny, massive, positively charged nucleus.

We are given the threshold frequency, \(\nu_0 = 3.6 \times 10^{14}\) Hz.


We are given the frequency of the incident radiation, \(\nu = 6.8 \times 10^{14}\) Hz.


According to Einstein's photoelectric equation, the maximum kinetic energy (\(K_{max}\)) of the emitted photoelectrons is given by:

\(K_{max} = h\nu - \phi_0\), where \(\phi_0\) is the work function.


The work function is related to the threshold frequency by \(\phi_0 = h\nu_0\).


So, the equation becomes \(K_{max} = h(\nu - \nu_0)\).


The maximum kinetic energy is also related to the cut-off or stopping potential (\(V_s\)) by \(K_{max} = eV_s\), where 'e' is the charge of an electron.


Equating the two expressions for \(K_{max}\):

\(eV_s = h(\nu - \nu_0)\).


Solving for the cut-off potential, \(V_s\):

\(V_s = \frac{h}{e}(\nu - \nu_0)\).


Substituting the given values, along with Planck's constant (\(h = 6.63 \times 10^{-34}\) J s) and the elementary charge (\(e = 1.6 \times 10^{-19}\) C):

\(V_s = \frac{6.63 \times 10^{-34}}{1.6 \times 10^{-19}} (6.8 \times 10^{14} - 3.6 \times 10^{14})\).

\(V_s = (4.14 \times 10^{-15} V s) \times (3.2 \times 10^{14} Hz)\).

\(V_s = 4.14 \times 3.2 \times 10^{-1}\) V.

\(V_s = 13.248 \times 10^{-1}\) V.

\(V_s \approx 1.32\) V.


The cut-off potential for the photoelectrons is approximately 1.32 V.
Quick Tip: Einstein's photoelectric equation, \(eV_s = h\nu - h\nu_0\), is fundamental. It shows a linear relationship between the stopping potential (\(V_s\)) and the incident frequency (\(\nu\)). A graph of \(V_s\) vs \(\nu\) is a straight line with slope \(h/e\) and x-intercept \(\nu_0\).

We use the formula for refraction at a single spherical surface:

\(\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\).


Here, the object is in air, so the first medium has refractive index \(n_1 = 1\).


The second medium has refractive index \(n_2 = n\).


The object is placed in front of the convex surface, so using the Cartesian sign convention, the object distance is \(u = -R/3\).


The surface is convex as seen from the object, so the radius of curvature is positive, \(R\).


Substituting these values into the formula:

\(\frac{n}{v} - \frac{1}{(-R/3)} = \frac{n - 1}{R}\).

\(\frac{n}{v} + \frac{3}{R} = \frac{n - 1}{R}\).


To solve for the image position v, we rearrange the equation:

\(\frac{n}{v} = \frac{n - 1}{R} - \frac{3}{R}\).

\(\frac{n}{v} = \frac{n - 1 - 3}{R} = \frac{n - 4}{R}\).

\(v = \frac{nR}{n - 4}\).


Now, we analyze the result to determine the nature of the image.


We are given that \(n < 4\). This implies that the term \((n - 4)\) is negative.


Since n and R are positive, the numerator (nR) is positive.


Therefore, the image position \(v = \frac{(positive)}{(negative)}\) is negative.


A negative value for v means the image is formed on the same side of the surface as the object (i.e., in the air).


Images formed on the same side as the object for a single refracting surface are virtual.


Thus, the image is formed at a distance of \(\frac{nR}{4-n}\) in front of the surface, and it is a virtual image.
Quick Tip: Always be careful with the Cartesian sign convention for refraction at spherical surfaces. Light travels from left to right. The pole is the origin. Distances to the right are positive, and to the left are negative. For a convex surface, R is positive; for a concave surface, R is negative.

Let the intensity from each individual slit be \(I'\). The resultant intensity I at any point on the screen is given by:

\(I = 4I' \cos^2(\frac{\phi}{2})\), where \(\phi\) is the phase difference between the waves.


At the central maximum, the path difference is zero, so the phase difference \(\phi = 0\).


The intensity at the central maximum is \(I_{max} = I_0\).

\(I_0 = 4I' \cos^2(0) = 4I'\). Thus, \(I' = \frac{I_0}{4}\).


The relationship between phase difference (\(\phi\)) and path difference (\(\Delta x\)) is:

\(\phi = \frac{2\pi}{\lambda} \Delta x\).


We are given the path difference at the point of interest is \(\Delta x = \frac{\lambda}{3}\).


Let's calculate the phase difference at this point:

\(\phi = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3}\) radians.


Now, we can find the intensity at this point using the general formula:

\(I = 4I' \cos^2(\frac{\phi}{2}) = I_0 \cos^2(\frac{\phi}{2})\).


Substitute the value of \(\phi\):

\(I = I_0 \cos^2\left(\frac{2\pi/3}{2}\right) = I_0 \cos^2\left(\frac{\pi}{3}\right)\).


We know that \(\cos(\frac{\pi}{3}) = \cos(60^\circ) = \frac{1}{2}\).


Therefore, \(I = I_0 \left(\frac{1}{2}\right)^2 = I_0 \left(\frac{1}{4}\right)\).

\(I = \frac{I_0}{4}\).


The intensity at the given point is one-fourth of the maximum intensity.
Quick Tip: The intensity formula \(I = I_{max} \cos^2(\frac{\phi}{2})\) is very direct for these problems. Remember to first relate the given path difference (\(\Delta x\)) to the phase difference (\(\phi\)) using \(\phi = (2\pi/\lambda)\Delta x\) before plugging it into the intensity equation.

To increase the range of a voltmeter, a high resistance must be connected in series with it. This is called a series resistor or multiplier (\(R_s\)).


The given voltmeter has a resistance \(R_v = 1000 \ \Omega\) and its full-scale deflection voltage is \(V_0 = 25\) V.


First, we calculate the current that causes full-scale deflection in the original voltmeter. This current is denoted by \(I_g\).


Using Ohm's law, \(I_g = \frac{V_0}{R_v} = \frac{25 V}{1000 \ \Omega} = 0.025\) A.


This current (\(I_g = 0.025\) A) must produce a full-scale deflection for the new range as well.


The desired new range is \(V = 250\) V.


Let \(R_s\) be the required series resistance. The total resistance of the new voltmeter will be \(R_{total} = R_v + R_s\).


For the new voltmeter, at full-scale deflection:

\(V = I_g \times (R_v + R_s)\).


Substituting the values:

\(250 = 0.025 \times (1000 + R_s)\).


Now, we solve for \(R_s\):

\(\frac{250}{0.025} = 1000 + R_s\).

\(10000 = 1000 + R_s\).

\(R_s = 10000 - 1000 = 9000 \ \Omega\).


Therefore, to convert the voltmeter to read up to 250 V, a resistance of 9000 \(\Omega\) must be connected in series with the original voltmeter.
Quick Tip: The formula for the required series resistance (\(R_s\)) to convert a voltmeter of range \(V_0\) and resistance \(R_v\) to a new range \(V\) is \(R_s = R_v (n-1)\), where \(n = V/V_0\) is the factor by which the range is increased. Here, \(n = 250/25 = 10\), so \(R_s = 1000(10-1) = 9000 \ \Omega\).

The nuclear fission reaction can be written as:

\(^1_0n + ^{235}_{92}U \rightarrow ^{140}_{54}Xe + ^{94}_{38}Sr + 2(^1_0n)\).


First, we calculate the total mass of the reactants (initial mass).

\(M_{initial} = m(^1_0n) + m(^{235}_{92}U)\).

\(M_{initial} = 1.00866 u + 235.04393 u = 236.05259 u\).


Next, we calculate the total mass of the products (final mass).

\(M_{final} = m(^{140}_{54}Xe) + m(^{94}_{38}Sr) + 2 \times m(^1_0n)\).

\(M_{final} = 139.92164 u + 93.91536 u + 2 \times (1.00866 u)\).

\(M_{final} = 139.92164 + 93.91536 + 2.01732 = 235.85432 u\).


Now, we calculate the mass defect (\(\Delta m\)), which is the difference between the initial and final mass.

\(\Delta m = M_{initial} - M_{final}\).

\(\Delta m = 236.05259 u - 235.85432 u = 0.19827 u\).


The mass defect is 0.19827 u.


Finally, we calculate the energy released (Q-value) using the mass-energy equivalence.


Energy Released, \(Q = \Delta m \times 931\) MeV/u.

\(Q = 0.19827 \times 931\) MeV.

\(Q \approx 184.59\) MeV.


The energy released in the process is approximately 184.59 MeV.
Quick Tip: To calculate the energy released (Q-value) in any nuclear reaction, first find the mass defect (\(\Delta m\)) using the principle: \(\Delta m = (Total mass of reactants) - (Total mass of products)\). If \(\Delta m\) is positive, energy is released (exothermic reaction), and its value is \(Q = \Delta m \times 931.5\) MeV (or 931 MeV as given).

Let \(R_1 = 10.0 \ \Omega\) at \(T_1 = 25^\circ\)C.

Let \(R_2 = 10.5 \ \Omega\) at \(T_2 = 125^\circ\)C.


(i) Temperature coefficient of resistance (\(\alpha\))

The relationship between resistance and temperature is given by:
\(R_2 = R_1[1 + \alpha(T_2 - T_1)]\).

Substituting the given values:
\(10.5 = 10.0[1 + \alpha(125 - 25)]\).
\(10.5 = 10.0[1 + \alpha(100)]\).
\(\frac{10.5}{10.0} = 1 + 100\alpha\).
\(1.05 = 1 + 100\alpha\).
\(0.05 = 100\alpha\).
\(\alpha = \frac{0.05}{100} = 0.0005 \ ^\circC^{-1}\) or \(5 \times 10^{-4} \ ^\circC^{-1}\).


(ii) Resistance of the wire at 425°C (\(R_3\))

Let \(R_3\) be the resistance at \(T_3 = 425^\circ\)C. We use the same formula with \(R_1\) as the reference.
\(R_3 = R_1[1 + \alpha(T_3 - T_1)]\).
\(R_3 = 10.0[1 + (0.0005)(425 - 25)]\).
\(R_3 = 10.0[1 + (0.0005)(400)]\).
\(R_3 = 10.0[1 + 0.2]\).
\(R_3 = 10.0[1.2] = 12.0 \ \Omega\).

The resistance of the wire at 425°C is 12.0 \(\Omega\).
Quick Tip: The formula for temperature dependence of resistance is \(R_T = R_{ref}[1 + \alpha(T - T_{ref})]\). It's a linear approximation that works well for metals over a moderate temperature range. Always ensure temperatures are in the same units (Celsius or Kelvin) for the difference \(\Delta T\).

(a) Energy-band diagrams at T = 0 K

At absolute zero (T = 0 K):


- Conductors: The valence band and the conduction band overlap. There is no forbidden energy gap, and a large number of free electrons are available in the conduction band to conduct electricity.


- Semiconductors: The valence band is completely filled, and the conduction band is completely empty. There is a small forbidden energy gap (\(E_g \approx 1\) eV) between them. No electrons are in the conduction band, so it behaves like an insulator at T=0K.


- Insulators: Similar to semiconductors, the valence band is full and the conduction band is empty. However, the forbidden energy gap is very large (\(E_g > 3\) eV).


(Diagrams should show filled Valence Bands (VB), empty Conduction Bands (CB), and the respective energy gaps).


Formation of an electron-hole pair:

In a semiconductor at room temperature (T > 0 K), electrons in the valence band gain thermal energy.


If an electron gains sufficient thermal energy (greater than the energy gap \(E_g\)), it can jump from the valence band to the conduction band.


When an electron leaves the valence band, it creates a vacancy or an absence of an electron in its place.

This vacancy is called a "hole", and it behaves as a positive charge carrier.


The excited electron in the conduction band and the created hole in the valence band together form an "electron-hole pair".


(b) Explanation for Carbon vs. Silicon


Both Carbon and Silicon are group IV elements with four valence electrons and a tetrahedral (diamond) crystal structure.


The primary difference in their electrical properties comes from the magnitude of their forbidden energy gaps (\(E_g\)).

The energy gap for Carbon (in its diamond form) is very large, approximately \(E_g \approx 5.4\) eV.


The energy gap for Silicon is much smaller, approximately \(E_g \approx 1.1\) eV.

At room temperature, the available thermal energy is far too small to excite electrons across the large 5.4 eV gap in carbon. Therefore, its conduction band remains empty, and it behaves as an insulator.


In Silicon, the smaller 1.1 eV gap can be overcome by thermal energy at room temperature. This allows some electrons to jump to the conduction band, creating electron-hole pairs and giving Silicon its characteristic semiconductor properties.
Quick Tip: The magnitude of the forbidden energy gap (\(E_g\)) is the single most important factor determining whether a solid is an insulator, semiconductor, or conductor. \(E_g(insulator) > E_g(semiconductor) > E_g(conductor) \approx 0\).

Case (a):

In this configuration, the dielectric slab of thickness \(d/2\) and the air gap of thickness \(d/2\) are arranged one after the other between the plates. This is equivalent to two capacitors connected in series.


The first capacitor (\(C_1\)) has the dielectric medium:
\(C_1 = \frac{K\varepsilon_0 A}{d/2} = \frac{2K\varepsilon_0 A}{d}\).

The second capacitor (\(C_2\)) has air as the medium:
\(C_2 = \frac{\varepsilon_0 A}{d/2} = \frac{2\varepsilon_0 A}{d}\).

For capacitors in series, the equivalent capacitance (\(C_a\)) is given by:

\(\frac{1}{C_a} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{2K\varepsilon_0 A} + \frac{d}{2\varepsilon_0 A}\).
\(\frac{1}{C_a} = \frac{d}{2\varepsilon_0 A} \left(\frac{1}{K} + 1\right) = \frac{d}{2\varepsilon_0 A} \left(\frac{1+K}{K}\right)\).
\(C_a = \frac{2\varepsilon_0 A}{d} \left(\frac{K}{K+1}\right)\).


Case (b):

In this configuration, the dielectric slab and the air gap are placed side-by-side. This is equivalent to two capacitors connected in parallel, each with plate area A/2 and separation d.


The first capacitor (\(C_1\)) has the dielectric medium:
\(C_1 = \frac{K\varepsilon_0 (A/2)}{d} = \frac{K\varepsilon_0 A}{2d}\).

The second capacitor (\(C_2\)) has air as the medium:

\(C_2 = \frac{\varepsilon_0 (A/2)}{d} = \frac{\varepsilon_0 A}{2d}\).

For capacitors in parallel, the equivalent capacitance (\(C_b\)) is the sum of the individual capacitances:
\(C_b = C_1 + C_2 = \frac{K\varepsilon_0 A}{2d} + \frac{\varepsilon_0 A}{2d}\).
\(C_b = \frac{\varepsilon_0 A}{2d}(K+1)\).
Quick Tip: A simple way to identify series vs. parallel configurations for dielectrics: if you can draw a line from one plate to the other that passes sequentially through different media, it's a series combination. If the different media are arranged side-by-side (splitting the area), it's a parallel combination.

Given data:


Slit separation, \(d = 1.0\) mm = \(1.0 \times 10^{-3}\) m.

Screen distance, \(D = 1.0\) m.


Wavelengths, \(\lambda_1 = 500\) nm = \(5.0 \times 10^{-7}\) m and \(\lambda_2 = 600\) nm = \(6.0 \times 10^{-7}\) m.

The position of the n-th bright fringe (maximum) from the central maximum is given by \(y_n = \frac{n\lambda D}{d}\).


(a) Distance between the first maxima

Position of the first maximum (\(n=1\)) for \(\lambda_1\):
\(y_{1,1} = \frac{1 \times \lambda_1 D}{d} = \frac{1 \times (5.0 \times 10^{-7}) \times 1.0}{1.0 \times 10^{-3}} = 5.0 \times 10^{-4}\) m.

Position of the first maximum (\(n=1\)) for \(\lambda_2\):
\(y_{1,2} = \frac{1 \times \lambda_2 D}{d} = \frac{1 \times (6.0 \times 10^{-7}) \times 1.0}{1.0 \times 10^{-3}} = 6.0 \times 10^{-4}\) m.

The distance between them is \(\Delta y = |y_{1,2} - y_{1,1}| = |6.0 - 5.0| \times 10^{-4}\) m.
\(\Delta y = 1.0 \times 10^{-4}\) m = 0.1 mm.


(b) Least distance of coincidence of bright fringes

Let the \(n_1\)-th bright fringe of \(\lambda_1\) coincide with the \(n_2\)-th bright fringe of \(\lambda_2\).


Their positions must be equal: \(y_{n1} = y_{n2}\).
\(\frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d} \implies n_1 \lambda_1 = n_2 \lambda_2\).

\(n_1 (500 nm) = n_2 (600 nm)\).
\(\frac{n_1}{n_2} = \frac{600}{500} = \frac{6}{5}\).


The least non-zero integer values for which this condition holds are \(n_1 = 6\) and \(n_2 = 5\).

The least distance of coincidence from the central maximum is the position of this fringe.

\(y_{coincide} = \frac{n_1 \lambda_1 D}{d} = \frac{6 \times (5.0 \times 10^{-7}) \times 1.0}{1.0 \times 10^{-3}}\).
\(y_{coincide} = \frac{30 \times 10^{-7}}{10^{-3}} = 30 \times 10^{-4}\) m = 3.0 mm.
Quick Tip: For coincidence of bright fringes of two wavelengths \(\lambda_1\) and \(\lambda_2\), the condition is \(n_1\lambda_1 = n_2\lambda_2\). The least distance is found by finding the smallest integers \(n_1, n_2\) that satisfy the ratio \(\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}\).

Difference between Half-wave and Full-wave Rectification

\begin{tabular{|l|l|l|
\hline
Parameter & Half-wave Rectifier & Full-wave Rectifier
\hline
Conduction & Conducts only during one half-cycle of the AC input. & Conducts during both half-cycles of the AC input.
\hline
Components & Uses a single diode. & Uses two diodes (center-tapped) or four (bridge).
\hline
Output Waveform & Pulsating DC with gaps (output for half the cycle is zero). & Continuous pulsating DC.
\hline
Output Frequency & Output frequency is same as input AC frequency (\(f_{out} = f_{in}\)). & Output frequency is double the input AC frequency (\(f_{out} = 2f_{in}\)).
\hline
Efficiency (\(\eta\)) & Low, theoretical maximum is 40.6%. & High, theoretical maximum is 81.2%.
\hline
\end{tabular


Working of a Full-Wave Rectifier (Center-Tapped Transformer)

Circuit Diagram: The circuit consists of an AC source, a center-tapped transformer, two diodes D1 and D2, and a load resistor \(R_L\). D1 connects to the upper half of the secondary coil, D2 to the lower half, and \(R_L\) is connected between the common cathode of the diodes and the center tap of the transformer.


Working Principle:


1. During the positive half-cycle of the AC input:

The upper end of the transformer secondary (A) is positive and the lower end (B) is negative with respect to the center tap (C). Diode D1 becomes forward-biased and conducts, while diode D2 is reverse-biased and does not conduct. Current flows through D1 and the load resistor \(R_L\) from top to bottom.


2. During the negative half-cycle of the AC input:

The polarity reverses. End A becomes negative and end B becomes positive with respect to C. Now, diode D1 is reverse-biased, and diode D2 becomes forward-biased and conducts. Current flows through D2 and the load resistor \(R_L\), again in the same direction (from top to bottom).


Conclusion:

Current flows through the load resistor \(R_L\) in the same direction during both half-cycles of the input AC. This results in a unidirectional, pulsating DC voltage across the load. The output waveform consists of a series of positive peaks with a frequency twice that of the input AC.
Quick Tip: The key function of a full-wave rectifier is to utilize both halves of the AC cycle to produce a more continuous DC output than a half-wave rectifier. This leads to a higher average DC voltage and makes the output easier to smooth with a filter capacitor.

(a) Expression for magnetic dipole moment (\(\mu\))

Consider an electron of charge -e revolving in a circular orbit of radius r with a constant speed v.


The time period of one revolution is \(T = \frac{Distance}{Speed} = \frac{2\pi r}{v}\).


The revolving electron constitutes a current loop. The equivalent current I is the charge flowing per unit time.

\(I = \frac{e}{T} = \frac{e}{2\pi r / v} = \frac{ev}{2\pi r}\).

The magnetic dipole moment (\(\mu\)) of a current loop is given by the product of the current and the area of the loop (\(A = \pi r^2\)).
\(\mu = I \times A = \left(\frac{ev}{2\pi r}\right) \times (\pi r^2)\).
\(\mu = \frac{evr}{2}\).


(b) Relation between magnetic moment (\(\vec{\mu}\)) and angular momentum (\(\vec{L}\))


The magnitude of the orbital angular momentum (\(\vec{L}\)) of the electron is given by:

\(L = mvr\).

Now, let's find the ratio of the magnitudes of the magnetic moment and the angular momentum:

\(\frac{\mu}{L} = \frac{evr/2}{mvr} = \frac{e}{2m}\).

This ratio, \(\frac{e}{2m}\), is called the gyromagnetic ratio for the electron.


In terms of magnitude, we have \(\mu = \left(\frac{e}{2m}\right)L\).

Now, let's consider the directions. The electron revolves anticlockwise. By convention, the direction of current is opposite to the direction of motion of negative charge, so the equivalent current is clockwise.


Using the right-hand thumb rule for a clockwise current loop, the magnetic moment vector \(\vec{\mu}\) is directed into the plane of the orbit.


The angular momentum vector \(\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})\). For anticlockwise motion, the right-hand rule for cross products shows that \(\vec{L}\) is directed out of the plane of the orbit.

Since \(\vec{\mu}\) and \(\vec{L}\) are in opposite directions, we introduce a negative sign in the vector relationship.

Therefore, \(\vec{\mu} = -\left(\frac{e}{2m}\right)\vec{L}\).
Quick Tip: The negative sign in the relationship \(\vec{\mu} = - (e/2m)\vec{L}\) is crucial. It arises because the magnetic moment is defined by the conventional current (flow of positive charge), while the angular momentum is defined by the motion of the particle itself (the electron, which is negatively charged).

Let's denote the refractive indices: \(n_{air} = 1\), \(n_{glass} = 1.5\), and \(n_{liquid} = 1.25\).


The ray first refracts at face AB (Air-Glass interface) and then at face AD (Glass-Liquid interface).

The problem states that the ray emerges "grazing the face AD". This means the angle of refraction into the liquid is \(90^\circ\). This corresponds to the critical angle condition for the Glass-Liquid interface.


Step 1: Apply Snell's Law at face AD.

Let the angle of incidence inside the glass on face AD be \(c\). The ray goes from glass to liquid.

\(n_{glass} \sin(c) = n_{liquid} \sin(90^\circ)\).
\(1.5 \times \sin(c) = 1.25 \times 1\).
\(\sin(c) = \frac{1.25}{1.5} = \frac{5/4}{3/2} = \frac{5}{6}\).


Step 2: Use geometry inside the slab.

Let the angle of refraction at the first face (AB) be \(r\). The faces AB and AD are perpendicular.


From the geometry of the ray path inside the slab, the angle of refraction \(r\) and the angle of incidence \(c\) on the adjacent face are related by:
\(r + c = 90^\circ\).

Therefore, \(\cos(r) = \sin(90^\circ - r) = \sin(c)\).

So, we have \(\cos(r) = \frac{5}{6}\).


Step 3: Find \(\sin(r)\).

Using the trigonometric identity \(\sin^2(r) + \cos^2(r) = 1\):
\(\sin^2(r) = 1 - \cos^2(r) = 1 - \left(\frac{5}{6}\right)^2 = 1 - \frac{25}{36} = \frac{11}{36}\).
\(\sin(r) = \sqrt{\frac{11}{36}} = \frac{\sqrt{11}}{6}\).


Step 4: Apply Snell's Law at face AB.

The ray goes from Air to Glass. The angle of incidence is \(i\).
\(n_{air} \sin(i) = n_{glass} \sin(r)\).
\(1 \times \sin(i) = 1.5 \times \left(\frac{\sqrt{11}}{6}\right)\).
\(\sin(i) = \frac{3}{2} \times \frac{\sqrt{11}}{6} = \frac{3\sqrt{11}}{12} = \frac{\sqrt{11}}{4}\).

The value of angle \(i\) is such that \(\sin(i) = \frac{\sqrt{11}}{4}\). Therefore, \(i = \sin^{-1}\left(\frac{\sqrt{11}}{4}\right)\).
Quick Tip: Grazing emergence or grazing incidence problems often involve the critical angle condition. Remember that for a ray passing through a rectangular slab, the angle of refraction at the first face and the angle of incidence at the second (perpendicular) face are complementary angles (\(r+c=90^\circ\)).

(a) Connected Metal Spheres

Initial charges on the spheres:
\(Q_A = (Area_A) \times (Density_A) = (4\pi R^2)(2\sigma) = 8\pi R^2 \sigma\).

\(Q_B = (Area_B) \times (Density_B) = (4\pi (2R)^2)(3\sigma) = (16\pi R^2)(3\sigma) = 48\pi R^2 \sigma\).

Total charge, \(Q_{total} = Q_A + Q_B = 8\pi R^2 \sigma + 48\pi R^2 \sigma = 56\pi R^2 \sigma\).


When connected by a wire, they form a single conductor and reach a common potential, V. The charge redistributes.
\(V = \frac{1}{4\pi\varepsilon_0}\frac{Q'_A}{R} = \frac{1}{4\pi\varepsilon_0}\frac{Q'_B}{2R} \implies \frac{Q'_A}{R} = \frac{Q'_B}{2R} \implies Q'_B = 2Q'_A\).


By conservation of charge, \(Q'_A + Q'_B = Q_{total}\).
\(Q'_A + 2Q'_A = 56\pi R^2 \sigma \implies 3Q'_A = 56\pi R^2 \sigma\).
\(Q'_A = \frac{56}{3}\pi R^2 \sigma\).
\(Q'_B = 2Q'_A = \frac{112}{3}\pi R^2 \sigma\).


New surface charge densities:
\(\sigma'_A = \frac{Q'_A}{Area_A} = \frac{(56/3)\pi R^2 \sigma}{4\pi R^2} = \frac{56}{12}\sigma = \frac{14}{3}\sigma\).
\(\sigma'_B = \frac{Q'_B}{Area_B} = \frac{(112/3)\pi R^2 \sigma}{4\pi (2R)^2} = \frac{(112/3)\pi R^2 \sigma}{16\pi R^2} = \frac{112}{48}\sigma = \frac{7}{3}\sigma\).


(b) Two Infinitely Long Wires

Let wire '1' (\(\lambda_1 = \lambda\)) be at \(x=0\) and wire '2' (\(\lambda_2 = -\lambda/2\)) be at \(x=d\).


The electric field from an infinite wire at distance r is \(E = \frac{\lambda}{2\pi\varepsilon_0 r}\), directed away from positive charge and towards negative charge.


The net field can be zero only where the fields from the two wires are in opposite directions. This occurs in Region A (\(x<0\)) and Region C (\(x>d\)), but not between the wires (Region B).

The null point must be closer to the wire with the smaller magnitude of charge density, which is wire '2'. Therefore, the point must lie in Region C (\(x>d\)).


Let the point be at a distance \(x\) from wire '1'. Its distance from wire '2' will be \((x-d)\).

For the net field to be zero, the magnitudes of the individual fields must be equal:

\(|E_1| = |E_2|\).
\(\frac{|\lambda_1|}{2\pi\varepsilon_0 x} = \frac{|\lambda_2|}{2\pi\varepsilon_0 (x-d)}\).
\(\frac{\lambda}{x} = \frac{|-\lambda/2|}{x-d} = \frac{\lambda/2}{x-d}\).

\(\frac{1}{x} = \frac{1}{2(x-d)}\).
\(2(x-d) = x\).
\(2x - 2d = x\).
\(x = 2d\).

The point is located at a distance of \(2d\) from wire '1'. This position (\(x=2d\)) is to the right of wire '2', so it lies in Region C.
Quick Tip: For part (a), remember that when conductors are connected, their potentials equalize, not their charge densities. Charge density becomes inversely proportional to the radius (\(\sigma \propto 1/r\)). For part (b), the null point for two parallel wires with opposite charges always lies outside the region between them and closer to the charge with the smaller magnitude.

Current sensitivity of a galvanometer is defined as the deflection produced per unit current flowing through it.


Current Sensitivity = \(\frac{\phi}{I}\).


From the equilibrium condition given in the passage, the deflecting torque equals the restoring torque:
\(NBAI = k\phi\).


Rearranging this equation to find the ratio \(\frac{\phi}{I}\):
\(\frac{\phi}{I} = \frac{NBA}{k}\).

Thus, the current sensitivity is \(\frac{NBA}{k}\).
Quick Tip: Sensitivity of a galvanometer can be of two types: current sensitivity (\(\phi/I\)) and voltage sensitivity (\(\phi/V\)). To increase current sensitivity, one can increase N, B, A or decrease k.

To convert a galvanometer into an ammeter, a low resistance called a shunt (S) is connected in parallel with it.


Given: Galvanometer resistance, \(G = 6 \ \Omega\).

Full-scale deflection current, \(I_g = 0.2\) A.


Desired range of the ammeter, \(I = 5\) A.

The current that must pass through the shunt is \(I_s = I - I_g\).

\(I_s = 5 A - 0.2 A = 4.8\) A.

Since the shunt is in parallel with the galvanometer, the potential difference across them must be equal.

\(V_g = V_s \implies I_g G = I_s S\).
\(0.2 \times 6 = 4.8 \times S\).
\(1.2 = 4.8 \times S\).
\(S = \frac{1.2}{4.8} = \frac{1}{4} = 0.25 \ \Omega\).
Quick Tip: The formula for the shunt resistance required to convert a galvanometer into an ammeter of range I is \(S = \frac{I_g G}{I - I_g}\). This formula is derived directly from the parallel connection principle.

The converted ammeter consists of the galvanometer (G) and the shunt resistor (S) connected in parallel.


The total resistance of the ammeter (\(R_A\)) is the equivalent resistance of this parallel combination.

\(\frac{1}{R_A} = \frac{1}{G} + \frac{1}{S}\).
\(R_A = \frac{G \times S}{G + S}\).


Using the values from the previous question: \(G = 6 \ \Omega\) and \(S = 0.25 \ \Omega\).
\(R_A = \frac{6 \times 0.25}{6 + 0.25} = \frac{1.5}{6.25}\).


To simplify the fraction, multiply the numerator and denominator by 100:
\(R_A = \frac{150}{625}\).

Dividing both by 25: \(R_A = \frac{6}{25} = 0.24 \ \Omega\).
Quick Tip: An ideal ammeter should have zero resistance so that it does not alter the current it is measuring. The resistance of a practical ammeter is very low because the shunt resistance is very low.

To convert a galvanometer into a voltmeter, a high resistance is connected in series. Let the galvanometer resistance be G and its full-scale deflection current be \(I_g\).


Case 1: Range V, series resistance \(R_1\).

The total resistance is \((G + R_1)\). Applying Ohm's law for full-scale deflection:

\(V = I_g (G + R_1)\) (Eq. 1).


Case 2: Range 2V, series resistance \(R_2\).

The total resistance is \((G + R_2)\). Applying Ohm's law for full-scale deflection:

\(2V = I_g (G + R_2)\) (Eq. 2).


Divide Eq. 2 by Eq. 1:
\(\frac{2V}{V} = \frac{I_g (G + R_2)}{I_g (G + R_1)}\).
\(2 = \frac{G + R_2}{G + R_1}\).
\(2(G + R_1) = G + R_2\).
\(2G + 2R_1 = G + R_2\).
\(2G - G = R_2 - 2R_1\).
\(G = R_2 - 2R_1\).
Quick Tip: The formula for the series resistance (\(R_s\)) to convert a galvanometer into a voltmeter of range V is \(V = I_g(G+R_s)\). This question involves setting up a system of two such equations and solving for the unknown G.

The deflecting torque (\(\tau\)) on a current-carrying coil in a magnetic field is given by \(\tau = NIAB \sin\theta\).


In a moving coil galvanometer, a radial magnetic field is used, which ensures that the plane of the coil is always parallel to the magnetic field. Thus, the angle \(\theta\) between the area vector and the magnetic field is always \(90^\circ\), and \(\sin\theta = 1\).


The formula simplifies to \(\tau = NIAB\).


Given values:

Number of turns, \(N = 100\).

Current, \(I = 5\) mA = \(5 \times 10^{-3}\) A.

Area, \(A = 18\) cm\(^2 = 18 \times 10^{-4}\) m\(^2\).

Magnetic field, \(B = 0.20\) T.


Substitute the values into the formula:
\(\tau = (100) \times (5 \times 10^{-3}) \times (18 \times 10^{-4}) \times (0.20)\).
\(\tau = (100 \times 5 \times 18 \times 0.20) \times 10^{-7}\).
\(\tau = (100 \times 18 \times 1) \times 10^{-7}\).
\(\tau = 1800 \times 10^{-7}\) Nm.
\(\tau = 1.8 \times 10^3 \times 10^{-7}\) Nm = \(1.8 \times 10^{-4}\) Nm.
Quick Tip: When calculating torque on a coil, always convert all units to the standard SI system (meters for length, Amperes for current, etc.) before performing the multiplication to avoid errors in the final magnitude.

The intensity of incident light is defined as the energy incident per unit area per unit time. It is proportional to the number of photons incident per second.


According to the quantum theory of the photoelectric effect, one incident photon typically ejects one electron (if its energy is sufficient).


Therefore, the number of photoelectrons ejected per second is directly proportional to the number of photons incident per second.


The photoelectric current (I) is the rate of flow of these ejected electrons, so it is also directly proportional to the number of electrons ejected per second.


Combining these, the photoelectric current (I) is directly proportional to the intensity of the incident light.


A graph of I versus Intensity will be a straight line passing through the origin with a positive slope. In the provided figures, Graph (C) depicts this linear relationship.
Quick Tip: Remember the key dependencies in the photoelectric effect: 1. Photocurrent \(\propto\) Intensity. 2. Kinetic Energy of photoelectrons \(\propto\) Frequency (and is independent of intensity). 3. There is a minimum (threshold) frequency below which no emission occurs.

The saturation current in a photoelectric experiment corresponds to the state where all emitted photoelectrons are collected by the anode.


The magnitude of the saturation current is determined by the number of photoelectrons emitted from the cathode per second.


The number of photoelectrons emitted per second is directly proportional to the number of photons incident per second.


The number of incident photons per second is a measure of the intensity of the light, not its frequency.


If the intensity is kept constant, the number of photons incident per second does not change, even if their individual energy (frequency) increases.


Therefore, increasing the frequency while keeping the intensity constant does not change the number of emitted photoelectrons per second.


Consequently, the saturation current remains the same.
Quick Tip: Think of light intensity as the "number of bullets" (photons) and frequency as the "power of each bullet" (energy). The saturation current depends on the number of bullets, while the stopping potential (a measure of max K.E.) depends on the power of each bullet.

Einstein's photoelectric equation relates the maximum kinetic energy (\(K_{max}\)) of photoelectrons to the frequency of incident light (\(\nu\)) and the work function (\(\phi_0\)).

\(K_{max} = h\nu - \phi_0\).

The maximum kinetic energy is related to the cut-off (or stopping) potential, \(V_s\), by \(K_{max} = eV_s\).


Substituting this into the equation, we get:
\(eV_s = h\nu - \phi_0\).


Rearranging this to express \(V_s\) as a function of \(\nu\):
\(V_s = \left(\frac{h}{e}\right)\nu - \frac{\phi_0}{e}\).


This equation is in the form of a straight line, \(y = mx + c\), where \(y = V_s\) and \(x = \nu\).


The slope of the graph of \(V_s\) versus \(\nu\) is \(m = \frac{h}{e}\).

Since the charge of an electron, 'e', is a known constant, Planck's constant 'h' can be determined by measuring the slope of this graph.
Quick Tip: The plot of stopping potential (\(V_s\)) vs. frequency (\(\nu\)) is a crucial experimental verification of the photoelectric equation. Its slope is a universal constant (\(h/e\)), and its x-intercept gives the threshold frequency (\(\nu_0\)).

According to Einstein's photoelectric equation, the maximum kinetic energy of an ejected photoelectron is given by:

\(K_{max} = h\nu - \phi_0\), where \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, and \(\phi_0\) is the work function of the metal.


For a given metal surface, the work function \(\phi_0\) is constant.

Therefore, the maximum kinetic energy \(K_{max}\) is directly proportional to the frequency \(\nu\) of the incident light.


The order of frequencies for the colors of the visible spectrum is:

Frequency(Red) < Frequency(Yellow) < Frequency(Blue).
\(\nu_R < \nu_Y < \nu_B\).


Since the kinetic energy is proportional to the frequency, the kinetic energies will follow the same order:
\(K_R < K_Y < K_B\).

This can also be written as \(K_B > K_Y > K_R\).
Quick Tip: Remember the order of colors in the visible spectrum by the acronym VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red). This order represents decreasing frequency and increasing wavelength.

For the photoelectric effect to occur, the energy of the incident photon (\(E = h\nu\)) must be greater than or equal to the work function (\(\phi_0\)) of the metal.

Visible light has a photon energy range from approximately 1.8 eV (for red light) to 3.1 eV (for violet light).

A metal will exhibit the photoelectric effect with visible light only if its work function is less than the energy of the incident photons. This means the work function must be low, preferably less than 3.1 eV.

Let's compare the work functions of the given metals:

- Caesium (Cs): \(\phi_0 \approx 2.14\) eV.

- Zinc (Zn): \(\phi_0 \approx 4.31\) eV.

- Cadmium (Cd): \(\phi_0 \approx 4.22\) eV.

- Magnesium (Mg): \(\phi_0 \approx 3.66\) eV.

Only Caesium has a work function (2.14 eV) that is low enough to be overcome by the energy of photons in the visible spectrum (1.8 eV to 3.1 eV).

The other metals require higher energy photons, such as those in the ultraviolet (UV) range, to cause photoemission.
Quick Tip: Alkali metals (like Sodium, Potassium, Caesium) are known for their very low work functions, making them ideal materials for demonstrating the photoelectric effect with visible light in phototubes and other photosensitive devices.

(a) (i) Kirchhoff's Laws Application


Let the currents flowing from the positive terminals of \(E_1, E_2,\) and \(E_3\) be \(I_1, I_2,\) and \(I_3\) respectively. The diagram shows \(E_1\) and \(E_2\) in parallel, and this combination in series with \(E_3\). However, the standard diagram for this problem is a simple parallel combination. Let's assume three batteries are connected in parallel to a common pair of nodes. Let the top node be P and the bottom node be Q.


Applying Kirchhoff's Current Law (KCL) at node P: \(I_1 + I_2 + I_3 = 0\). (Eq. 1)


Applying Kirchhoff's Voltage Law (KVL) to the loop containing \(E_1\) and \(E_2\):

\(E_1 - I_1r_1 - (E_2 - I_2r_2) = 0 \implies 4 - 2I_1 - (2 - 4I_2) = 0 \implies 2 - 2I_1 + 4I_2 = 0 \implies I_1 - 2I_2 = 1\). (Eq. 2)


Applying KVL to the loop containing \(E_2\) and \(E_3\):

\(E_2 - I_2r_2 - (E_3 - I_3r_3) = 0 \implies 2 - 4I_2 - (6 - 2I_3) = 0 \implies -4 - 4I_2 + 2I_3 = 0 \implies I_3 - 2I_2 = 2\). (Eq. 3)


From (Eq. 2), \(I_1 = 1 + 2I_2\).

From (Eq. 3), \(I_3 = 2 + 2I_2\).


Substitute these into (Eq. 1): \((1 + 2I_2) + I_2 + (2 + 2I_2) = 0 \implies 3 + 5I_2 = 0 \implies I_2 = -0.6\) A.


The negative sign means the current flows in the opposite direction to our assumption.


Now find \(I_1\) and \(I_3\):
\(I_1 = 1 + 2(-0.6) = 1 - 1.2 = -0.2\) A.
\(I_3 = 2 + 2(-0.6) = 2 - 1.2 = 0.8\) A.


So, currents are: through \(E_1\): 0.2 A, through \(E_2\): 0.6 A, through \(E_3\): 0.8 A. The directions need to be specified based on the signs.


(a) (ii) Resistance of a Wire Frame


The given figure is a Wheatstone bridge with an additional resistor between A and B. Let the top vertex be C and the bottom vertex be D. The resistors are AC, AD, CB, DB, CD, and AB. This is a complex network.


The structure forms a triangular prism. Let's assume the question refers to the common Wheatstone bridge diagram shown. This diagram is a balanced Wheatstone bridge. Let vertices be P (top), Q (bottom), A (left), B (right). Resistors are AP, AQ, PB, QB, and PQ.

\(R_{AP} = R_{AQ} = R_{PB} = R_{QB} = R_{PQ} = R=10\Omega\).

Since \(R_{AP}/R_{AQ} = R/R = 1\) and \(R_{PB}/R_{QB} = R/R = 1\), the bridge is balanced.


No current flows through the central resistor \(R_{PQ}\). It can be removed.


The upper arm resistance is \(R_{APB} = R_{AP} + R_{PB} = R+R = 2R\).

The lower arm resistance is \(R_{AQB} = R_{AQ} + R_{QB} = R+R = 2R\).

The equivalent resistance between A and B is the parallel combination of the upper and lower arms.

\(R_{eff} = \frac{(2R)(2R)}{2R+2R} = \frac{4R^2}{4R} = R\).

Given \(R=10 \Omega\), so \(R_{eff} = 10 \ \Omega\).


OR


(b) (i) Current in a Copper Rod

(I) The electric field E inside a conductor is related to current density J and resistivity \(\rho\) by \(E = \rho J\).


Current density is \(J = I/A\). So, \(E = \rho \frac{I}{A}\).

Since the current I and resistivity \(\rho\) are constant throughout the rod, the electric field is inversely proportional to the cross-sectional area A.

\(E \propto \frac{1}{A}\).

The ratio of the electric fields at points A and B is:

\(\frac{E_A}{E_B} = \frac{\rho I / A_A}{\rho I / A_B} = \frac{A_B}{A_A}\).
\(\frac{E_A}{E_B} = \frac{2.0 \times 10^{-7} m^2}{1.0 \times 10^{-7} m^2} = 2\).


The ratio is 2:1.

(II) The drift velocity (\(v_d\)) is related to the current by \(I = n e A v_d\).


We need to find the drift velocity at point B, \(v_{d,B}\).
\(v_{d,B} = \frac{I}{n e A_B}\).

Given: \(I=1\) A, \(n = 8.5 \times 10^{28}\) m\(^{-3}\), \(e = 1.6 \times 10^{-19}\) C, \(A_B = 2.0 \times 10^{-7}\) m\(^2\).

\(v_{d,B} = \frac{1}{(8.5 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (2.0 \times 10^{-7})}\).
\(v_{d,B} = \frac{1}{8.5 \times 1.6 \times 2.0 \times 10^{28-19-7}} = \frac{1}{27.2 \times 10^2} = \frac{1}{2720}\).

\(v_{d,B} \approx 3.67 \times 10^{-4}\) m/s.


(b) (ii) Net Electric Field

The point P where we need to find the field is \(\vec{r} = (3\hat{i} + 4\hat{j})\).


The position of charge \(q_1 = 16\ \mu\)C is \(\vec{r_1} = 3\hat{i}\).

The position of charge \(q_2 = 1\ \mu\)C is \(\vec{r_2} = 4\hat{j}\).

The vector from \(q_1\) to P is \(\vec{r}_{1P} = \vec{r} - \vec{r_1} = (3\hat{i} + 4\hat{j}) - (3\hat{i}) = 4\hat{j}\). The distance is \(|\vec{r}_{1P}| = 4\) m.


The electric field at P due to \(q_1\) is \(\vec{E_1} = k \frac{q_1}{r_{1P}^2} \hat{r}_{1P} = (9 \times 10^9) \frac{16 \times 10^{-6}}{4^2} \hat{j} = (9 \times 10^9) \frac{16 \times 10^{-6}}{16} \hat{j} = 9 \times 10^3 \hat{j}\) N/C.


The vector from \(q_2\) to P is \(\vec{r}_{2P} = \vec{r} - \vec{r_2} = (3\hat{i} + 4\hat{j}) - (4\hat{j}) = 3\hat{i}\). The distance is \(|\vec{r}_{2P}| = 3\) m.


The electric field at P due to \(q_2\) is \(\vec{E_2} = k \frac{q_2}{r_{2P}^2} \hat{r}_{2P} = (9 \times 10^9) \frac{1 \times 10^{-6}}{3^2} \hat{i} = (9 \times 10^9) \frac{1 \times 10^{-6}}{9} \hat{i} = 1 \times 10^3 \hat{i}\) N/C.


The net electric field \(\vec{E}\) at P is the vector sum of \(\vec{E_1}\) and \(\vec{E_2}\).
\(\vec{E} = \vec{E_1} + \vec{E_2} = (1 \times 10^3 \hat{i} + 9 \times 10^3 \hat{j})\) N/C.
Quick Tip: For part (a)(ii), recognizing a balanced Wheatstone bridge is a major shortcut. If the ratio of resistances in the two arms is equal, the galvanometer/central resistor can be ignored. For part (b)(i), remember that for a conductor of non-uniform cross-section in steady state, the current I is constant everywhere, but current density J, electric field E, and drift velocity \(v_d\) are all inversely proportional to the area A.

(a) (i) Self-Inductance and Energy Stored


Definition of Self-Inductance:
Self-inductance of a coil is the property by which it opposes any change in the strength of the current flowing through it by inducing an electromotive force (emf) in itself. It is numerically equal to the magnetic flux linked with the coil when unit current flows through it (\(\Phi = LI\)), or the induced emf when the rate of change of current is unity (\(\mathcal{E} = -L \frac{dI}{dt}\)). Its SI unit is the Henry (H).


Derivation of Energy Stored: To build up a current in an inductor, work must be done against the back emf. The instantaneous power required is \(P = |\mathcal{E}| I = (L \frac{dI}{dt})I\).


The small amount of work done (\(dW\)) in a small time \(dt\) to increase the current by \(dI\) is:

\(dW = P dt = (L \frac{dI}{dt})I dt = LI dI\).

The total work done (which is stored as magnetic potential energy U) in building the current from 0 to I is:

\(U = \int dW = \int_0^I LI dI = L \int_0^I I dI\).
\(U = L \left[\frac{I^2}{2}\right]_0^I = \frac{1}{2}LI^2\).


(a) (ii) Graphs for Two Inductors

Let \(L_1 = 10\) mH and \(L_2 = 20\) mH.


(I) EMF vs. Rate of Change of Current: The magnitude of induced emf is \(|\mathcal{E}| = L \frac{dI}{dt}\). This is an equation of a straight line of the form \(y = mx\), where \(y=|\mathcal{E}|\), \(x = \frac{dI}{dt}\), and the slope is the inductance L. Both graphs will be straight lines passing through the origin. The graph for \(L_2\) will have a steeper slope than the graph for \(L_1\) because \(L_2 > L_1\).


(II) Energy Stored vs. Current: The energy stored is \(U = \frac{1}{2}LI^2\). This is an equation of a parabola of the form \(y = kx^2\), where \(y=U\), \(x=I\), and \(k = L/2\). Both graphs will be parabolas opening upwards, starting from the origin. The parabola for \(L_2\) will be steeper (open up faster) than the parabola for \(L_1\) because for the same current I, the energy stored is greater for the larger inductance.


OR


(b) (i) Mutual Inductance


Definition of Mutual Inductance: Mutual inductance between two coils is the property by which a change of current in one coil (primary) induces an emf in the other coil (secondary). It is numerically equal to the magnetic flux linked with the secondary coil when unit current flows through the primary coil (\(\Phi_2 = MI_1\)). Its SI unit is the Henry (H).


Derivation for Coaxial Solenoids: Consider two long coaxial solenoids of the same length \(l\). Let the inner solenoid (S1) have radius \(r_1\), \(N_1\) turns, and number of turns per unit length \(n_1 = N_1/l\). Let the outer solenoid (S2) have radius \(r_2\), \(N_2\) turns, and \(n_2 = N_2/l\).


Let a current \(I_1\) flow through the outer solenoid S2. The magnetic field inside S2 is uniform and is given by \(B_2 = \mu_0 n_2 I_1\). This field is zero outside S2.


The flux through each turn of the inner solenoid S1 is \(\Phi_{turn} = B_2 \times A_1 = (\mu_0 n_2 I_1)(\pi r_1^2)\).

The total flux linked with the inner solenoid S1 is \(\Phi_1 = N_1 \times \Phi_{turn} = N_1 (\mu_0 n_2 I_1)(\pi r_1^2)\).


By definition, \(\Phi_1 = M_{12} I_1\). Comparing the two expressions:
\(M_{12} I_1 = N_1 (\mu_0 n_2 I_1)(\pi r_1^2)\).
\(M_{12} = \mu_0 N_1 n_2 \pi r_1^2 = \mu_0 (n_1 l) n_2 \pi r_1^2 = \mu_0 n_1 n_2 A_1 l\).


(b) (ii) Flux Calculation

The magnitude of the induced emf is given by \(|\mathcal{E}| = L \frac{dI}{dt}\).


The current increases uniformly, so the rate of change of current is constant.
\(\frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{2 A - 0 A}{40 s} = \frac{2}{40} = 0.05\) A/s.


We are given the induced emf, \(|\mathcal{E}| = 5\) mV = \(5 \times 10^{-3}\) V.

We can find the self-inductance L of the inductor.

\(L = \frac{|\mathcal{E}|}{dI/dt} = \frac{5 \times 10^{-3} V}{0.05 A/s} = \frac{5 \times 10^{-3}}{5 \times 10^{-2}} = 0.1\) H.

The current at time t is given by \(I(t) = (\frac{dI}{dt})t = 0.05 t\).


We need to find the flux linked with the inductor at \(t = 10\) s.

First, find the current at \(t=10\) s: \(I(10) = 0.05 \times 10 = 0.5\) A.

The flux linked with the inductor is given by \(\Phi = LI\).
\(\Phi = (0.1 H) \times (0.5 A) = 0.05\) Wb.
Quick Tip: For part (a)(i), the energy stored in an inductor (\(U = \frac{1}{2}LI^2\)) is the magnetic analogue of the energy stored in a capacitor (\(U = \frac{1}{2}CV^2\)). For part (b)(i), the mutual inductance \(M_{12}\) always equals \(M_{21}\). The formula \(M = \mu_0 n_1 n_2 A_1 l\) shows that M depends on the geometry and the medium, and crucially, on the area of the smaller coil, as that is the area through which the flux is linked.

(a) (i) Reflecting Telescope (Cassegrain)


Ray Diagram and Explanation: A Cassegrain telescope consists of a large concave parabolic primary mirror and a small convex secondary mirror. Parallel rays from a distant object enter the telescope and are reflected by the primary mirror. Before these rays can converge to a focus, they are intercepted by the convex secondary mirror placed on the axis of the telescope. The secondary mirror reflects the light back through a hole in the center of the primary mirror. An eyepiece is placed behind this hole, where the final image is formed. The final image is inverted.


Advantages over Refracting Telescope:


1. No Chromatic Aberration: Reflecting telescopes use mirrors instead of lenses. Mirrors do not cause chromatic aberration (dispersion of light into colors) because the law of reflection is independent of wavelength. Refracting telescopes suffer from this defect, which blurs the image.


2. High Light-Gathering Power and Resolution: It is mechanically easier to build very large mirrors than large lenses. A large aperture (diameter) of the primary mirror allows the telescope to gather more light from faint objects and provides a higher resolving power (ability to distinguish fine details), as resolution is proportional to the aperture diameter.


(a) (ii) Refracting Telescope Calculations


Given: Focal length of objective, \(f_o = 50 f_e\), where \(f_e\) is the focal length of the eyepiece.

For a telescope in normal adjustment (final image at infinity), the tube length (L) is the sum of the focal lengths of the objective and the eyepiece.

\(L = f_o + f_e\).

We are given \(L = 102\) cm.

Substituting the given relation into the tube length equation:
\(102 = (50 f_e) + f_e\).
\(102 = 51 f_e\).
\(f_e = \frac{102}{51} = 2\) cm.

Now, find the focal length of the objective lens:
\(f_o = 50 f_e = 50 \times 2 = 100\) cm.

The focal lengths are: Objective = 100 cm, Eyepiece = 2 cm.


OR


(b) (i) Compound Microscope

Advantages over Simple Microscope:


1. Higher Magnification: A compound microscope uses two lenses (objective and eyepiece) to achieve a much higher magnification than a simple microscope (which is just a single convex lens). The total magnification is the product of the magnifications of the two lenses.


2. Greater Resolving Power: The resolving power of a microscope is its ability to distinguish between two closely spaced points. The resolving power of a compound microscope is significantly higher than that of a simple microscope, allowing for the observation of much finer details.


Ray Diagram and Explanation (Image at Near Point):

An object is placed just outside the focal point of the objective lens. The objective forms a real, inverted, and magnified intermediate image. This intermediate image is formed within the focal length of the eyepiece. The eyepiece then acts like a simple microscope, using this intermediate image as its object to form a final, virtual, and highly magnified image at the near point (D) of the observer. (The ray diagram should show these two stages of magnification).


(b) (ii) Power of Combination Lens

We use the Lens Maker's Formula for each lens: \(\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).


Lens 1: Planoconcave

Refractive index \(n_1\). For the planar face, \(R_1 = \infty\). For the concave face, \(R_2 = -R\) (using sign convention, light from left to right).

\(\frac{1}{f_1} = (n_1-1)\left(\frac{1}{\infty} - \frac{1}{-R}\right) = (n_1-1)\left(\frac{1}{R}\right) = \frac{n_1-1}{R}\).


Lens 2: Equiconvex

Refractive index \(n_2\). Both faces are convex with the same radius R. For the first face, \(R_1 = +R\). For the second face, \(R_2 = -R\).

\(\frac{1}{f_2} = (n_2-1)\left(\frac{1}{R} - \frac{1}{-R}\right) = (n_2-1)\left(\frac{1}{R} + \frac{1}{R}\right) = (n_2-1)\left(\frac{2}{R}\right) = \frac{2(n_2-1)}{R}\).


Power of the Combination:

When lenses are in contact, the equivalent focal length \(F\) is given by \(\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}\).


The power of the combination is \(P = \frac{1}{F}\).
\(P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2}\).
\(P = \frac{n_1-1}{R} + \frac{2(n_2-1)}{R}\).
\(P = \frac{(n_1-1) + 2(n_2-1)}{R} = \frac{n_1 - 1 + 2n_2 - 2}{R} = \frac{n_1 + 2n_2 - 3}{R}\).
Quick Tip: For optical instruments: For a telescope in normal adjustment, tube length \(L=f_o+f_e\) and magnification \(M=-f_o/f_e\). For a microscope, magnification \(M \approx (-L/f_o)(D/f_e)\). For lens combinations, remember that powers add directly: \(P_{total} = P_1 + P_2 + ...\)