Step 1: Concept of Dielectric Constant.

The dielectric constant (or relative permittivity) of a material indicates its capacity to insulate electric charges from each other. Materials with high dielectric constants typically function as better insulators. However, metals exhibit a distinct behavior when it comes to their dielectric constants.

Step 2: Evaluation of Options.

- (A) \( k = 0 \): The dielectric constant of metals such as silver is zero, as metals contain free electrons that shield electric fields, preventing the material from polarizing in response to an external electric field.

- (B) \( k = +1 \): A dielectric constant of +1 is typical of vacuum or air, not silver.

- (C) \( k = -1 \): The dielectric constant cannot be negative for a stable, passive material. Hence, a negative value is unfeasible for silver or any other material in this context.

- (D) \( k = \infty \) (infinity): A dielectric constant of infinity is theoretical, often used to describe ideal dielectric materials that are able to polarize perfectly in an electric field, which does not apply to silver.


Step 3: Conclusion.

The correct value of the dielectric constant for silver is \( k = 0 \), as it is a metal and does not exhibit polarization in response to an electric field. Therefore, option (A) is the correct answer. Quick Tip: The dielectric constant of metals is effectively zero because metals have free electrons that screen electric fields.

Step 1: Recognizing the Transistor Configuration.

In a common-emitter amplifier setup, the input signal \( V_i \) is applied to the base of the N-P-N transistor, while the output voltage \( V_o \) is taken from the collector. The emitter is usually grounded.


Step 2: Circuit Analysis.

Based on the provided diagram, the input voltage \( V_i \) is applied to the base of the N-P-N transistor, and the output voltage \( V_o \) is taken from the collector. This matches the typical configuration of a common-emitter amplifier.


Step 3: Conclusion.

Thus, the configuration depicted in the diagram corresponds to a common-emitter amplifier, and option (B) is the correct choice. Quick Tip: In a common-emitter amplifier, the input is applied to the base, and the output is taken from the collector.

Step 1: Understanding the Relationship Between Refracting Angle and Deviation.

For a thin prism, the angle of deviation \( \delta \) is related to the refracting angle \( A \) and the refractive index \( n \) by the following equation: \[ \delta = (n - 1)A \]
If the refracting angle \( A \) is equal to the angle of deviation \( \delta \), we set \( \delta = A \), yielding: \[ A = (n - 1)A \]

Step 2: Solving for the Refractive Index.

By canceling \( A \) from both sides (assuming \( A \neq 0 \)), we obtain: \[ 1 = n - 1 \]
Solving for \( n \), we get: \[ n = 2 \]

Step 3: Conclusion.

The refractive index of the thin prism material is \( n = 2 \), which makes option (B) the correct answer. Quick Tip: The refractive index of a thin prism can be determined using the relationship between the refracting angle and the angle of deviation.

Step 1: Understanding the Equation.

The given equation represents a nuclear reaction in which two light nuclei combine to form a heavier nucleus, releasing energy. The equation is: \[ ^4_1H \rightarrow ^4_2He + 2 \beta^0 + Q \]
This reaction is characteristic of nuclear fusion, where two hydrogen nuclei (protons) fuse to form a helium nucleus, releasing energy and emitting a beta particle (\(\beta^0\)).

Step 2: Evaluating the Options.

- (A) Nuclear fission: Fission refers to the splitting of a heavy nucleus into lighter nuclei, which is not described by this equation.

- (B) Nuclear fusion: This is the correct option. The equation illustrates nuclear fusion, where lighter nuclei combine to form a heavier nucleus.

- (C) Nuclear disintegration: Disintegration involves the decay of a nucleus, whereas this reaction involves the fusion of nuclei, not decay.

- (D) None of these: This is incorrect, as nuclear fusion is the correct answer.


Step 3: Conclusion.

The equation depicts a nuclear fusion reaction, so option (B) is the correct answer. Quick Tip: Nuclear fusion involves the combination of lighter nuclei to form a heavier nucleus, releasing energy in the process.

Step 1: Understanding Fundamental Particles.

Fundamental particles are those that are not composed of smaller constituents. Examples include the electron, positron, and neutrino, which are all considered elementary particles. On the other hand, the \( \alpha \)-particle is not fundamental because it consists of two protons and two neutrons, making it a composite particle.

Step 2: Analysis of Options.

- (A) Positron: The positron is the antiparticle of the electron and is considered a fundamental particle.

- (B) Electron: The electron is an elementary particle and therefore fundamental.

- (C) \( \alpha \)-particle: This particle is made up of two protons and two neutrons, so it is a composite particle and not fundamental.

- (D) Neutrino: A neutrino is an elementary particle and thus fundamental.


Step 3: Conclusion.

The \( \alpha \)-particle is not a fundamental particle, so option (C) is the correct answer. Quick Tip: Fundamental particles are indivisible and not made up of smaller particles, such as electrons, neutrinos, and positrons.

Step 1: Understanding the RLC Series Circuit.

In a series RLC circuit, the phase relationship between current and voltage depends on the driving frequency \( n \) relative to the resonant frequency \( n_r \). The resonant frequency is given by: \[ n_r = \frac{1}{2 \pi \sqrt{LC}} \]
When the driving frequency is less than the resonant frequency, the circuit behaves as inductive, causing the current to lag behind the voltage.

Step 2: Analysis of Options.

- (A) \( n = 0 \): At zero frequency, there is no alternating current, so the concept of lagging does not apply meaningfully.

- (B) \( n < n_r \): At frequencies below resonance, inductive reactance dominates, leading to the current lagging behind the voltage. This is the correct option.

- (C) \( n = n_r \): At resonance, inductive and capacitive reactances cancel out, and the current is in phase with the voltage.

- (D) \( n > n_r \): At frequencies above resonance, capacitive reactance dominates, causing the current to lead the voltage rather than lag.


Step 3: Conclusion.

The current lags the voltage when the source frequency is less than the resonant frequency, so option (B) is the correct answer. Quick Tip: In an RLC circuit, when the driving frequency is less than the resonant frequency, the current lags behind the voltage.

1. Transverse Nature: Electromagnetic waves are transverse in nature, meaning that the oscillations of the electric and magnetic fields occur perpendicular to the direction in which the wave propagates. The electric field oscillates in one plane, while the magnetic field oscillates in a plane perpendicular to both the electric field and the direction of wave propagation. This transverse property is a fundamental characteristic of electromagnetic waves.

2. Propagation Through a Vacuum: Unlike mechanical waves, electromagnetic waves do not require a material medium for propagation. They can travel through the vacuum of space, which explains how sunlight reaches Earth through the emptiness of space. This ability to propagate without a medium makes electromagnetic waves essential for applications such as space communication and astronomical observations. Quick Tip: Electromagnetic waves can travel through space without any medium, unlike sound waves that need a medium to propagate.

The threshold wavelength refers to the maximum wavelength of light or electromagnetic radiation that can cause the emission of electrons from a material. This concept is particularly relevant in the context of the photoelectric effect. When light strikes the surface of a material, it can transfer energy to the electrons. However, electrons will only be emitted if the energy of the incident photons is equal to or greater than the material’s work function.

Since the energy of a photon is inversely proportional to its wavelength, shorter wavelengths correspond to higher-energy photons. The threshold wavelength represents the specific wavelength at which a photon's energy is just sufficient to overcome the work function of the material. If the incoming light has a longer wavelength than this threshold, its photons do not have enough energy to eject electrons from the material.

Mathematically, the threshold wavelength \( \lambda_{th} \) is related to the work function \( \phi \) by the equation: \[ \phi = \frac{hc}{\lambda_{th}} \]
where:
- \( \phi \) is the work function (in joules),
- \( h \) is Planck’s constant (\( 6.626 \times 10^{-34} \, J \cdot s \)),
- \( c \) is the speed of light in vacuum (\( 3 \times 10^8 \, m/s \)),
- \( \lambda_{th} \) is the threshold wavelength (in meters). Quick Tip: For the photoelectric effect, only photons with a wavelength shorter than the threshold wavelength can eject electrons from a material.

When a convex lens with a refractive index \( n_1 = \frac{3}{2} \) is immersed in a liquid with a refractive index \( n_2 = \frac{3}{2} \), the refractive index difference between the lens and the surrounding medium becomes zero.

The focal length \( f \) of a lens is given by the formula: \[ \frac{1}{f} = (n_1 - n_2) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \( n_1 \) is the refractive index of the lens and \( n_2 \) is the refractive index of the surrounding medium.

Since \( n_1 = n_2 \), the refractive index difference \( (n_1 - n_2) \) becomes zero. This implies that the lens loses its focusing power, and the focal length becomes infinite. The lens effectively behaves as if it is not present in the system, and no image will be formed. Therefore, the nature of the lens becomes neutral, and it no longer functions as a converging lens. Quick Tip: When the refractive indices of the lens and the surrounding medium are equal, the lens loses its ability to focus light, and its focal length becomes infinite.

The current gain (\(\beta\)) of a transistor is the ratio of the output current (collector current) to the input current (base current) in a transistor. It is a measure of how much the transistor amplifies the input current.


Mathematically, \[ \beta = \frac{I_C}{I_B} \]
where:
- \(I_C\) is the collector current,

- \(I_B\) is the base current.


The value of \(\beta\) depends on the type of transistor (NPN or PNP) and its material properties. It is typically a large number, ranging from 20 to 1000, which means that a small input current can generate a large output current.
Quick Tip: The current gain (\(\beta\)) is a key parameter in transistor amplifiers as it shows how much the input current is amplified.

To calculate the charge on a nucleus, we need to consider the number of protons in the nucleus. The atomic number of nitrogen (denoted by \(Z\)) is 7, meaning that the nitrogen nucleus \({}^7_{}N^{14}\) contains 7 protons.


Each proton has a charge of \(+1.602 \times 10^{-19}\) coulombs (the elementary charge).


Thus, the total charge \(Q\) on the nitrogen nucleus is: \[ Q = Z \times e = 7 \times (1.602 \times 10^{-19} \, C) \] \[ Q = 1.1214 \times 10^{-18} \, C \]

So, the charge on the nucleus \({}^7_{}N^{14}\) is \(1.1214 \times 10^{-18}\) coulombs.
Quick Tip: The charge on a nucleus is calculated by multiplying the atomic number by the elementary charge, as each proton carries the elementary charge.

1 henry (1 H) of self-inductance is the inductance of a coil in which a change in current of 1 ampere per second produces an electromotive force (EMF) of 1 volt.


Mathematically, the self-inductance \(L\) is related to the induced voltage \(V\) and the rate of change of current \(\frac{dI}{dt}\) by the equation: \[ V = L \times \frac{dI}{dt} \]
Where:
- \(V\) is the induced voltage in volts,

- \(L\) is the self-inductance in henries,

- \(\frac{dI}{dt}\) is the rate of change of current in amperes per second.


Therefore, 1 henry is defined as the self-inductance of a coil where an increase of 1 ampere per second in current results in an induced voltage of 1 volt.
Quick Tip: 1 henry is the self-inductance of a coil that induces 1 volt for every 1 ampere per second of current change.

The kinetic energy \(K.E.\) of the electron is given as 45 eV. We can convert this into joules: \[ K.E. = 45 \, eV = 45 \times 1.602 \times 10^{-19} \, J = 7.209 \times 10^{-18} \, J \]

The expression for the kinetic energy of an electron is also: \[ K.E. = \frac{1}{2} m v^2 \]
Where:
- \(m\) is the mass of the electron,

- \(v\) is the velocity of the electron.


Rearranging for \(v\), we get: \[ v = \sqrt{\frac{2 K.E.}{m}} = \sqrt{\frac{2 \times 7.209 \times 10^{-18}}{9.11 \times 10^{-31}}} = 1.268 \times 10^7 \, m/s \]

Now, using the formula for the radius of the circular path in a magnetic field: \[ r = \frac{mv}{qB} \]
Where:
- \(r\) is the radius,

- \(m\) is the mass of the electron,

- \(v\) is the velocity of the electron,

- \(q\) is the charge of the electron,

- \(B\) is the magnetic field strength.


Substituting the known values:
- \(q = 1.602 \times 10^{-19} \, C\),

- \(B = 9 \times 10^{-5} \, T\),


We get: \[ r = \frac{9.11 \times 10^{-31} \times 1.268 \times 10^7}{1.602 \times 10^{-19} \times 9 \times 10^{-5}} = 0.079 \, m \]

Thus, the radius of the circular path is \(0.079 \, m\).
Quick Tip: The radius of the circular path of an electron in a magnetic field depends on its velocity, mass, and charge, as well as the strength of the magnetic field.

When a charged particle of charge \(q\) and mass \(m\) moves in a uniform electric field \(E\), the work done by the electric field is equal to the change in kinetic energy of the particle.


The work done \(W\) by the electric field is given by: \[ W = F \times d = qE \times d \]
where \(F = qE\) is the force acting on the particle.


Since the particle starts from rest, the work done is equal to the kinetic energy gained by the particle: \[ K.E. = \frac{1}{2} m v^2 \]

Equating the work done to the kinetic energy: \[ qEd = \frac{1}{2} m v^2 \]

Rearranging for \(v\), we get: \[ v = \sqrt{\frac{2qEd}{m}} \]

Thus, the velocity \(v\) attained by the particle is: \[ v = \sqrt{\frac{2qEd}{m}} \] Quick Tip: The velocity of a particle in an electric field depends on its charge, mass, the electric field strength, and the distance it travels.

The magnifying power \( M \) of a telescope is given by the formula: \[ M = \frac{f_o}{f_e} \]
where \( f_o \) is the focal length of the objective lens and \( f_e \) is the focal length of the eye lens.

We are given that the magnifying power \( M = 9 \) and the distance between the eye lens and the objective lens \( d = 20 \, cm \). This distance is the sum of the focal lengths of the two lenses: \[ d = f_o + f_e \]
Thus, we have the system of equations:
1. \( M = 9 = \frac{f_o}{f_e} \)
2. \( d = f_o + f_e = 20 \, cm \)

From the first equation, we can express \( f_o \) in terms of \( f_e \): \[ f_o = 9 f_e \]
Substitute this into the second equation: \[ 9 f_e + f_e = 20 \] \[ 10 f_e = 20 \] \[ f_e = 2 \, cm \]

Now substitute \( f_e = 2 \, cm \) into \( f_o = 9 f_e \): \[ f_o = 9 \times 2 = 18 \, cm \]

Thus, the focal lengths of the lenses are: \[ \boxed{f_o = 18 \, cm} \quad and \quad \boxed{f_e = 2 \, cm} \] Quick Tip: For telescopes, the magnifying power is the ratio of the focal lengths of the objective lens to the eye lens. The total distance between the lenses is the sum of their focal lengths.

- **Ohmic Resistance:**
Ohmic resistance refers to the type of resistance exhibited by materials that follow Ohm's Law, which states that the current passing through a conductor is directly proportional to the voltage applied across it, with the proportionality constant being the resistance. Mathematically, Ohm’s Law is represented as: \[ V = I R \]
where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. In ohmic materials, the resistance remains constant regardless of changes in the applied voltage or current. The graph of voltage \( V \) versus current \( I \) for an ohmic material is a straight line, indicating a constant resistance. For example, metals like copper and aluminum are ohmic resistors, as they maintain a linear relationship between voltage and current.

The key point is that for ohmic resistors, the resistance does not change with temperature or voltage, as long as the material remains within its linear region. In other words, if we apply a higher voltage to an ohmic material, the current will increase proportionally, and the ratio of voltage to current will stay the same.

- **Non-Ohmic Resistance:**
Non-ohmic resistance refers to the resistance shown by materials that do not obey Ohm’s Law. In these materials, the relationship between the voltage and the current is not linear, and the resistance changes with varying voltage or current. Non-ohmic materials exhibit a variable resistance that depends on factors such as temperature, the direction of current, or the applied voltage.

For example, semiconductors like diodes or light bulbs show non-ohmic behavior. A diode, for instance, has a very small current when the voltage is below a certain threshold (reverse or forward bias), but once the threshold is exceeded, the current increases exponentially with voltage, which results in a nonlinear \( V \)-\( I \) characteristic. The graph of voltage vs current for a non-ohmic resistor is not a straight line but a curve, reflecting that the resistance is changing as voltage and current vary.

Other examples of non-ohmic resistances include light bulbs, where the filament's resistance increases as the temperature rises due to an increase in current.


Dynamic Resistance:

Dynamic resistance, also called incremental resistance, is the resistance of a non-ohmic component at a particular point on its current-voltage curve. It is defined as the ratio of the change in voltage (\( \Delta V \)) to the change in current (\( \Delta I \)) for small variations around a particular operating point on the curve. Mathematically, dynamic resistance is expressed as: \[ r_d = \frac{\Delta V}{\Delta I} \]
This expression allows us to find the resistance at a specific point on the curve where the voltage and current are not constant. For instance, in a diode, the dynamic resistance changes depending on the voltage and current applied. At very small voltages (below the threshold), the dynamic resistance is very high, but after the threshold voltage is surpassed, the resistance drops significantly, and current increases more rapidly.

Dynamic resistance is particularly useful for analyzing devices where the current-voltage relationship is nonlinear, such as in diodes or transistors. In contrast to static resistance, which is a fixed value in ohmic materials, dynamic resistance provides an instantaneous measure of how the component behaves under varying conditions. Quick Tip: Ohmic resistors obey Ohm's Law, and their resistance is constant regardless of the voltage or current. Non-ohmic resistors, however, exhibit a variable resistance that depends on factors like voltage and temperature. Dynamic resistance measures the instantaneous resistance of non-ohmic materials.

Nuclei can be classified based on their atomic number (\(Z\)) and mass number (\(A\)) as follows:

1. **Atomic Number (\(Z\))**: The atomic number represents the number of protons in the nucleus and is denoted by \(Z\). It determines the chemical properties of the element and its position in the periodic table. Nuclei with the same atomic number belong to the same element.

2. **Mass Number (\(A\))**: The mass number is the total number of protons and neutrons in the nucleus and is denoted by \(A\). It gives the approximate mass of the nucleus. Different isotopes of the same element have the same atomic number (\(Z\)) but different mass numbers (\(A\)) because they have different numbers of neutrons.

Types of Nuclei Based on \(Z\) and \(A\):

- **Isotopes**: Nuclei with the same atomic number (\(Z\)) but different mass numbers (\(A\)) are called isotopes. They have the same number of protons but different numbers of neutrons. Example: \({}^1_1H\) (protium) and \({}^2_1H\) (deuterium).

- **Isobars**: Nuclei with the same mass number (\(A\)) but different atomic numbers (\(Z\)) are called isobars. Example: \({}^4_2He\) (helium) and \({}^4_1H\) (hydrogen).

- **Isotones**: Nuclei with the same number of neutrons but different atomic numbers and mass numbers are called isotones. Example: \({}^4_2He\) and \({}^3_1H\).

- **Isomers**: Nuclei with the same atomic number (\(Z\)) and mass number (\(A\)) but different energy states are called isomers. They are the same element but in different nuclear states. Quick Tip: The atomic number determines the chemical identity of the element, while the mass number helps to classify different isotopes and their stability.

A **NOR gate** is a digital logic gate that outputs true (1) only when both of its inputs are false (0). It is the inverse of the OR gate.

### Logic Symbol of NOR Gate:
The logic symbol of a NOR gate is the same as the OR gate with a small circle (representing NOT) at the output. The symbol for a 2-input NOR gate is:
\[ \begin{array}{|c|} \hline NOR Gate Symbol
\hline \end{array} \]

### Truth Table for NOR Gate:
The truth table for a 2-input NOR gate is:
\[ \begin{array}{|c|c|c|} \hline Input A & Input B & Output (A NOR B)
\hline 0 & 0 & 1
0 & 1 & 0
1 & 0 & 0
1 & 1 & 0
\hline \end{array} \]

- When both inputs \(A\) and \(B\) are 0, the output is 1.
- In all other cases, the output is 0. Quick Tip: The NOR gate is a universal gate because any other logic gate can be implemented using just NOR gates.

The equations of current and voltage are given as: \[ i = I_{max} \sin(\omega t + \phi_i) \] \[ v = V_{max} \sin(\omega t + \phi_v) \]

Where:
- \(I_{max} = 3.5\) A (maximum current),

- \(V_{max} = 28\) V (maximum voltage),

- \(\omega = 628\) rad/s (angular frequency),

- \(\phi_i = \frac{\pi}{6}\) (phase of current),

- \(\phi_v = -\frac{\pi}{6}\) (phase of voltage).


(i) Root Mean Square Value of Current:

The root mean square (rms) value of current is given by: \[ I_{rms} = \frac{I_{max}}{\sqrt{2}} = \frac{3.5}{\sqrt{2}} = 2.475 \, A \]


(ii) Time Period:

The time period \(T\) is related to the angular frequency \(\omega\) by the formula: \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{628} \approx 0.010 \, seconds \]


(iii) Phase Difference Between Current and Voltage:

The phase difference \(\Delta \phi\) between current and voltage is the difference between their phase angles: \[ \Delta \phi = \phi_v - \phi_i = \left(-\frac{\pi}{6}\right) - \left(\frac{\pi}{6}\right) = -\frac{\pi}{3} \]

Thus, the phase difference between current and voltage is \(-\frac{\pi}{3}\) radians (or -60°).
Quick Tip: The root mean square value of a sinusoidal quantity is \(\frac{Maximum value}{\sqrt{2}}\). The time period is the inverse of the frequency, and the phase difference is the difference between the phase angles of current and voltage.

Properties of \(\gamma\)-rays:

1. High Energy: \(\gamma\)-rays are the most energetic form of electromagnetic radiation, with energies higher than X-rays.

2. Short Wavelength: \(\gamma\)-rays have very short wavelengths, typically less than \(10^{-12}\) m.

3. Penetrating Power: Due to their high energy, \(\gamma\)-rays have great penetrating power and can pass through most materials, including concrete and lead.

4. Ionizing Radiation: \(\gamma\)-rays can ionize atoms and molecules, which can damage biological tissues and DNA. They are used in cancer treatment (radiotherapy) to destroy cancer cells.


Wavelength Comparison:

The wavelength of \(\gamma\)-rays is much smaller than that of ultraviolet rays. \(\gamma\)-rays have wavelengths in the range of \(10^{-12}\) m or smaller, while ultraviolet rays have wavelengths in the range of \(10^{-8}\) m. Therefore, **\(\gamma\)-rays have a shorter wavelength than ultraviolet rays**. Quick Tip: \(\gamma\)-rays are highly energetic and ionizing, making them useful in medical treatments but also hazardous due to their ability to penetrate and damage living tissues.

The actual angle of dip \( \delta \) can be determined using the relationship between the angle of suspension \( \theta \), the magnetic declination (the angle with the magnetic meridian), and the angle of dip. The formula to calculate the actual angle of dip is given by:
\[ \tan(\delta) = \tan(\theta) \times \cos(\alpha) \]
where:

- \( \delta \) is the actual angle of dip,

- \( \theta = 45^\circ \) is the angle between the magnet and the horizontal,

- \( \alpha = 30^\circ \) is the angle the magnet makes with the magnetic meridian.


Substituting the values into the equation:
\[ \tan(\delta) = \tan(45^\circ) \times \cos(30^\circ) \]

We know that \( \tan(45^\circ) = 1 \) and \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \). Therefore:
\[ \tan(\delta) = 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]

Now, to find \( \delta \), we take the inverse tangent of both sides:
\[ \delta = \tan^{-1}\left( \frac{\sqrt{3}}{2} \right) \]

Using a calculator:
\[ \delta \approx 60^\circ \]

Thus, the actual angle of dip is approximately:
\[ \boxed{60^\circ} \] Quick Tip: The angle of dip can be determined using the angle of suspension and the angle with the magnetic meridian. The formula involves the tangent and cosine of the given angles.

The power \(P\) of a lens is given by the formula: \[ P = \frac{1}{f} \]
Where \(f\) is the focal length of the lens.

The lens formula for a lens in air is given by: \[ \frac{1}{f} = (n - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Where:
- \(n = 1.65\) (refractive index of crown glass),
- \(R_1\) and \(R_2\) are the radii of curvature of the two surfaces of the lens.

For an equiconcave lens, both radii of curvature are equal in magnitude but opposite in sign. Thus, we have: \[ \frac{1}{f} = (1.65 - 1)\left( \frac{1}{R_1} - \frac{1}{-R_1} \right) = 0.65 \times \frac{2}{R_1} \]

Now, using the formula for power: \[ P = \frac{1}{f} = -2.5 \, D \]
Thus, \[ -2.5 = 0.65 \times \frac{2}{R_1} \]
Solving for \(R_1\): \[ R_1 = \frac{0.65 \times 2}{2.5} = 0.52 \, m \]

Thus, the radius of curvature of each surface of the lens should be \(0.52 \, m\).
Quick Tip: For an equiconcave lens, the radii of curvature are equal in magnitude but opposite in sign.

A **transistor** is a semiconductor device used to amplify or switch electronic signals and electrical power. It consists of three layers of semiconductor material, known as the emitter, base, and collector.

The common emitter configuration is the most commonly used configuration in amplifiers. It has the following characteristics:
- The emitter is common to both the input and output.
- The base controls the current between the collector and emitter.
- It provides high voltage gain.

### Circuit Diagram of Common Emitter Configuration: \[ [Insert diagram of common emitter configuration here] \]

### Output Characteristics:
In the common emitter configuration, the output characteristics are a plot of collector current (\(I_C\)) versus collector-emitter voltage (\(V_{CE}\)) for different values of base current (\(I_B\)).

The graph typically shows three regions:
1. **Cutoff region**: Where \(I_C = 0\).
2. **Active region**: Where \(I_C\) increases linearly with \(V_{CE}\).
3. **Saturation region**: Where the collector current becomes almost constant, irrespective of further increases in \(V_{CE}\).

The graph demonstrates the transistor's ability to amplify the input signal. Quick Tip: The common emitter configuration provides voltage amplification and is widely used in signal amplification.

In a series L-C-R circuit, resonance occurs when the inductive reactance (\( X_L \)) and capacitive reactance (\( X_C \)) are equal in magnitude but opposite in phase. The formula for the resonant frequency \( f_0 \) of a series L-C-R circuit is given by:
\[ f_0 = \frac{1}{2 \pi \sqrt{LC}} \]
where:

- \( L \) is the inductance in henries (H),

- \( C \) is the capacitance in farads (F).


At this frequency, the impedance of the circuit is purely resistive, and the current reaches its maximum value.
Quick Tip: At resonance in a series L-C-R circuit, the inductive and capacitive reactances cancel each other, resulting in a minimum impedance equal to the resistance.

Bohr's model of the atom was proposed by Niels Bohr in 1913 to explain the structure of the hydrogen atom and the discrete energy levels of electrons. According to Bohr's model:


1. **Electron Orbitals:** Electrons revolve around the nucleus in fixed, circular orbits without radiating energy. These orbits are called "stationary orbits" or "energy levels."

2. **Quantized Energy Levels:** The energy associated with each orbit is quantized, meaning that only certain energy levels are allowed. The energy of an electron in the \( n \)-th orbit is given by:

\[ E_n = - \frac{13.6 \, eV}{n^2} \]
where \( n \) is the principal quantum number (1, 2, 3, ...).

3. **Electron Transitions:** Electrons can absorb or emit energy in discrete amounts when they jump between these orbits. The energy of the emitted or absorbed radiation is equal to the difference in energy between the two orbits:

\[ \Delta E = E_{higher} - E_{lower} \]
4. **Angular Momentum Quantization:** The angular momentum of an electron in the \( n \)-th orbit is quantized and is an integer multiple of \( \hbar = \frac{h}{2\pi} \) (where \( h \) is Planck’s constant):

\[ L = n \hbar \]

Bohr's model successfully explained the spectral lines of hydrogen and laid the foundation for quantum mechanics.
Quick Tip: Bohr’s model introduced the concept of quantized energy levels, which helped explain the discrete spectral lines observed in atomic spectra.

According to Bohr's model, the energy of the electron in the \( n \)-th orbit is given by:
\[ E_n = - \frac{13.6 \, eV}{n^2} \]
We are given the energy \( E = -3.4 \, eV \). To find \( n \), we solve for \( n \) in the equation:
\[ -3.4 = - \frac{13.6}{n^2} \] \[ n^2 = \frac{13.6}{3.4} = 4 \] \[ n = 2 \]

Thus, the electron is in the second energy level (\( n = 2 \)).


The angular momentum \( L \) of the electron in the \( n \)-th orbit is given by:
\[ L = n \hbar \]
where \( \hbar = \frac{h}{2\pi} \) is the reduced Planck’s constant.


For \( n = 2 \):
\[ L = 2 \hbar \]

Thus, the angular momentum of the electron is:
\[ \boxed{L = 2 \hbar} \] Quick Tip: The angular momentum of an electron in Bohr’s model is quantized and directly proportional to the principal quantum number \( n \).

Consider a prism with refracting angle \( A \), and let \( \delta_m \) be the angle of minimum deviation. When light passes symmetrically through the prism at minimum deviation, the angle of incidence \( i \) and the angle of emergence \( e \) are equal, i.e.,
\[ i = e \]

From the geometry of the prism, the relation between angle of incidence \( i \), angle of refraction \( r \), and prism angle \( A \) is:
\[ A = r_1 + r_2 \]
But under minimum deviation condition, the path of the ray inside the prism is symmetrical, so:
\[ r_1 = r_2 = r \Rightarrow A = 2r \Rightarrow r = \frac{A}{2} \]

The total deviation \( \delta \) is given by:
\[ \delta = i + e - A \Rightarrow \delta_m = 2i - A \Rightarrow i = \frac{A + \delta_m}{2} \]

Now apply Snell's law at the first surface of the prism:
\[ n = \frac{\sin i}{\sin r} \]

Substitute the values of \( i \) and \( r \):
\[ n = \frac{\sin\left( \frac{A + \delta_m}{2} \right)}{\sin\left( \frac{A}{2} \right)} \]


Final Formula:
\[ \boxed{n = \frac{\sin\left( \frac{A + \delta_m}{2} \right)}{\sin\left( \frac{A}{2} \right)}} \]

This is the required expression for the refractive index \( n \) of the material of the prism in terms of the prism angle \( A \) and angle of minimum deviation \( \delta_m \).
Quick Tip: At minimum deviation, light travels symmetrically through the prism. Use geometry and Snell’s law to derive the relation between refractive index, angle of prism, and angle of minimum deviation.

Kirchhoff’s First Law (Junction Rule):

Kirchhoff's first law states that the total current entering a junction (or node) in an electric circuit is equal to the total current leaving the junction.

This law is based on the principle of conservation of electric charge.

\[ \sum I_{in} = \sum I_{out} \]

Where:

- \( I_{in} \): Current flowing into the junction

- \( I_{out} \): Current flowing out of the junction


Kirchhoff’s Second Law (Loop Rule):

Kirchhoff’s second law states that the algebraic sum of the potential differences (voltages) in any closed loop or mesh of a circuit is always zero.

This law is based on the conservation of energy.

\[ \sum E = \sum IR \]

Where:

- \( E \): Electromotive force (emf)

- \( I \): Current flowing through elements of the loop

- \( R \): Resistance in each segment of the loop


This means that the total energy supplied in a closed loop is equal to the total energy used.
Quick Tip: Kirchhoff's laws are essential tools in solving complex electrical circuits involving multiple branches and loops.

Louis de Broglie proposed that every moving particle is associated with a wave. This wave is known as the matter wave or de Broglie wave. He suggested that particles such as electrons, protons, and even macroscopic objects exhibit both wave-like and particle-like behavior, which is a fundamental principle of quantum mechanics.


de Broglie Hypothesis:

The wavelength \( \lambda \) associated with a particle of momentum \( p \) is given by:
\[ \lambda = \frac{h}{p} \]
where:
\( h \) = Planck’s constant \( (6.626 \times 10^{-34} \, Js) \)
\( p \) = momentum of the particle = \( mv \)


Now, in terms of kinetic energy:

Kinetic energy \( K.E. = \frac{1}{2}mv^2 \)

Solving for velocity:
\[ v = \sqrt{\frac{2K.E.}{m}} \]
Now, \( p = mv = m \cdot \sqrt{\frac{2K.E.}{m}} = \sqrt{2mK.E.} \)


Substitute into de Broglie equation:
\[ \lambda = \frac{h}{\sqrt{2mK.E.}} \]


Final Result:
\[ \boxed{\lambda = \frac{h}{\sqrt{2mK.E.}}} \] Quick Tip: According to de Broglie, matter exhibits wave-particle duality. This is the foundation for quantum mechanics and was confirmed experimentally by the Davisson-Germer experiment.

Consider a parallel plate capacitor of plate area \( A \) and separation \( d \). Let the space between the plates be partially filled with a dielectric of dielectric constant \( K \) up to a thickness \( t \), and the remaining space \( (d - t) \) is air (or vacuum).


We can treat the system as two capacitors in series:
1. Capacitor with dielectric of thickness \( t \) and dielectric constant \( K \),
2. Capacitor with air gap of thickness \( d - t \) and dielectric constant 1.


Capacitance of each section:

- With dielectric: \[ C_1 = \frac{K \varepsilon_0 A}{t} \]
- With air: \[ C_2 = \frac{\varepsilon_0 A}{d - t} \]

Since they are in series, the equivalent capacitance \( C \) is given by:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{t}{K \varepsilon_0 A} + \frac{d - t}{\varepsilon_0 A} = \frac{1}{\varepsilon_0 A} \left( \frac{t}{K} + (d - t) \right) \]
\[ C = \frac{\varepsilon_0 A}{\left( \frac{t}{K} + d - t \right)} \]


Final Result:
\[ \boxed{C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}} \] Quick Tip: When a dielectric partially fills a capacitor, treat the regions as separate capacitors in series, and apply the series formula to compute net capacitance.

Energy Level Diagram:

The hydrogen atom has discrete energy levels given by the formula: \[ E_n = -13.6 \, \frac{1}{n^2} \, eV \quad where \quad n = 1, 2, 3, \ldots \]

The electron transitions from a higher energy level \(n_i\) to a lower level \(n_f\) result in the emission of photons of specific wavelengths. Different spectral series correspond to transitions ending at specific lower energy levels:


Lyman series: Transitions to \(n = 1\)
Balmer series: Transitions to \(n = 2\)
Paschen series: Transitions to \(n = 3\)
Brackett series: Transitions to \(n = 4\)
Pfund series: Transitions to \(n = 5\)


Regions of the Electromagnetic Spectrum:


Lyman series – Ultraviolet region
Balmer series – Visible region
Paschen, Brackett, Pfund – Infrared region


Quick Tip: The energy of emitted light depends on the difference in energy levels. Smaller transitions (e.g., Balmer) lie in the visible range, while larger ones (e.g., Lyman) lie in the UV region.

Electromagnetic Induction:

Electromagnetic induction is the process of generating an electromotive force (emf) or current in a conductor when the magnetic flux linked with it changes. This phenomenon was discovered by Michael Faraday in 1831.

Faraday's Laws of Electromagnetic Induction:


First Law:

Whenever the magnetic flux through a circuit changes, an emf is induced in the circuit. If the circuit is closed, a current is also induced.


Second Law:

The magnitude of the induced emf is directly proportional to the rate of change of magnetic flux through the circuit. \[ Induced emf (\varepsilon) = -\frac{d\Phi}{dt} \]
Where:

\(\Phi\) is the magnetic flux,
\(\frac{d\Phi}{dt}\) is the rate of change of magnetic flux,
The negative sign indicates the direction of induced emf (Lenz's Law). Quick Tip: Electromagnetic induction is the basic working principle behind generators, transformers, and many electrical machines.

Definition of Wavefront:

A wavefront is defined as the continuous locus of all the particles of a medium that are vibrating in the same phase.

It is a surface over which the phase of the wave is constant.


Depending on the source of disturbance, wavefronts may be:

Spherical wavefronts – from a point source.

Cylindrical wavefronts – from a linear source.

Plane wavefronts – when the source is very far away.


Huygens' Principle:

According to Huygens’ Principle:

Every point on a given wavefront acts as a fresh source of secondary wavelets.

The secondary wavelets travel in all directions with the same speed as the original wave.

The new wavefront is the forward envelope of all these secondary wavelets.



Laws of Reflection Using Huygens' Principle:

Let a plane wavefront strike a reflecting surface. According to Huygens:

Each point on the incident wavefront acts as a source of secondary wavelets.

The reflected wavefront is formed by drawing tangents to all secondary wavelets.



By geometry, we find that: \[ \angle incidence = \angle reflection \]
Thus, Huygens’ construction proves the laws of reflection. Quick Tip: Huygens’ principle is useful in explaining wave phenomena like reflection, refraction, and diffraction geometrically.

LED (Light Emitting Diode):

An LED is a semiconductor device that emits light when an electric current flows through it. It is a \(p\)-\(n\) junction diode that emits photons during the process of electron-hole recombination.


Principle:

The working of an LED is based on the principle of electroluminescence, where light is emitted when a forward-biased \(p\)-\(n\) junction diode recombines electrons and holes near the junction.

The energy released during this recombination is emitted in the form of photons (light).


Construction:


LED is made using semiconductor materials like gallium arsenide (GaAs), gallium phosphide (GaP), etc.

It consists of a \(p\)-type and an \(n\)-type semiconductor.

The LED is housed in a transparent plastic casing to allow emitted light to escape.

The longer lead is the anode (+), and the shorter lead is the cathode (–).




Working:


When a forward voltage is applied, electrons from the \(n\)-side and holes from the \(p\)-side move toward the junction.

They recombine near the junction. The electrons lose energy.

This energy is released as photons (light).

The color of light depends on the bandgap energy of the material used.
Quick Tip: LEDs are energy-efficient and used widely in indicators, displays, and lighting due to their low power consumption and long life.