Step 1: Initial Setup with Coulomb's Law

Let the two point charges be \(q_1\) and \(q_2\), and the distance between them be \(r\). The electric force in a vacuum (or air) is given by Coulomb's Law: \[ F_{vacuum} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \]
We are given that \(F_{vacuum} = 80 N\).


Step 2: Force in the Dielectric Medium

When these charges are placed in a dielectric medium, the permittivity of the space changes from \(\epsilon_0\) (permittivity of free space) to \(\epsilon_m\) (permittivity of the medium). The force in the medium is: \[ F_{medium} = \frac{1}{4\pi\epsilon_m} \frac{q_1 q_2}{r^2} \]
We are given that \(F_{medium} = 8 N\).


Step 3: Relating the Forces using the Dielectric Constant

The dielectric constant (\(K\)) is defined as the ratio of the permittivity of the medium to the permittivity of free space: \(K = \frac{\epsilon_m}{\epsilon_0}\).
From this, we can write \(\epsilon_m = K \cdot \epsilon_0\).
Now, let's establish a ratio between the force in the vacuum and the force in the medium: \[ \frac{F_{vacuum}}{F_{medium}} = \frac{\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}}{\frac{1}{4\pi\epsilon_m} \frac{q_1 q_2}{r^2}} \]
The terms \(\frac{q_1 q_2}{4\pi r^2}\) cancel out, leaving: \[ \frac{F_{vacuum}}{F_{medium}} = \frac{\epsilon_m}{\epsilon_0} \]
Since \(K = \frac{\epsilon_m}{\epsilon_0}\), we have: \[ \frac{F_{vacuum}}{F_{medium}} = K \]

Step 4: Calculation and Final Answer

Substitute the given values into the derived relationship: \[ K = \frac{80 N}{8 N} \] \[ K = 10 \]
The dielectric constant of the medium is 10. Therefore, option (B) is correct.
Quick Tip: For any medium other than a vacuum, the dielectric constant \(K\) is always greater than 1. This means the electric force between charges is always weakened when they are placed in a dielectric material. If you get an answer for K that is less than 1, you have likely inverted the formula.

Step 1: Principle of Balancing Nuclear Reactions

A nuclear reaction must be balanced. This means that the total number of protons (which determines the atomic number, Z) and the total number of neutrons must be the same on both the reactant and product sides of the equation. The mass number (A) is the sum of protons and neutrons.

Step 2: Tally of Protons and Neutrons on the Reactant Side

Let's analyze the particles on the left side of the reaction:

For Helium (\(_{2}^{4}He\)): It has 2 protons (Z=2) and \(4-2=2\) neutrons.
For Nitrogen (\(_{7}^{14}N\)): It has 7 protons (Z=7) and \(14-7=7\) neutrons.

Total on the reactant side:

Total Protons = \(2 + 7 = 9\)
Total Neutrons = \(2 + 7 = 9\)


Step 3: Tally of Protons and Neutrons on the Product Side

Now let's analyze the particles on the right side of the reaction:

For Oxygen (\(_{8}^{17}O\)): It has 8 protons (Z=8) and \(17-8=9\) neutrons.
For the unknown particle X: Let it have 'p' protons and 'n' neutrons.

To balance the reaction, the totals must match the reactant side:

Total Protons must be 9: \(8 + p = 9 \implies p = 1\)
Total Neutrons must be 9: \(9 + n = 9 \implies n = 0\)


Step 4: Identifying Particle X and Final Answer

The unknown particle X is composed of 1 proton and 0 neutrons. A particle with these properties is a proton.
The mass number (A) for X is the sum of its protons and neutrons: \(A = p + n = 1 + 0 = 1\).
The atomic number (Z) for X is its number of protons: \(Z = p = 1\).
So, X is \(_{1}^{1}p\), a proton. Therefore, option (A) is correct.
Quick Tip: Familiarize yourself with the notations for common particles in nuclear physics: Proton: \(_{1}^{1}p\) or \(_{1}^{1}H\) Neutron: \(_{0}^{1}n\) Electron (Beta-minus particle): \(_{-1}^{0}e\) or \(_{-1}^{0}\beta\) Positron (Beta-plus particle): \(_{+1}^{0}e\) or \(_{+1}^{0}\beta\) Alpha particle (Helium nucleus): \(_{2}^{4}He\) or \(_{2}^{4}\alpha\) Deuteron (Deuterium nucleus): \(_{1}^{2}H\) or \(_{1}^{2}D\)

Step 1: Foundational Principle - Bohr's Quantization Rule

According to Niels Bohr's model of the atom, the angular momentum of an electron is quantized, meaning it can only take on discrete, specific values. The allowed value of angular momentum (\(L\)) for an electron in the \(n^{th}\) orbit is given by the formula: \[ L_n = n \frac{h}{2\pi} \]
where \(n\) is the principal quantum number (an integer like 1, 2, 3, ...) and \(h\) is Planck's constant.


Step 2: Calculating Angular Momentum for Specific Orbits

To find the difference between successive orbits, let's calculate the angular momentum for the first few orbits by substituting values of \(n\).

For the first orbit (\(n=1\)):
\[ L_1 = (1) \frac{h}{2\pi} = \frac{h}{2\pi} \]
For the second orbit (\(n=2\)):
\[ L_2 = (2) \frac{h}{2\pi} = \frac{h}{\pi} \]
For the third orbit (\(n=3\)):
\[ L_3 = (3) \frac{h}{2\pi} \]


Step 3: Determining the Difference Between Successive Orbits

Now, let's compute the difference in angular momentum between consecutive orbits using the values from Step 2.

Difference between the 2nd and 1st orbit:
\[ \Delta L = L_2 - L_1 = \frac{2h}{2\pi} - \frac{h}{2\pi} = \frac{h}{2\pi} \]
Difference between the 3rd and 2nd orbit:
\[ \Delta L = L_3 - L_2 = \frac{3h}{2\pi} - \frac{2h}{2\pi} = \frac{h}{2\pi} \]

We can observe a consistent pattern. The difference between the angular momentum of any two successive orbits is always the same.


Step 4: Final Conclusion

The calculation shows that the difference in angular momentum between any two consecutive orbits is a constant value of \(\frac{h}{2\pi}\). This quantity is also known as the reduced Planck constant, \(\hbar\). Therefore, option (A) is correct.
Quick Tip: Remember Bohr's three postulates:
(1) Electrons revolve in stable circular orbits without radiating energy.
(2) Angular momentum is quantized: \(L = n\hbar\).
(3) Energy is radiated/absorbed when an electron jumps between orbits, with frequency \(f = \frac{\Delta E}{h}\).
The difference in angular momentum between any two consecutive orbits is always \(\hbar\) or \(\frac{h}{2\pi}\).

Step 1: Understanding the Concept:

Rutherford's gold foil experiment (or \(\alpha\)-particle scattering experiment) involved bombarding a thin sheet of gold foil with positively charged alpha particles. The scattering pattern of these particles led to the discovery of the atomic nucleus. The question asks for the fundamental force responsible for this scattering.


Step 2: Detailed Explanation:

Let's analyze the forces involved:


Nature of Particles: The \(\alpha\)-particle is a helium nucleus (\(_{2}^{4}He^{2+}\)), which is positively charged. The target, the nucleus of a gold atom, is also positively charged (as it contains protons).

Coulomb's force: This is the electrostatic force of repulsion between two like charges (positive \(\alpha\)-particle and positive nucleus). This force is long-range and follows an inverse-square law (\(F \propto 1/r^2\)), which perfectly explains why most particles passed straight through (far from a nucleus) while a few were deflected at large angles (passed close to a nucleus).

Gravitational force: This is the force of attraction between masses. While it exists, it is extremely weak compared to the electrostatic force at the atomic scale and is completely negligible in this context.

Nuclear force: This is a very strong, short-range attractive force that holds protons and neutrons together within the nucleus. It only acts over distances of about \(10^{-15}\) m. The \(\alpha\)-particles are repelled by the Coulomb force long before they get close enough for the nuclear force to be significant, except in very high-energy head-on collisions. The scattering itself is governed by the electrostatic repulsion.

Magnetic force: This force acts on moving charges within a magnetic field. There is no external magnetic field applied in this experiment, so this force is not the cause of the scattering.


The primary interaction causing the \(\alpha\)-particles to deviate from their straight paths is the electrostatic repulsion between the positively charged \(\alpha\)-particles and the positively charged gold nuclei. This force is known as Coulomb's force.


Step 3: Final Answer:

The force responsible for the scattering is Coulomb's force. Therefore, option (B) is correct.
Quick Tip: Rutherford's experiment is a cornerstone of modern physics. Remember its key conclusion: the atom consists of a tiny, dense, positively charged nucleus surrounded by a large volume of mostly empty space containing electrons. The scattering is a classic example of an interaction governed by Coulomb's law.

Step 1: Understanding the Concept:

The electromagnetic (EM) spectrum is the range of all types of EM radiation. Radiation is classified by wavelength or frequency. The order of the spectrum in terms of increasing frequency (and decreasing wavelength) is: Radio waves, Microwaves, Infrared, Visible light, Ultraviolet, X-rays, Gamma rays.


Step 2: Detailed Explanation:

Let's analyze the given frequency ranges:


Visible Light: The visible spectrum, which our eyes can see, ranges from approximately \(4.3 \times 10^{14}\) Hz (red) to \(7.5 \times 10^{14}\) Hz (violet).

Ultraviolet (UV) Radiation: UV radiation has a frequency just higher than that of violet light. Its range typically starts from the end of the visible spectrum and extends upwards. The accepted range for UV is roughly \(8 \times 10^{14}\) Hz to \(3 \times 10^{16}\) Hz.



(A) \(3 \times 10^{10} - 3 \times 10^{12}\) Hz: This corresponds to the microwave and far-infrared region.

(B) \(3 \times 10^{12} - 3 \times 10^{14}\) Hz: This corresponds to the infrared region.

(C) \(3 \times 10^{14} - 3 \times 10^{16}\) Hz: This range begins at the boundary of infrared/visible light and extends through the entire ultraviolet region, up to the start of the X-ray region. This is the most appropriate range among the choices.

(D) \(3 \times 10^{16}\) and above: Frequencies above \(3 \times 10^{16}\) Hz are generally classified as X-rays and Gamma rays.


Step 3: Final Answer:

The frequency range that best represents ultraviolet radiation from the given options is \(3 \times 10^{14} - 3 \times 10^{16}\) Hz. Therefore, option (C) is correct.
Quick Tip: To remember the order of the electromagnetic spectrum, use a mnemonic like "Roman Men Invented Very Unusual X-ray Guns" for Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma Ray. Remember that frequency increases and wavelength decreases as you move from left to right in this mnemonic.

Step 1: Understanding the Concept:

When a lens is used to view a very distant object like the sun, the parallel rays of light from the object converge to form an image at the focal plane of the lens. The size of this image is related to the focal length of the lens and the angular size of the object.


Step 2: Key Formula or Approach:

For small angles, the relationship between the angular size (\(\theta\)), the image diameter (\(d\)), and the focal length (\(f\)) can be approximated by: \[ \theta \approx \tan(\theta) = \frac{Opposite}{Adjacent} = \frac{d}{f} \]
This formula requires the angle \(\theta\) to be in radians.

The conversion from degrees to radians is: \( angle in radians = angle in degrees \times \frac{\pi}{180} \).


Step 3: Detailed Explanation:

We are given:

Angular size of the sun, \(\theta = 0.5^\circ\).

Focal length of the convex lens, \(f = 50\) cm.


First, we must convert the angular size from degrees to radians: \[ \theta_{rad} = 0.5 \times \frac{\pi}{180} \approx 0.5 \times \frac{3.14159}{180} \approx 0.008727 radians \]

Now, we can use the formula \(d = f \times \theta_{rad}\) to find the diameter of the image (\(d\)): \[ d = 50 cm \times 0.008727 \] \[ d \approx 0.43635 cm \]

Step 4: Final Answer:

The diameter of the image is approximately 0.43 cm. Comparing this with the given options, option (D) is the closest match.
Quick Tip: A common mistake in problems involving angles in physics formulas (like in optics, simple harmonic motion, or circular motion) is forgetting to convert the angle from degrees to radians. Always check the requirements of the formula. For the small angle approximation (\(\tan \theta \approx \sin \theta \approx \theta\)), the angle \(\theta\) must be in radians.

Step 1: Understanding the Concept:

The de Broglie hypothesis proposes that all matter, such as electrons and protons, exhibits wave-like properties. A particle with momentum \(p\) has an associated wavelength \(\lambda\), known as the de Broglie wavelength. Kinetic energy (K) is the energy a particle possesses due to its motion. The task is to derive an expression for the wavelength \(\lambda\) in terms of the particle's kinetic energy \(K\).


Step 2: Key Formula or Approach:

The de Broglie wavelength is fundamentally defined as: \[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the particle (\(p=mv\)).

The classical kinetic energy (K) of a particle of mass \(m\) is given by \(K = \frac{1}{2}mv^2\). This can be related to momentum by the formula: \[ K = \frac{p^2}{2m} \]

Step 3: Detailed Explanation:

Our goal is to substitute momentum \(p\) in the de Broglie equation with an expression involving kinetic energy \(K\). We start with the kinetic energy formula and solve for \(p\): \[ K = \frac{p^2}{2m} \]
Multiplying both sides by \(2m\): \[ p^2 = 2mK \]
Taking the square root of both sides gives the momentum: \[ p = \sqrt{2mK} \]
Now, we substitute this expression for \(p\) into the de Broglie wavelength formula: \[ \lambda = \frac{h}{\sqrt{2mK}} \]
This equation gives the de Broglie wavelength directly from the particle's kinetic energy and mass.


Step 4: Final Answer:

The de Broglie wavelength (\(\lambda\)) in terms of kinetic energy (\(K\)) for a particle of mass \(m\) is given by the expression: \[ \lambda = \frac{h}{\sqrt{2mK}} \] Quick Tip: This formula is extremely important in quantum mechanics, especially for problems involving electrons accelerated through a potential difference V. In that case, the kinetic energy gained is \(K = qV\) (where q is the charge), so the wavelength becomes \(\lambda = \frac{h}{\sqrt{2mqV}}\).

Step 1: Understanding the Concept:

Total Internal Reflection (TIR) is a phenomenon where a light ray traveling from a denser medium (e.g., prism with refractive index \(n\)) to a rarer medium (e.g., air with refractive index \(\approx 1\)) is completely reflected back into the denser medium. This occurs only if the angle of incidence in the denser medium is greater than a specific angle called the critical angle, C. The question asks for the condition on the prism's apex angle A for TIR to be possible at the second face.


Step 2: Key Formula or Approach:

1. Critical Angle (C): The critical angle is defined by Snell's Law when the angle of refraction is 90\(^\circ\). For a prism in air, \( n \sin(C) = 1 \sin(90^\circ) \), which gives \(\sin(C) = \frac{1}{n}\) or \(C = \sin^{-1}\left(\frac{1}{n}\right)\).

2. Condition for TIR: For TIR to occur at the second face of the prism, the angle of incidence there, \(r_2\), must be greater than or equal to the critical angle C. That is, \(r_2 \ge C\).

3. Prism Angle Formula: In a prism, the apex angle \(A\) is related to the angle of refraction at the first face (\(r_1\)) and the angle of incidence at the second face (\(r_2\)) by the geometric relation \(A = r_1 + r_2\).


Step 3: Detailed Explanation:

Let a ray of light enter the first face of the prism and refract at an angle \(r_1\). This ray then travels to the second face, striking it at an angle of incidence \(r_2\).

The condition for TIR to take place at this second face is: \[ r_2 \ge C \]
Using the prism angle formula, we can express \(r_2\) as \(r_2 = A - r_1\). Substituting this into the TIR condition: \[ A - r_1 \ge C \]
Rearranging the inequality to find the condition on A: \[ A \ge C + r_1 \]
This inequality shows that for TIR to be possible, the prism angle A must be at least \(C + r_1\). The value of \(r_1\) depends on the initial angle of incidence \(i\) on the first face. To determine if TIR is possible at all, we must check if this condition can be met for any valid \(r_1\).

The minimum possible value for \(r_1\) is 0, which occurs when the light ray enters the prism at normal incidence (\(i = 0\)). If we substitute this minimum value of \(r_1\) into our condition, we get the minimum requirement for A: \[ A \ge C + 0 \] \[ A \ge C \]
This means that for TIR to be even a possibility, the prism angle A must be at least equal to the critical angle C. If \(A < C\), then \(A - r_1\) will always be less than C (since \(r_1 \ge 0\)), and TIR can never occur.


Step 4: Final Answer:

Thus, total internal reflection is possible only if the angle of the prism \(A\) is greater than or equal to the critical angle \(C\). Substituting the expression for the critical angle, \(C = \sin^{-1}\left(\frac{1}{n}\right)\), we prove that TIR is possible only when \(A \ge \sin^{-1}\left(\frac{1}{n}\right)\). The question's use of "more than" implies the strict condition for practical observation.
Quick Tip: This principle is fundamental to the design of optical instruments. For example, 45\(^\circ\)-90\(^\circ\)-45\(^\circ\) prisms are used in binoculars and periscopes. For typical glass (\(n \approx 1.5\)), the critical angle is \(C \approx 42^\circ\). Since \(45^\circ > 42^\circ\), these prisms are designed to produce total internal reflection.

Step 1: Understanding the Concept:

The electrical resistance (\(R\)) of a conductor depends on its intrinsic property (resistivity, \(\rho\)), its length (\(L\)), and its cross-sectional area (\(A\)). Since the problem specifies two copper wires of the same length, their resistivity and length are identical. Therefore, the difference in their resistance will solely depend on their cross-sectional areas, which are determined by their diameters.


Step 2: Key Formula or Approach:

The formula for the resistance of a wire is: \[ R = \rho \frac{L}{A} \]
The cross-sectional area \(A\) of a wire with diameter \(d\) is circular, so its area is: \[ A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \]
Substituting the area into the resistance formula shows that resistance is inversely proportional to the square of the diameter: \[ R = \rho \frac{L}{(\pi d^2 / 4)} \implies R \propto \frac{1}{d^2} \]

Step 3: Detailed Explanation:

Let the two wires be designated as Wire 1 and Wire 2.
We are given the following information:

Same material (copper): \(\rho_1 = \rho_2 = \rho\)
Same length: \(L_1 = L_2 = L\)
Ratio of diameters: \(\frac{d_1}{d_2} = \frac{2}{1}\)

We want to find the ratio of their resistances, \(\frac{R_1}{R_2}\).
Using the resistance formula for both wires: \[ \frac{R_1}{R_2} = \frac{\rho_1 \frac{L_1}{A_1}}{\rho_2 \frac{L_2}{A_2}} \]
Since \(\rho\) and \(L\) are the same, they cancel out, leaving: \[ \frac{R_1}{R_2} = \frac{A_2}{A_1} \]
Now, substitute the formula for area in terms of diameter: \[ \frac{R_1}{R_2} = \frac{\pi d_2^2 / 4}{\pi d_1^2 / 4} = \frac{d_2^2}{d_1^2} = \left(\frac{d_2}{d_1}\right)^2 \]
We are given \(\frac{d_1}{d_2} = \frac{2}{1}\), which means \(\frac{d_2}{d_1} = \frac{1}{2}\).
Substituting this value into our ratio: \[ \frac{R_1}{R_2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

Step 4: Final Answer:

The ratio of the resistances \(R_1 : R_2\) is 1 : 4. This means the thicker wire (\(d_1\)) has one-fourth the resistance of the thinner wire (\(d_2\)).
Quick Tip: Remember the quick relationship: resistance is inversely proportional to the area (\(R \propto 1/A\)) and inversely proportional to the square of the diameter or radius (\(R \propto 1/d^2\)). If the diameter is doubled, the resistance becomes one-fourth.

Step 1: Understanding the Concept:

The photoelectric effect is the phenomenon where electrons are ejected from a metal surface when light shines on it. According to quantum theory, light consists of packets of energy called photons. For an electron to be ejected, it must absorb a photon with enough energy to overcome the binding force holding it to the metal. This minimum energy required for ejection is called the work function (\(\Phi\)) of the material.


Step 2: Key Formula or Approach:

The energy of a single photon (\(E\)) is given by the Planck-Einstein relation: \[ E = hf = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant, \(c\) is the speed of light, \(f\) is the frequency, and \(\lambda\) is the wavelength of the light.
For the photoelectric effect to occur, the energy of the incident photon must be greater than or equal to the work function: \[ E \ge \Phi \quad or \quad \frac{hc}{\lambda} \ge \Phi \]

Step 3: Detailed Explanation:

The condition for photoemission, \(\frac{hc}{\lambda} \ge \Phi\), reveals an important relationship between wavelength and electron emission. Since wavelength \(\lambda\) is in the denominator, a larger wavelength corresponds to lower photon energy. This implies that there is a maximum wavelength beyond which the photon energy will be insufficient to overcome the work function.
This maximum possible wavelength for which photoemission can occur is called the threshold wavelength, denoted by \(\lambda_0\).

The threshold wavelength corresponds to the exact point where the photon energy is just equal to the work function: \[ E = \Phi \implies \frac{hc}{\lambda_0} = \Phi \]
If the incident light has a wavelength \(\lambda > \lambda_0\), its photons will have less energy than the work function (\(E < \Phi\)), and no electrons will be emitted, no matter how intense the light is.


Step 4: Final Answer:

The threshold wavelength (\(\lambda_0\)) is defined as the maximum wavelength of incident radiation that can cause the photoelectric effect for a specific material. It is a characteristic property of the material and is related to its work function by \(\lambda_0 = hc/\Phi\).
Quick Tip: A common point of confusion is the difference between threshold frequency and threshold wavelength. Remember their inverse relationship: \textbf{Threshold Frequency (\(f_0\))}: The minimum frequency required for photoemission. \textbf{Threshold Wavelength (\(\lambda_0\))}: The maximum wavelength that allows photoemission.

The total magnetic field at point O is the vector sum of the magnetic fields produced by the three sections of the wire: the two straight portions and the semi-circular portion.


(i) Magnetic Field due to Straight Portions


Step 1: Understanding the Concept:

The magnetic field produced by a current-carrying wire can be calculated using the Biot-Savart Law. This law relates the magnetic field to the current element, its length, and the position vector to the point of interest.


Step 2: Key Formula or Approach:

The Biot-Savart Law is given by: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{i(d\vec{l} \times \hat{r})}{r^2} \]
Here, \(d\vec{l}\) is the vector representing the current element and \(\hat{r}\) is the unit vector pointing from the current element to the point O where the field is being calculated.


Step 3: Detailed Explanation:

For both straight portions of the wire, the point O lies on the axis of the wire. This means for any current element \(d\vec{l}\) on the straight wires, the position vector \(\vec{r}\) pointing to O is either parallel (angle of 0\(^\circ\)) or anti-parallel (angle of 180\(^\circ\)) to \(d\vec{l}\).

The cross product \(d\vec{l} \times \hat{r}\) has a magnitude of \(|d\vec{l}| |\hat{r}| \sin\theta = dl \cdot 1 \cdot \sin\theta\).

Since \(\theta = 0^\circ\) or \(\theta = 180^\circ\), \(\sin\theta = 0\).

Therefore, the cross product is zero, and the magnetic field contribution from both straight portions at point O is zero.
\[ \vec{B}_{straight} = 0 \]

(ii) Magnetic Field due to Semi-circular Portion


Step 1: Understanding the Concept:

A current flowing through a circular arc produces a magnetic field at the center of the arc. The magnitude of this field is proportional to the current and the angle subtended by the arc at the center.


Step 2: Key Formula or Approach:

The magnetic field at the center of a full circular loop of radius R carrying current i is: \[ B_{full circle} = \frac{\mu_0 i}{2R} \]
A semi-circle is half of a full circle, so it subtends an angle of \(\pi\) radians at the center. The field it produces will be half of that of a full circle.


Step 3: Detailed Explanation:

The magnetic field due to the semi-circular portion of radius R is: \[ B_{semi-circle} = \frac{1}{2} \times B_{full circle} = \frac{1}{2} \left( \frac{\mu_0 i}{2R} \right) \] \[ B_{semi-circle} = \frac{\mu_0 i}{4R} \]
Direction: To find the direction of the magnetic field, we use the Right-Hand Grip Rule. If we curl the fingers of our right hand in the direction of the current in the semi-circle (counter-clockwise in the diagram provided, although the arrows suggest a left-to-right flow), the thumb points in the direction of the magnetic field. For the current shown flowing from left to right, the thumb points into the plane of the page.


Step 4: Final Answer:

(i) The magnetic field at O due to each straight portion is zero.

(ii) The value of the magnetic field at O due to the semi-circle of radius R is \(\mathbf{\frac{\mu_0 i}{4R}}\), directed into the plane of the page.
Quick Tip: When calculating magnetic fields from complex wire shapes, always break the wire into simpler geometric segments (straight lines, arcs). Calculate the field from each segment and then add them vectorially. Remember that for any point lying on the extended line of a straight current-carrying wire, the magnetic field is zero.

Step 1: Understanding the Concept:

Coherent sources are a fundamental requirement for observing a stable interference pattern when two or more waves superpose. Interference is the phenomenon where waves combine to form a resultant wave of greater, lower, or the same amplitude. For this pattern to be stationary and observable, the sources of the waves must be "coherent."


Step 2: Detailed Explanation:

Two sources of waves are said to be coherent if they satisfy two conditions:


Same Frequency: The waves emitted by the sources must have the same frequency and wavelength. If the frequencies are different, the phase relationship between the waves will change rapidly, and a stable interference pattern will not be formed.

Constant Phase Difference: The waves emitted by the sources must have a phase difference that is constant over time. It does not have to be zero, but it must not change. If the phase difference varies randomly, the positions of maximum and minimum intensity (crests and troughs) will shift randomly, and the interference pattern will be washed out.


In summary, coherent sources are sources that emit waves of the same frequency and with a constant phase difference between them.


Step 3: Importance and Examples:

Importance: Coherence is essential for producing a stable, visible interference pattern. Incoherent sources, like two independent light bulbs, emit waves with random and rapidly changing phase differences, so no sustained interference pattern can be observed.


Examples: In practice, coherent sources are usually derived from a single parent source. For instance, in Young's Double-Slit Experiment, a single light source illuminates two narrow, closely spaced slits. These two slits then act as two new, perfectly coherent sources because the light waves emerging from them originate from the same initial wavefront. Lasers are also excellent examples of sources that produce highly coherent light.


Step 4: Final Answer:

Coherent sources are two or more sources of waves that emit waves having the same frequency and a constant phase difference. These two conditions are necessary for the formation of a sustained and observable interference pattern.
Quick Tip: A simple way to remember the conditions for coherence is "Same Frequency, Constant Phase". These two properties ensure that the interference pattern (the bright and dark fringes in the case of light) does not move or flicker, making it observable.

Step 1: Understanding the Concept:

Electromagnetic (EM) waves are a form of energy that propagates through space. They are produced by the acceleration of electric charges. As predicted by Maxwell's equations, these waves consist of synchronized oscillations of electric and magnetic fields.


Step 2: Detailed Explanation:

Definition:

Electromagnetic waves are disturbances consisting of time-varying electric and magnetic fields that are perpendicular to each other and also perpendicular to the direction of propagation of the wave. This makes them transverse waves.


Characteristics of Electromagnetic Waves:


Transverse Nature: The electric field (\(\vec{E}\)) and magnetic field (\(\vec{B}\)) vectors are mutually perpendicular to each other and also perpendicular to the direction of wave propagation.

No Medium Required: They do not require a material medium for their propagation and can travel through a vacuum.

Speed in Vacuum: All electromagnetic waves travel through a vacuum at a constant speed, known as the speed of light, \(c \approx 3 \times 10^8\) m/s. The speed is given by the relation \(c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}\), where \(\mu_0\) is the permeability and \(\epsilon_0\) is the permittivity of free space.

Energy and Momentum: They carry energy and momentum. The energy is shared equally between the oscillating electric and magnetic fields. When they strike a surface, they exert pressure, known as radiation pressure.

In-Phase Oscillation: The electric and magnetic fields oscillate in the same phase. They reach their maximum and minimum values at the same time and place.

Unaffected by External Fields: Since they are composed of oscillating fields and are not charged particles, they are not deflected by external static electric or magnetic fields.

Wave Phenomena: They exhibit all the properties of waves, such as reflection, refraction, interference, diffraction, and polarization.
Quick Tip: A key takeaway is the transverse nature: E, B, and the direction of travel are all mutually perpendicular. This is a defining feature that distinguishes EM waves from longitudinal waves like sound.

Step 1: Understanding the Concept:

This problem combines two key principles of the Bohr model for the hydrogen atom: the formula for quantized energy levels and the postulate for quantized angular momentum. First, we must identify the principal quantum number (\(n\)) of the electron's orbit from its given energy. Then, we can use this value of \(n\) to calculate the angular momentum.


Step 2: Key Formula or Approach:

1. The energy (\(E_n\)) of an electron in the \(n^{th}\) orbit of a hydrogen atom is given by: \[ E_n = -\frac{13.6}{n^2} eV \]
2. The angular momentum (\(L_n\)) of an electron in the \(n^{th}\) orbit is quantized according to Bohr's second postulate: \[ L_n = n \frac{h}{2\pi} \]
where \(h\) is Planck's constant (\(h \approx 6.626 \times 10^{-34}\) J·s).


Step 3: Detailed Explanation:

Part 1: Find the Principal Quantum Number (n)

We are given the energy of the electron, \(E_n = -3.4\) eV. We can use the energy formula to find \(n\): \[ -3.4 = -\frac{13.6}{n^2} \] \[ n^2 = \frac{13.6}{3.4} = 4 \]
Taking the square root, we get: \[ n = 2 \]
This means the electron is in the first excited state (\(n=2\)).


Part 2: Calculate the Angular Momentum (L)

Now that we have \(n=2\), we can calculate the angular momentum using Bohr's quantization rule: \[ L_2 = 2 \times \frac{h}{2\pi} = \frac{h}{\pi} \]
Substituting the value of Planck's constant: \[ L_2 = \frac{6.626 \times 10^{-34} J·s}{3.14159} \] \[ L_2 \approx 2.11 \times 10^{-34} J·s \]

Step 4: Final Answer:

The angular momentum of the electron is approximately \(2.11 \times 10^{-34}\) J·s.
Quick Tip: It is useful to memorize the energies of the first few levels of the hydrogen atom: \(E_1 = -13.6\) eV, \(E_2 = -3.4\) eV, \(E_3 = -1.51\) eV. Recognizing that -3.4 eV corresponds to the n=2 level can save you time on calculations.

Step 1: Understanding the Concept:

Plane polarized light is light in which the electric field vibrations are confined to a single plane. Unpolarized light has vibrations in all possible planes perpendicular to the direction of propagation. A Polaroid is an optical filter that has a "transmission axis" and only allows the component of light polarized parallel to this axis to pass through. This property can be used to analyze an unknown light beam.


Step 2: Key Formula or Approach:

The detection method is based on Malus's Law, which describes the intensity of light transmitted through a Polaroid (acting as an analyzer): \[ I = I_{max} \cos^2\theta \]
where \(I\) is the transmitted intensity, \(I_{max}\) is the maximum transmitted intensity (when the analyzer's axis is aligned with the light's polarization), and \(\theta\) is the angle between the analyzer's transmission axis and the plane of polarization of the incident light.


Step 3: Detailed Explanation:

The procedure to detect plane polarized light is as follows:


Setup: Place a Polaroid, which will serve as an analyzer, in the path of the light beam being tested.
Action: Rotate the Polaroid analyzer slowly through a full 360\(^\circ\) rotation about the axis of the light beam.
Observation: Observe the intensity of the light transmitted through the Polaroid.
Conclusion:

If the light is plane polarized, the intensity of the transmitted light will vary significantly during the rotation. It will change from a maximum value to a minimum value of zero, twice in one full rotation. The intensity is maximum when the Polaroid's axis is parallel to the light's polarization plane (\(\theta = 0^\circ\)) and zero when it is perpendicular (\(\theta = 90^\circ\)).
If the light is unpolarized, the intensity of the transmitted light will remain constant as the Polaroid is rotated.
If the light is partially polarized, the intensity will vary between a maximum and a non-zero minimum during rotation.



Step 4: Final Answer:

To detect plane polarized light, one must rotate a Polaroid in its path. If the transmitted light intensity varies and becomes zero at two positions during one complete rotation, the incident light is confirmed to be plane polarized.
Quick Tip: The key to this detection method is the \textbf{rotation} of the analyzer. The signature of plane polarized light is the complete extinction (intensity becomes zero) at two specific orientations of the analyzer, 180\(^\circ\) apart.

Step 1: Understanding the Concept:

An electron revolving in a circular orbit is equivalent to a tiny current loop. According to Ampere's law, any current loop produces a magnetic field and has an associated magnetic dipole moment. The magnitude of this magnetic moment depends on the equivalent current and the area of the loop.


Step 2: Key Formula or Approach:

1. The magnetic dipole moment (\(\mu\)) of a current loop is given by \(\mu = iA\), where \(i\) is the current and \(A\) is the area of the loop.

2. The equivalent current \(i\) produced by an electron (charge \(e\)) revolving with a time period \(T\) is \(i = \frac{e}{T}\).

3. The time period of revolution is the circumference of the orbit divided by the speed: \(T = \frac{2\pi r}{v}\).

Combining these formulas, we get an expression for the magnetic moment: \[ \mu = iA = \left(\frac{e}{T}\right)(\pi r^2) = \left(\frac{e}{2\pi r/v}\right)(\pi r^2) = \frac{evr}{2} \]

Step 3: Detailed Explanation:

We are given the following values:

Speed of the electron, \(v = 2 \times 10^6\) m/s.
Radius of the orbit, \(r = 5.0 \times 10^{-11}\) m.
Charge of an electron, \(e = 1.6 \times 10^{-19}\) C.

Now, we substitute these values into the derived formula for magnetic moment: \[ \mu = \frac{evr}{2} \] \[ \mu = \frac{(1.6 \times 10^{-19} C) \times (2 \times 10^6 m/s) \times (5.0 \times 10^{-11} m)}{2} \] \[ \mu = \frac{1.6 \times 2 \times 5.0}{2} \times 10^{-19 + 6 - 11} A·m^2 \] \[ \mu = 8.0 \times 10^{-24} A·m^2 \]

Step 4: Final Answer:

The magnetic moment of the rotating electron is \(8.0 \times 10^{-24}\) A·m\(^2\).
Quick Tip: The formula \(\mu = \frac{evr}{2}\) is very useful. It can also be related to the orbital angular momentum \(L = mvr\). The ratio \(\frac{\mu}{L} = \frac{evr/2}{mvr} = \frac{e}{2m}\) is called the gyromagnetic ratio, a fundamental constant for an electron's orbital motion.

Step 1: Understanding the Concept:

A parallel plate capacitor consists of two large, flat, parallel conducting plates, each with an area A, separated by a small distance d. Capacitance (C) is a measure of its ability to store electric charge, defined as the ratio of the charge (Q) on one plate to the potential difference (V) between the plates.


Step 2: Derivation of the Expression:

Let the two plates of the capacitor have a surface charge density of \(+\sigma\) and \(-\sigma\). The charge on each plate is \(Q = \sigma A\).

Assuming the distance d is small compared to the dimensions of the plates, the electric field (\(E\)) between them is uniform and is given by: \[ E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0} \]
where \(\epsilon_0\) is the permittivity of free space.

The potential difference (\(V\)) between the plates is the work done in moving a unit charge from one plate to another, which is \(V = E \times d\). \[ V = \left(\frac{Q}{A\epsilon_0}\right)d = \frac{Qd}{A\epsilon_0} \]
The capacitance C is defined as \(C = \frac{Q}{V}\). Substituting the expression for V: \[ C = \frac{Q}{\frac{Qd}{A\epsilon_0}} = \frac{A\epsilon_0}{d} \]
This is the expression for the capacitance of a parallel plate capacitor with a vacuum (or air) between the plates.

If a dielectric material with a dielectric constant K is inserted between the plates, the capacitance becomes: \[ C = \frac{KA\epsilon_0}{d} \]

Step 3: How to Increase the Capacity:

Based on the formula \(C = \frac{KA\epsilon_0}{d}\), the capacitance can be increased in the following ways:

By increasing the area of the plates (A): Capacitance is directly proportional to the area of the plates.
By decreasing the distance between the plates (d): Capacitance is inversely proportional to the separation between the plates.
By introducing a dielectric medium: Inserting a dielectric material with a high dielectric constant (K > 1) between the plates increases the capacitance by a factor of K. Quick Tip: Remember the three factors affecting capacitance: Area, Distance, and Medium. An easy way to recall this is the formula \(C = \frac{KA\epsilon_0}{d}\). To increase C, you increase the numerator factors (K, A) or decrease the denominator factor (d).

Step 1: Understanding the Concept:

This problem describes the phenomenon of motional electromotive force (EMF). When a conductor moves through a magnetic field, an EMF is induced across its ends. Here, the axle of the train acts as the conductor, moving horizontally, and it cuts through the vertical component of the Earth's magnetic field. This induced EMF will be measured by the milivoltmeter.


Step 2: Key Formula or Approach:

The motional EMF (\(\epsilon\)) induced in a conductor of length \(l\) moving with velocity \(v\) perpendicular to a magnetic field \(B\) is given by: \[ \epsilon = B_v l v \]
where \(B_v\) is the component of the magnetic field perpendicular to both the length and the velocity. In this case, it is the vertical component of the Earth's magnetic field.


Step 3: Detailed Explanation:

First, we need to ensure all units are in the SI system.

Velocity (v): The speed of the train is given as 180 km/hour.
\[ v = 180 \, \frac{km}{hour} \times \frac{1000 \, m}{1 \, km} \times \frac{1 \, hour}{3600 \, s} = 50 \, m/s \]
Magnetic Field (\(B_v\)): The vertical component is \(B_v = 0.2 \times 10^{-4}\) weber/m\(^2\) = \(0.2 \times 10^{-4}\) Tesla (T).
Length (l): The distance between the tracks, which is the length of the conductor (axle), is \(l = 1\) m.

Now, we can calculate the induced EMF: \[ \epsilon = (0.2 \times 10^{-4} \, T) \times (1 \, m) \times (50 \, m/s) \] \[ \epsilon = (0.2 \times 50) \times 10^{-4} \, V \] \[ \epsilon = 10 \times 10^{-4} \, V = 1.0 \times 10^{-3} \, V \]
Since \(1 millivolt (mV) = 10^{-3} V\), the induced EMF is: \[ \epsilon = 1.0 \, mV \]

Step 4: Final Answer:

The reading in the milivoltmeter will be 1.0 mV.
Quick Tip: In motional EMF problems (\(\epsilon = Blv\)), always ensure that B, l, and v are mutually perpendicular. If they are not, you must use the appropriate components. In this problem, the train's velocity is horizontal, the axle is horizontal, and the vertical component of the magnetic field is used, fulfilling the perpendicularity condition.

Meaning of Wavefront:

A wavefront is defined as the locus of all points in a medium that are vibrating in the same phase. For example, if you drop a stone in a pond, the ripples spread out in circles; each circle represents a wavefront where all particles are at the same stage of their oscillation.

For a point source, the wavefronts are spherical.
For a line source, the wavefronts are cylindrical.
At a large distance from a source, a small portion of a spherical or cylindrical wavefront can be considered a plane wavefront.

The direction of wave propagation is always perpendicular to the wavefront.


Explanation of Reflection using Huygen's Theory:

Huygen's principle states that every point on a wavefront acts as a source of secondary spherical wavelets. The new wavefront at any later time is the forward envelope (tangent) of these secondary wavelets. Let's use this to prove the law of reflection (\(i=r\)).


Explanation:

Let XY be a reflecting surface and AB be an incident plane wavefront making an angle of incidence \(i\) with the surface.
At time \(t=0\), the wavefront touches the surface at point A. According to Huygen's principle, A becomes a source of secondary wavelets.
The point B on the wavefront will travel a distance BC to reach the surface at point C. Let the time taken be \(t\). Then, \(BC = vt\), where \(v\) is the speed of light.
In the same time \(t\), the secondary wavelet from A will spread out in the same medium as a hemisphere of radius \(AD = vt\).
The reflected wavefront is the common tangent CD drawn from point C to the wavelet originating from A. The angle of reflection is \(r\).

Proof of Law of Reflection:
Consider the two right-angled triangles, \(\triangle ABC\) and \(\triangle ADC\).

The side AC is common to both triangles.
The side \(BC = vt\) and the radius of the wavelet \(AD = vt\). Thus, \(BC = AD\).
Both are right-angled triangles (\(\angle B = \angle D = 90^\circ\)).

By the Right-angle-Hypotenuse-Side (RHS) congruence criterion, \(\triangle ABC \cong \triangle ADC\).
Therefore, the corresponding angles must be equal. \[ \angle BAC = \angle DCA \]
From the geometry of the diagram, \(\angle BAC = i\) (angle of incidence) and \(\angle DCA = r\) (angle of reflection). \[ i = r \]
This proves the law of reflection.
Quick Tip: When drawing the diagram for Huygen's proof of reflection or refraction, the key is to correctly identify the distances traveled by the wavelets in the given time. For reflection, the speed is the same, so \(BC = AD\). For refraction, the speeds are different, so the radii of the wavelets will be different.

Step 1: Understanding the Concept:

The resistance of a p-n junction diode in forward bias mentioned here is the dynamic or AC resistance (\(R_f\)). It is not the static resistance (\(V/I\)), but rather the ratio of a small change in voltage across the diode to the resulting small change in the current flowing through it.


Step 2: Key Formula or Approach:

The dynamic forward resistance (\(R_f\)) is defined by the relation: \[ R_f = \frac{\Delta V_f}{\Delta I_f} \]
where \(\Delta V_f\) is the change in forward bias voltage and \(\Delta I_f\) is the corresponding change in forward current. We need to find \(\Delta V_f\).


Step 3: Detailed Explanation:

We are given the following values:

Forward bias resistance, \(R_f = 25 \, \Omega\).
Change in current, \(\Delta I_f = 2 \, mA\). We must convert this to Amperes:
\[ \Delta I_f = 2 \times 10^{-3} \, A \]

Rearranging the formula to solve for the change in voltage \(\Delta V_f\): \[ \Delta V_f = R_f \times \Delta I_f \]
Substituting the given values: \[ \Delta V_f = (25 \, \Omega) \times (2 \times 10^{-3} \, A) \] \[ \Delta V_f = 50 \times 10^{-3} \, V \] \[ \Delta V_f = 0.05 \, V \]

Step 4: Final Answer:

The forward bias voltage should be changed by 0.05 Volts.
Quick Tip: For diode problems, always be careful to distinguish between static resistance (\(R = V/I\)) and dynamic resistance (\(R = \Delta V / \Delta I\)). Dynamic resistance is relevant for small signal AC analysis, as in this question. Also, remember to convert units like mA and k\(\Omega\) to base SI units (A and \(\Omega\)) before calculation.

Meaning of Wattless Current:

Wattless current is the component of an alternating current in a circuit that does not contribute to the average power consumed over a full cycle. The average power in an AC circuit is given by \(P_{avg} = V_{rms} I_{rms} \cos\phi\), where \(\phi\) is the phase angle between the voltage and current.
If the phase angle \(\phi\) is 90\(^\circ\) (as in a purely inductive or purely capacitive circuit), then \(\cos\phi = \cos(90^\circ) = 0\). This makes the average power consumed zero. The current that flows in such a circuit is called wattless current because it flows without any net dissipation of power. It corresponds to the component of current that is in quadrature (90\(^\circ\) out of phase) with the voltage.


Calculations:


Step 1: Understanding the Concept and Formulas:

We have a purely capacitive AC circuit. We need to find the capacitive reactance (\(X_C\)), which is the opposition to the current flow, and then use Ohm's law for AC circuits to find the RMS current (\(I_{rms}\)).

Capacitive Reactance: \(X_C = \frac{1}{2\pi f C}\)
RMS Current: \(I_{rms} = \frac{V_{rms}}{X_C}\)


Step 2: Detailed Explanation:

We are given:

Capacitance, \(C = 15 \, \muF = 15 \times 10^{-6} \, F\).
RMS Voltage, \(V_{rms} = 220 \, V\).
Frequency, \(f = 50 \, Hz\).

Part 1: Calculate the Reactance (\(X_C\)) of the circuit \[ X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi (50)(15 \times 10^{-6})} \] \[ X_C = \frac{1}{100\pi \times 15 \times 10^{-6}} = \frac{1}{1500\pi \times 10^{-6}} = \frac{10^6}{1500\pi} \]
Using \(\pi \approx 3.14159\): \[ X_C \approx \frac{1,000,000}{1500 \times 3.14159} \approx \frac{1,000,000}{4712.385} \approx 212.2 \, \Omega \]
Part 2: Calculate the RMS value of AC current (\(I_{rms}\)) \[ I_{rms} = \frac{V_{rms}}{X_C} = \frac{220 \, V}{212.2 \, \Omega} \] \[ I_{rms} \approx 1.037 \, A \]

Step 3: Final Answer:

The reactance of the circuit is approximately 212.2 \(\Omega\), and the rms value of the AC current is approximately 1.04 A.
Quick Tip: Remember the mnemonics for phase in AC circuits: "ELI the ICE man". For an inductor (L), EMF (E) leads Current (I). For a capacitor (C), Current (I) leads EMF (E). This helps you remember that for a capacitor, current leads voltage by 90\(^\circ\), which is why the current is wattless in a purely capacitive circuit.

Step 1: Principle of an AC Generator:

The working principle of an AC generator is electromagnetic induction. It states that whenever the magnetic flux linked with a coil changes, an electromotive force (EMF) is induced in the coil. If the coil is part of a closed circuit, this induced EMF drives a current through it.


Step 2: Construction:

An AC generator consists of the following main parts:

Armature Coil (ABCD): A rectangular coil with a large number of turns of insulated copper wire wound on a soft iron core.
Field Magnets (N, S): Strong permanent magnets or electromagnets that produce a uniform magnetic field.
Slip Rings (\(R_1, R_2\)): Two hollow metallic rings to which the ends of the armature coil are connected. They rotate along with the coil.
Brushes (\(B_1, B_2\)): Two flexible carbon brushes that are kept in light contact with the rotating slip rings. They are used to pass the current from the coil to the external circuit.


Step 3: Working:


The armature coil is rotated mechanically in the magnetic field. As it rotates, the angle between the plane of the coil and the magnetic field lines changes continuously.
This continuous change in orientation causes the magnetic flux (\(\Phi = NBA\cos\theta\)) linked with the coil to change.
According to Faraday's law of induction, this change in flux induces an EMF in the coil. The direction of the induced current is given by Fleming's Right-Hand Rule.
First Half Rotation: As the coil rotates, let's say arm AB moves downwards and arm CD moves upwards. The induced current flows from A to B and from C to D. The current flows out through brush \(B_1\) and into the circuit, returning through brush \(B_2\).
Second Half Rotation: Arm AB now moves upwards, and arm CD moves downwards. The direction of the induced current in the arms reverses (now from B to A and D to C). The current now flows out through brush \(B_2\) and returns through brush \(B_1\).
Since the direction of the current in the external circuit reverses after every half rotation, an alternating current is produced. The magnitude of the induced EMF is sinusoidal, given by \(\epsilon = \epsilon_0 \sin(\omega t)\). Quick Tip: The key difference between an AC and a DC generator lies in how the current is extracted. AC generators use two separate slip rings, while DC generators use a single split-ring commutator to reverse the connection every half-turn, keeping the current in the external circuit flowing in one direction.

Step 1: Understanding the Concept and Variables:

We are considering a real image formed by a convex lens on a screen.

Let \(u\) be the magnitude of the object distance.
Let \(v\) be the magnitude of the image distance.
The total distance between the object and the screen is \(x\), so \(x = u + v\).
The magnification is \(m\). For a real, inverted image, the magnification is negative, but here \(m\) is treated as its magnitude, so \(m = \frac{v}{u}\).


Step 2: Key Formula or Approach:

The thin lens formula connects the object distance, image distance, and focal length. For a real image formed by a convex lens, using sign convention (\(u\) is negative, \(v\) is positive), the formula \(\frac{1}{v} - \frac{1}{(-u)} = \frac{1}{f}\) becomes: \[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \]
where \(u\) and \(v\) are magnitudes.


Step 3: Detailed Explanation (Derivation):

From the magnification formula, we have \(v = mu\).

Substitute this into the distance equation \(x = u + v\): \[ x = u + mu = u(1+m) \]
From this, we can express the object distance \(u\) in terms of \(x\) and \(m\): \[ u = \frac{x}{m+1} \]
Now, we can find the image distance \(v\): \[ v = mu = m\left(\frac{x}{m+1}\right) = \frac{mx}{m+1} \]
Now, substitute the expressions for \(u\) and \(v\) into the lens formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{\frac{mx}{m+1}} + \frac{1}{\frac{x}{m+1}} \] \[ \frac{1}{f} = \frac{m+1}{mx} + \frac{m+1}{x} \]
Take \(\frac{m+1}{x}\) as a common factor: \[ \frac{1}{f} = \frac{m+1}{x} \left(\frac{1}{m} + 1\right) \] \[ \frac{1}{f} = \frac{m+1}{x} \left(\frac{1+m}{m}\right) = \frac{(m+1)^2}{mx} \]
Finally, inverting the expression to find \(f\): \[ f = \frac{mx}{(m+1)^2} \]
Hence, the relation is proved.
Quick Tip: This is a standard result in optics. A key step in such derivations is to express the object distance (\(u\)) and image distance (\(v\)) in terms of the given variables (\(x\) and \(m\)) before substituting them into the lens formula.

Meaning of Polarization of Light:

Light is a transverse electromagnetic wave, which means the oscillations of its electric field vector occur in a plane perpendicular to the direction of wave propagation. Ordinary (unpolarized) light consists of waves with electric field vectors vibrating in all possible directions in this plane.
Polarization is the phenomenon of restricting these vibrations of the electric field vector to a single direction (or plane). The resulting light is called plane-polarized or linearly polarized light.


Method to Obtain Plane Polarized Light by Reflection:

Plane polarized light can be obtained from unpolarized light using the phenomenon of reflection, based on a principle discovered by David Brewster.

Principle (Brewster's Law): When unpolarized light is incident on the boundary between two transparent media (e.g., air and glass), the reflected light becomes completely plane-polarized if the angle of incidence is equal to a specific angle called the polarizing angle or Brewster's angle (\(i_p\)).


Description of the Method:

An unpolarized light beam is directed towards a smooth, transparent surface, such as a glass slab.
The angle of incidence (\(i\)) is adjusted by rotating the slab.
At a specific angle of incidence, \(i = i_p\), it is observed that the reflected light is completely plane-polarized. The vibrations of the electric field vector in this reflected polarized light are perpendicular to the plane of incidence (the plane containing the incident ray, reflected ray, and the normal).
At this polarizing angle, the reflected ray and the refracted ray are found to be perpendicular to each other.
Brewster's Law gives the relationship between the polarizing angle and the refractive index (\(n\)) of the medium:
\[ n = \tan(i_p) \]

Therefore, by making unpolarized light incident on a transparent surface at Brewster's angle, we can obtain a completely plane-polarized reflected beam.
Quick Tip: To remember the orientation of polarization by reflection, think of the electric field vibrations as arrows. The vibrations parallel to the surface are more likely to be transmitted/refracted, while those perpendicular to the plane of incidence are more easily reflected. This is why sunglasses with Polaroid filters are effective at reducing glare from horizontal surfaces like water or roads.

N/A Quick Tip: For the dipole problem, remember that work done equals the change in potential energy (\(W = \Delta U\)). Since \(U = -pE\cos\theta\), the work done from \(0\) to \(\theta\) is \(U(\theta) - U(0) = -pE\cos\theta - (-pE\cos 0) = pE(1-\cos\theta)\).
For the current problem, the concept is simple: current is charge flow per second. Find how many times the electron circles per second (the frequency \(f = v/2\pi r\)) and multiply by its charge (\(I = ef\)).

Gustav Kirchhoff formulated two fundamental laws that are essential for analyzing complex electrical circuits.


1. Kirchhoff's First Law (Kirchhoff's Current Law - KCL or the Junction Rule):


Statement: The algebraic sum of the electric currents meeting at any junction (or node) in an electrical circuit is zero.
Mathematical Form:
\[ \sum_{junction} I = 0 \]
Explanation: This law is a direct consequence of the law of conservation of electric charge. Charge cannot be created, destroyed, or accumulated at a junction. Therefore, the total rate at which charge enters a junction must be equal to the total rate at which charge leaves it.
Sign Convention: Currents entering a junction are typically taken as positive, while currents leaving the junction are taken as negative (or vice versa, as long as the convention is consistent).


2. Kirchhoff's Second Law (Kirchhoff's Voltage Law - KVL or the Loop Rule):


Statement: The algebraic sum of the changes in potential (or voltage drops and EMFs) around any closed loop or mesh in an electrical circuit is zero.
Mathematical Form:
\[ \sum_{closed loop} \Delta V = 0 \]
Explanation: This law is based on the law of conservation of energy. The electric force is a conservative force. This means that if an electric charge is moved around any closed path and returns to its starting point, the net work done on it is zero. Consequently, its net change in potential energy, and therefore electric potential, is also zero.
Sign Convention: When traversing a loop, the potential difference across a resistor is taken as negative if moving in the direction of the current and positive if moving against it. The EMF of a source is taken as positive if moving from the negative to the positive terminal and negative if moving from positive to negative. Quick Tip: A simple way to remember the basis of each law: \textbf{KCL (Junctions):} Conservation of \textbf{C}harge. \textbf{KVL (Loops):} Conservation of \textbf{E}nergy (Voltage is potential energy per unit charge). Always apply a consistent sign convention when using KVL to avoid errors.

Step 1: Understanding the Concept:

Rectification is the process of converting an alternating current (AC) into a direct current (DC). A half-wave rectifier uses a single p-n junction diode to achieve this. The basic principle is that a p-n diode allows current to flow in only one direction—when it is forward-biased—and blocks current flow in the opposite direction—when it is reverse-biased.


Step 2: Simple Circuit Diagram:

The circuit consists of an AC input source (usually connected via a step-down transformer), a p-n junction diode (D), and a load resistor (\(R_L\)) across which the DC output is obtained.

Circuit Diagram: An AC voltage source is connected to the primary coil of a transformer. The secondary coil is connected in series with a p-n diode (D) and a load resistor (\(R_L\)). The output voltage is measured across \(R_L\).


Step 3: Working:

The working can be explained by considering the two half-cycles of the input AC voltage.

During the Positive Half-Cycle of AC Input: The upper end of the transformer's secondary coil becomes positive with respect to the lower end. This makes the p-side of the diode positive relative to the n-side, putting the diode in forward bias. The diode conducts current, and a current flows through the load resistor \(R_L\). Consequently, a voltage (output) is developed across \(R_L\), which follows the shape of the positive half-cycle of the input.
During the Negative Half-Cycle of AC Input: The polarity of the secondary coil reverses. The upper end becomes negative with respect to the lower end. This makes the p-side of the diode negative relative to the n-side, putting the diode in reverse bias. In this state, the diode offers very high resistance and (ideally) does not conduct any current. Therefore, no current flows through \(R_L\), and the output voltage across it is zero.

This process repeats for every cycle, resulting in an output voltage that consists only of the positive half-cycles of the original AC input. This output is a pulsating, unidirectional (DC) voltage.


\hrule


Solution for the OR Part (Distance of Closest Approach):


Step 1: Understanding the Concept:

The proof is based on the principle of conservation of energy. An \(\alpha\)-particle (charge \(q_1 = +2e\)) is accelerated by a potential difference V, gaining kinetic energy. As it approaches a positively charged nucleus (charge \(q_2 = +Ze\)), the electrostatic repulsive force slows it down. The distance of closest approach (\(r\)) is the point where the \(\alpha\)-particle momentarily stops and turns back. At this point, all of its initial kinetic energy has been converted into electrostatic potential energy of the system.


Step 2: Key Formula or Approach:

1. Kinetic Energy (KE) gained by the \(\alpha\)-particle: \(KE = q_1 V = (2e)V\)

2. Electrostatic Potential Energy (PE) at separation \(r\): \(PE = \frac{1{4\pi\epsilon_0} \frac{q_1 q_2}{r} = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r}\)

By conservation of energy: \(KE_{initial} = PE_{final}\)


Step 3: Detailed Explanation (Derivation):

Equating the initial kinetic energy with the final potential energy: \[ (2e)V = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{r} \]
We can cancel \(2e\) from both sides: \[ V = \frac{1}{4\pi\epsilon_0} \frac{Ze}{r} \]
Now, we solve for the distance of closest approach, \(r\): \[ r = \left(\frac{1}{4\pi\epsilon_0}\right) \frac{Ze}{V} \]
To get the numerical value, we substitute the known constants:

Coulomb's constant, \(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\) N·m\(^2\)/C\(^2\).
Elementary charge, \(e = 1.6 \times 10^{-19}\) C.
\[ r = \frac{(9 \times 10^9) \times Z \times (1.6 \times 10^{-19})}{V} \] \[ r = \frac{14.4 \times 10^{-10} \times Z}{V} \quad (in meters) \]
Since 1 Angstrom (\(\AA\)) = \(10^{-10}\) meters, we can express the distance in Angstroms: \[ r = \frac{14.4 \times Z}{V} \times 10^{-10} m = 14.4 \frac{Z}{V} \AA \]
This proves the required relation.
Quick Tip: For the rectifier, remember that the key is the diode's one-way-street behavior for current. The output is "half" of the input wave. For the alpha particle problem, energy conservation is the central idea. The conversion of kinetic energy to potential energy is a common theme in collision and scattering problems in both mechanics and electromagnetism.

What is the Photoelectric Effect?

The photoelectric effect is the phenomenon of emission of electrons from a substance (usually a metal) when electromagnetic radiation, such as light, of a sufficiently high frequency is incident upon it. The emitted electrons are called photoelectrons.


Laws of Photoelectric Emission:
Based on experimental observations, the laws are:

For a given material, there exists a certain minimum frequency of incident radiation, called the threshold frequency (\(\nu_0\)), below which no photoelectric emission occurs, no matter how intense the light is.
The photoelectric emission is an instantaneous process. The time lag between the incidence of radiation and the emission of photoelectrons is very small (less than \(10^{-9}\) s).
The number of photoelectrons emitted per second (which determines the photoelectric current) is directly proportional to the intensity of the incident radiation, provided the frequency is above the threshold frequency.
The maximum kinetic energy of the emitted photoelectrons is independent of the intensity of the incident light but depends linearly on its frequency.


Einstein's Photoelectric Equation:

Step 1: Understanding the Concept (Einstein's Postulate):

To explain the photoelectric effect, Einstein proposed that light is quantized and consists of discrete packets of energy called photons. The energy of each photon is \(E = h\nu\), where \(h\) is Planck's constant and \(\nu\) is the frequency of light.


Step 2: Derivation:

When a photon of energy \(h\nu\) strikes the metal surface, its energy is used in two parts:

A portion of the energy is used to overcome the surface barrier and free the electron from the metal. This minimum energy required is called the work function (\(\Phi_0\)) of the metal.
The rest of the photon's energy is imparted to the emitted electron as its maximum kinetic energy (\(K_{max}\)).

By the law of conservation of energy: \[ Energy of Photon = Work Function + Max. Kinetic Energy of Electron \] \[ h\nu = \Phi_0 + K_{max} \]
The work function can be expressed in terms of the threshold frequency (\(\nu_0\)) as \(\Phi_0 = h\nu_0\). Substituting this: \[ h\nu = h\nu_0 + K_{max} \]
Rearranging the equation to solve for the kinetic energy: \[ K_{max} = h\nu - h\nu_0 = h(\nu - \nu_0) \]
Since \(K_{max} = \frac{1}{2}m v_{max}^2\), where \(m\) is the mass of the electron and \(v_{max}\) is its maximum velocity, the equation becomes: \[ \frac{1}{2}mv_{max}^2 = h(\nu - \nu_0) \]
This is Einstein's photoelectric equation.


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Solution for the OR Part (Force between Parallel Wires):


Step 1: Understanding the Concept:

Two long, parallel wires carrying currents exert a magnetic force on each other. The magnetic field produced by one wire interacts with the current in the other wire. When the currents are in the same direction, the force is attractive. The magnitude of this force per unit length is given by a standard formula.


Step 2: Key Formula or Approach:

The force per unit length (\(F/L\)) between two parallel wires carrying currents \(I_1\) and \(I_2\) and separated by a distance \(d\) is: \[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \]
In this problem, the currents are the same, so \(I_1 = I_2 = I\). The formula becomes: \[ \frac{F}{L} = \frac{\mu_0 I^2}{2\pi d} \]

Step 3: Detailed Explanation (Calculation):

We are given the following values:

Force per unit length, \(F/L = 3 \times 10^{-3}\) N/m.
Distance between wires, \(d = 0.06\) m.
Permeability of free space, \(\mu_0 = 4\pi \times 10^{-7}\) T·m/A.

Substitute these values into the formula and solve for \(I\): \[ 3 \times 10^{-3} = \frac{(4\pi \times 10^{-7}) I^2}{2\pi (0.06)} \]
Cancel \(2\pi\) from the numerator and denominator: \[ 3 \times 10^{-3} = \frac{2 \times 10^{-7} \times I^2}{0.06} \]
Rearrange to solve for \(I^2\): \[ I^2 = \frac{(3 \times 10^{-3}) \times 0.06}{2 \times 10^{-7}} \] \[ I^2 = \frac{0.18 \times 10^{-3}}{2 \times 10^{-7}} = 0.09 \times 10^4 = 900 \]
Take the square root to find the current \(I\): \[ I = \sqrt{900} = 30 \, A \]

Step 4: Final Answer:

The value of the current flowing in each wire is 30 A.
Quick Tip: For the photoelectric effect, remember the energy balance: \(E_{photon} = W_{eject} + KE_{electron}\). For the parallel wires, a helpful way to remember the force direction is "like currents attract, unlike currents repel," which is opposite to the rule for static charges.

Step 1: Understanding the Concept:

In Young's double-slit experiment (YDSE), the intensity of light at any point on the screen is a result of the interference of waves from the two slits. The intensity depends on the phase difference (\(\phi\)) between the two waves, which is directly related to their path difference (\(\Delta x\)).


Step 2: Key Formula or Approach:

The relationship between phase difference (\(\phi\)) and path difference (\(\Delta x\)) is: \[ \phi = \frac{2\pi}{\lambda} \Delta x \]
The resultant intensity (\(I\)) at a point is given by: \[ I = I_{max} \cos^2\left(\frac{\phi}{2}\right) \]
where \(I_{max}\) is the maximum possible intensity (at the central maximum).


Step 3: Detailed Explanation (Calculation):

Case 1: Given Information

The path difference is given as \(\Delta x_1 = \lambda\).
First, find the corresponding phase difference \(\phi_1\): \[ \phi_1 = \frac{2\pi}{\lambda} (\lambda) = 2\pi radians \]
Now, use the intensity formula. The intensity at this point (\(I_1\)) is given as K units. \[ I_1 = I_{max} \cos^2\left(\frac{2\pi}{2}\right) = I_{max} \cos^2(\pi) = I_{max}(-1)^2 = I_{max} \]
So, we have \(K = I_{max}\). The maximum intensity in the interference pattern is K units.


Case 2: Find the new intensity

The new path difference is given as \(\Delta x_2 = \frac{\lambda}{3}\).
Find the new phase difference \(\phi_2\): \[ \phi_2 = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3} radians \]
Now, calculate the new intensity (\(I_2\)) using the intensity formula: \[ I_2 = I_{max} \cos^2\left(\frac{\phi_2}{2}\right) = I_{max} \cos^2\left(\frac{2\pi/3}{2}\right) = I_{max} \cos^2\left(\frac{\pi}{3}\right) \]
We know that \(\cos\left(\frac{\pi}{3}\right) = \cos(60^\circ) = \frac{1}{2}\). \[ I_2 = I_{max} \left(\frac{1}{2}\right)^2 = \frac{I_{max}}{4} \]
Since we found that \(I_{max} = K\), the new intensity is: \[ I_2 = \frac{K}{4} \]

Step 4: Final Answer:

The intensity of light at the point where the path difference is \(\frac{\lambda}{3}\) is \(\frac{K}{4}\) units.


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Solution for the OR Part (Single Slit Diffraction):


Elucidation of Diffraction:

Diffraction is the phenomenon of the bending or spreading of waves as they pass through a narrow opening (aperture) or around an obstacle. When a monochromatic plane wavefront is incident on a narrow slit, each point on the portion of the wavefront that passes through the slit acts as a source of secondary spherical wavelets, according to Huygens' principle. These wavelets spread out in all directions and interfere with one another. This interference results in a characteristic diffraction pattern on a screen placed far from the slit. The pattern consists of a bright central maximum, which is flanked on both sides by a series of alternating dark bands (minima) and weaker bright bands (secondary maxima).


Expression for Angular Width of Central Maximum:

Step 1: Condition for the First Minimum

Consider a slit of width 'a'. The central maximum is spread between the first minima on either side of the center. To find the location of the first minimum, we consider the wavelets from the top and bottom edges of the slit. The first minimum occurs at an angle \(\theta\) where the path difference between these two extreme rays is exactly one wavelength (\(\lambda\)). \[ Path Difference = a \sin\theta \]
For the first minimum, this path difference is equal to \(\lambda\): \[ a \sin\theta = \lambda \]

Step 2: Derivation

From the condition for the first minimum, we have \(\sin\theta = \frac{\lambda}{a}\).

For typical diffraction setups, the angle \(\theta\) is very small, so we can use the small-angle approximation: \(\sin\theta \approx \theta\) (where \(\theta\) is in radians). \[ \theta = \frac{\lambda}{a} \]
This angle \(\theta\) represents the angular position of the first minimum from the center. It is also the angular half-width of the central maximum.

The total angular width of the central maximum is the angle between the first minimum on one side of the center and the first minimum on the other side. \[ Total Angular Width = \theta + \theta = 2\theta \]
Substituting the expression for \(\theta\): \[ Total Angular Width = \frac{2\lambda}{a} \]

Step 3: Final Answer:

The expression for the angular width of the central maximum in single-slit diffraction is \(\frac{2\lambda}{a}\), where \(\lambda\) is the wavelength of the light and 'a' is the width of the slit.
Quick Tip: For YDSE, the intensity is maximum when path difference is \(n\lambda\) and zero when it's \((n+1/2)\lambda\). For single-slit diffraction, the intensity is zero (minimum) when path difference is \(n\lambda\) (from slit edges). Don't confuse these conditions! The central maximum in diffraction is twice as wide as the secondary maxima.

Rutherford's Atomic Model (Nuclear Model):

Based on his famous alpha-particle scattering experiment, Ernest Rutherford proposed a model of the atom in 1911. Its main features are:

Nucleus: Almost all the mass and the entire positive charge of an atom are concentrated in a very small, dense region at the center called the nucleus.
Empty Space: Most of the atom is empty space.
Electrons: The negatively charged electrons revolve around the nucleus in circular paths called orbits, much like planets orbiting the sun. The electrostatic force of attraction between the nucleus and electrons provides the necessary centripetal force for their revolution.


Drawbacks of Rutherford's Model:
Rutherford's model was inconsistent with classical physics and experimental observations in two key ways:

Instability of the Atom: According to classical electromagnetic theory, an accelerating charged particle must radiate energy. An electron revolving in an orbit is constantly accelerating (due to the change in the direction of its velocity). Therefore, it should continuously lose energy and spiral into the nucleus, making the atom unstable. This contradicts the observed stability of atoms.
Inability to Explain Line Spectra: As the electron spirals inwards, its frequency of revolution would increase continuously. This means it should emit a continuous spectrum of radiation. However, atoms (like hydrogen) are observed to emit a discrete line spectrum.


How Bohr's Model Removed the Drawbacks:
Niels Bohr, in 1913, modified Rutherford's model by introducing quantum concepts through three postulates:

Postulate of Stationary Orbits: Bohr proposed that electrons can revolve only in certain specific, non-radiating orbits called stationary orbits. While in these orbits, electrons do not emit energy. This postulate directly contradicted classical theory and solved the problem of atomic instability.
Postulate of Quantization of Angular Momentum: The allowed stationary orbits are those for which the angular momentum of the electron is an integral multiple of \(h/2\pi\), where \(h\) is Planck's constant. (\(L = n \frac{h}{2\pi}\), where n=1, 2, 3...).
Postulate of Frequency Condition: An atom emits radiation (a photon) only when an electron jumps from a higher energy stationary orbit (\(E_2\)) to a lower energy one (\(E_1\)). The frequency (\(\nu\)) of the emitted photon is given by \(h\nu = E_2 - E_1\). Since only specific orbits and energy levels are allowed, only specific frequencies of light can be emitted, thus explaining the observed discrete line spectra.


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Solution for the OR Part (Classification of Nuclei):


Nuclei are classified based on the number of protons (atomic number, Z) and the number of neutrons (neutron number, N) they contain. The total number of protons and neutrons is the mass number (A = Z + N).


1. Isotopes

Definition: Nuclei of the same element that have the same number of protons (same Z) but a different number of neutrons (different N), and thus a different mass number (A).
Example: The isotopes of hydrogen are Protium (\(_{1}^{1}H\), Z=1, N=0), Deuterium (\(_{1}^{2}H\), Z=1, N=1), and Tritium (\(_{1}^{3}H\), Z=1, N=2). Another example is Carbon-12 (\(_{6}^{12}C\)) and Carbon-14 (\(_{6}^{14}C\)).
Characteristics: Isotopes have identical chemical properties because chemical behavior is determined by the electron configuration, which depends on Z. However, they have different physical properties such as mass, density, and nuclear stability (some isotopes may be radioactive while others are stable).


2. Isobars

Definition: Nuclei of different elements that have the same mass number (same A) but different atomic numbers (different Z) and neutron numbers (different N).
Example: Argon-40 (\(_{18}^{40}Ar\)) and Calcium-40 (\(_{20}^{40}Ca\)). Both have A=40, but Z and N are different. Another example is Carbon-14 (\(_{6}^{14}C\)) and Nitrogen-14 (\(_{7}^{14}N\)).
Characteristics: Isobars are atoms of different elements, so they have different chemical and physical properties.


3. Isotones

Definition: Nuclei of different elements that have the same number of neutrons (same N) but different atomic numbers (different Z) and mass numbers (different A).
Example: Chlorine-37 (\(_{17}^{37}Cl\)) and Potassium-39 (\(_{19}^{39}K\)). For both, the neutron number is N = A-Z = 20. Another example is Boron-12 (\(_{5}^{12}B\)) and Carbon-13 (\(_{6}^{13}C\)), both having N=7.
Characteristics: Isotones are atoms of different elements, and thus they have different chemical and physical properties. Quick Tip: For the Bohr model, remember that its genius was in defying classical physics with a few simple (but revolutionary) quantum rules to explain experimental facts. For classifying nuclei, use the letters in the names as a mnemonic: Isoto\textbf{p}es (same \textbf{P}rotons), Isob\textbf{a}rs (same \textbf{A} - mass number), Isoto\textbf{n}es (same \textbf{N}eutrons).