Step 1: Understanding Flux and Solid Angle

The total electric flux emanating from a point charge \(q\) is given by Gauss's Law as \(\Phi_{total} = \frac{q}{\epsilon_0}\). This total flux radiates outwards in all directions, covering a total solid angle of \(4\pi\) steradians. The flux through any specific surface is proportional to the solid angle that the surface subtends at the location of the charge.


Step 2: Geometric Analysis of the Situation

The charge \(q\) is placed at the center of the circular open end of the cylinder. The open end can be thought of as an infinite plane that cuts the space into two halves. The cylindrical vessel lies entirely in one half of this space.


Step 3: Calculating the Solid Angle

From the perspective of the charge \(q\) located on this plane, all the electric field lines that go into the cylinder are contained within a hemisphere of space. The solid angle of a hemisphere is exactly half the total solid angle around a point. \[ Solid angle subtended by the vessel = \frac{1}{2} \times (Total solid angle) \] \[ \Omega_{vessel} = \frac{1}{2} \times 4\pi = 2\pi steradians \]

Step 4: Calculating the Flux through the Vessel

The flux through the vessel is the fraction of the total flux corresponding to the solid angle it subtends. \[ \Phi_{vessel} = \Phi_{total} \times \frac{\Omega_{vessel}}{4\pi} \]
Substituting the known values: \[ \Phi_{vessel} = \left( \frac{q}{\epsilon_0} \right) \times \frac{2\pi}{4\pi} \] \[ \Phi_{vessel} = \left( \frac{q}{\epsilon_0} \right) \times \frac{1}{2} = \frac{q}{2\epsilon_0} \]

Step 5: Final Answer

Thus, exactly half of the total flux from the charge passes through the surface of the open cylindrical vessel. The flux is \(\frac{q}{2\epsilon_0}\). Therefore, option (C) is correct.
Quick Tip: Whenever you encounter problems with open surfaces and need to find the electric flux, try to create a symmetrical closed surface by adding an imaginary part. Apply Gauss's Law to the closed surface and then use symmetry to determine the flux through the original open part.

Step 1: Understanding the Concept and Relevant Rule

The direction of the magnetic force on a moving charge is determined by Fleming's Left-Hand Rule. This rule relates the directions of force, magnetic field, and conventional current.

Thumb: Represents the direction of the magnetic Force (\(F\)).
Forefinger: Represents the direction of the Magnetic Field (\(B\)).
Middle Finger: Represents the direction of the conventional Current (\(I\)).


Step 2: Determining the Direction of Conventional Current

Fleming's Left-Hand Rule is based on the flow of positive charge (conventional current). An electron is negatively charged. Therefore, the direction of conventional current is opposite to the direction of the electron's motion.

The electron is projected towards the East.
So, the conventional current (\(I\)) is directed towards the West.


Step 3: Applying Fleming's Left-Hand Rule

We are given the following directions:

Direction of Force (\(F\)): The electron is deflected towards the North.
Direction of Current (\(I\)): Determined to be towards the West.

Now, let's orient our left hand according to the rule:

Point your Middle Finger in the direction of the current, which is to the West.
Point your Thumb in the direction of the force, which is towards the North.
Your Forefinger will now naturally point in the direction of the magnetic field.

Upon doing this, you will find that your forefinger points downwards, into the plane.

Step 4: Final Answer

The direction of the magnetic field is perpendicular to the plane and directed downwards. Therefore, option (D) is correct.
Quick Tip: For questions involving forces on charged particles, always check the sign of the charge. For electrons (negative charge), the direction of the magnetic force is opposite to what the standard right-hand rule would predict for a positive charge. Using Fleming's Left-Hand Rule, point the current finger opposite to the electron's velocity.

Step 1: Analyzing the Forces from Each Field Component

An electromagnetic wave is composed of an oscillating electric field (\(\vec{E}\)) and an oscillating magnetic field (\(\vec{B}\)). We must consider the force each field exerts on the electron separately.


Step 2: The Electric Force (\(\vec{F}_E\))

The definition of an electric field is that it exerts a force on a charge, regardless of the charge's state of motion. The formula is: \[ \vec{F}_E = q\vec{E} \]
For an electron, \(q = -e\). Thus, the electric field will always exert a force \(\vec{F}_E = -e\vec{E}\) on the electron.


Step 3: The Magnetic Force (\(\vec{F}_B\))

A magnetic field exerts a force only on a moving charge. The formula for this force is: \[ \vec{F_B = q(\vec{v} \times \vec{B}) \]
where \(\vec{v}\) is the velocity of the charge. If the charge is stationary (\(\vec{v} = 0\)), then the magnetic force is zero.


Step 4: Applying to the Initial State of the Electron

The problem states that the electron is "placed" in the path of the wave, which means its initial velocity is zero (\(\vec{v}=0\)).

At this initial moment, the magnetic force is \(\vec{F}_B = -e(\vec{0} \times \vec{B}) = 0\).
At the same initial moment, the electric force is non-zero: \(\vec{F}_E = -e\vec{E}\).

Therefore, the force that initiates the motion of the electron is purely the electric force. The electron will start to accelerate in the direction of this force.


Step 5: Final Answer

Since the motion is initiated by the electric force (\(\vec{F}_E\)), the electron will start moving along the line of the electric field vector. As the E-field of the wave oscillates, the electron will be forced to oscillate along this direction. Therefore, option (A) is correct.
Quick Tip: In problems involving the interaction of EM waves with charges, remember that the force exerted by the electric field component is almost always dominant over the force from the magnetic field component, unless the particle is moving at relativistic speeds.

Step 1: Understanding Lens Power

An alternative to the Lens Maker's Formula is to consider the power of the lens. The total power of a thin lens is the sum of the powers of its two refracting surfaces. The power of a single spherical refracting surface is given by \( P = \frac{\mu_2 - \mu_1}{R} \), where light travels from a medium with refractive index \(\mu_1\) to a medium with \(\mu_2\).


Step 2: Calculating Power of the First Surface

We assume the lens is in air (\(\mu_1 = 1\)) and light travels from left to right.
For the first surface, light enters the lens from the air.

\(\mu_1 = 1\) (air)
\(\mu_2 = 1.5\) (lens material)
The first surface is convex to the incoming light, so its radius of curvature is positive: \(R_1 = +R\).

The power of this first surface is: \[ P_1 = \frac{1.5 - 1}{+R} = \frac{0.5}{R} \]

Step 3: Calculating Power of the Second Surface

For the second surface, light exits the lens into the air.

\(\mu_1 = 1.5\) (lens material)
\(\mu_2 = 1\) (air)
The second surface is also convex, but its center of curvature is on the left side, so its radius is negative: \(R_2 = -R\).

The power of this second surface is: \[ P_2 = \frac{1 - 1.5}{-R} = \frac{-0.5}{-R} = \frac{0.5}{R} \]

Step 4: Calculating Total Power and Focal Length

The total power (\(P\)) of the lens is the sum of the powers of the two surfaces: \[ P = P_1 + P_2 = \frac{0.5}{R} + \frac{0.5}{R} = \frac{1}{R} \]
The focal length (\(f\)) is the reciprocal of the total power (\(f = 1/P\)). \[ f = \frac{1}{1/R} = R \]

Step 5: Final Answer

The focal length of the double convex lens is equal to its radius of curvature, \(f = R\).
Quick Tip: Always be careful with the sign convention in optics problems. For radii of curvature, a common convention is to measure distances from the optical center. If the center of curvature is on the side where light emerges (right side for light incident from left), the radius is positive. If it's on the side where light is incident (left side), the radius is negative.

Step 1: Understanding the Concept:

In AC (Alternating Current) circuits, the standard voltage rating (like the 220 V for household supply) refers to the RMS (Root Mean Square) value, not the peak or maximum voltage. The RMS value is a kind of average voltage that gives the same heating effect as a DC voltage of the same value. The peak voltage is the maximum value the voltage reaches during its sinusoidal cycle.


Step 2: Key Formula or Approach:

The relationship between the peak voltage (\(V_p\) or \(V_0\)) and the RMS voltage (\(V_{rms}\)) for a sinusoidal AC source is given by: \[ V_p = V_{rms} \times \sqrt{2} \]
or \[ V_{rms} = \frac{V_p}{\sqrt{2}} \]

Step 3: Detailed Explanation:

We are given the RMS voltage of the AC source: \[ V_{rms} = 220 V \]
We need to find the peak voltage, \(V_p\).

Using the formula: \[ V_p = V_{rms} \times \sqrt{2} \]
Substitute the given value: \[ V_p = 220 \times \sqrt{2} \]
We know that the value of \(\sqrt{2}\) is approximately 1.414.
\[ V_p \approx 220 \times 1.414 \] \[ V_p \approx 311.08 V \]
This value is approximately 310 V.


Step 4: Final Answer:

The peak voltage in a 220 volt A.C. source is about 310 V.
Quick Tip: Unless specified otherwise, any AC voltage or current value given in a problem is the RMS value. To find the peak value, multiply the RMS value by \(\sqrt{2}\) (\(\approx 1.414\)). To find the RMS value from the peak, divide by \(\sqrt{2}\).

Step 1: Understanding the Concept:

This question relates to the fundamentals of semiconductor physics. The conductivity of a material depends on the number of free charge carriers available to conduct electricity. A pure semiconductor (intrinsic semiconductor) has a relatively low number of charge carriers (electrons and holes, created by thermal excitation). Doping is the process of intentionally adding impurities to a pure semiconductor to significantly increase the number of charge carriers.


Step 2: Detailed Explanation:

The conductivity (\(\sigma\)) of a semiconductor is given by the formula: \[ \sigma = e(n_e\mu_e + n_h\mu_h) \]
where:
- \(e\) is the elementary charge.

- \(n_e\) is the concentration of free electrons.

- \(n_h\) is the concentration of holes.

- \(\mu_e\) and \(\mu_h\) are the mobilities of electrons and holes, respectively.


In a pure (intrinsic) semiconductor at room temperature, the concentration of electrons is equal to the concentration of holes (\(n_e = n_h = n_i\)), and this intrinsic carrier concentration \(n_i\) is very small.


Doping involves adding impurity atoms:

1. N-type Doping: If a pentavalent impurity (e.g., Phosphorus, Arsenic) is added to a tetravalent semiconductor (e.g., Silicon), each impurity atom donates one free electron. This drastically increases the electron concentration (\(n_e \gg n_i\)). Electrons become the majority charge carriers, and conductivity increases significantly.

2. P-type Doping: If a trivalent impurity (e.g., Boron, Aluminum) is added, each impurity atom creates a 'hole' (an absence of an electron). This drastically increases the hole concentration (\(n_h \gg n_i\)). Holes become the majority charge carriers, and conductivity increases significantly.


In both cases, doping substantially increases the concentration of majority charge carriers (\(n_e\) or \(n_h\)), which directly leads to a large increase in the overall conductivity (\(\sigma\)) of the semiconductor.


Step 3: Final Answer:

When an impurity is doped in a pure semiconductor, the number of charge carriers increases, and therefore, the conductivity of the semiconductor increases.
Quick Tip: Remember that the entire purpose of doping a semiconductor is to control and increase its conductivity. This process is fundamental to creating all modern semiconductor devices like diodes, transistors, and integrated circuits. Doping turns a poor conductor (intrinsic semiconductor) into a useful one (extrinsic semiconductor).

Step 1: Understanding the Concept:

Specific resistance, also known as resistivity, is an intrinsic property of a material that quantifies how strongly it resists the flow of electric current. It is denoted by the Greek letter \(\rho\) (rho).


Step 2: Key Formula or Approach:

The resistance \(R\) of a uniform conductor is related to its resistivity (\(\rho\)), length (\(L\)), and cross-sectional area (\(A\)) by the formula: \[ R = \rho \frac{L}{A} \]
We can rearrange this formula to solve for resistivity \(\rho\): \[ \rho = \frac{R \cdot A}{L} \]

Step 3: Detailed Explanation:

To find the unit of resistivity, we can substitute the SI units for the quantities on the right side of the rearranged formula:

The unit of resistance (\(R\)) is the ohm (\(\Omega\)).

The unit of area (\(A\)) is the square meter (\(m^2\)).

The unit of length (\(L\)) is the meter (\(m\)).


Substituting these units into the equation for \(\rho\): \[ Unit of \rho = \frac{Unit of R \times Unit of A}{Unit of L} = \frac{\Omega \cdot m^2}{m} \]
Simplifying the expression, we get: \[ Unit of \rho = \Omega \cdot m \]

Step 4: Final Answer:

The SI unit of specific resistance (resistivity) is the ohm-meter (\(\Omega \cdot m\)).
Quick Tip: Remember the distinction: Resistance (\(R\)) is a property of a specific object and depends on its shape and size, measured in ohms (\(\Omega\)). Resistivity (\(\rho\)) is a property of the material itself, measured in ohm-meters (\(\Omega \cdot m\)).

Step 1: Understanding the Concept:

A long, straight wire carrying a current produces a magnetic field in the form of concentric circles around the wire. The magnitude of this field can be calculated using Ampere's Law or the Biot-Savart Law.


Step 2: Key Formula or Approach:

For an infinitely long, straight wire carrying a current \(I\), the magnitude of the magnetic field \(B\) at a perpendicular distance \(r\) from the wire is given by: \[ B = \frac{\mu_0 I}{2 \pi r} \]
where \(\mu_0\) is the permeability of free space, with a value of \(4\pi \times 10^{-7} \, T \cdot m/A\).


Step 3: Detailed Explanation:

First, identify the given values and convert them to SI units:

Current, \(I = 10\) A.

The wire is along the Z-axis.

The point is P(10 cm, 0, 0).

The perpendicular distance \(r\) of the point P from the Z-axis is its x-coordinate, which is 10 cm.

Convert the distance to meters: \(r = 10 cm = 0.1 m\).


Now, substitute these values into the formula for the magnetic field: \[ B = \frac{(4\pi \times 10^{-7} \, T \cdot m/A) \times (10 \, A)}{2 \pi \times (0.1 \, m)} \]
Cancel out \(2\pi\) from the numerator and denominator: \[ B = \frac{2 \times 10^{-7} \times 10}{0.1} \] \[ B = \frac{2 \times 10^{-6}}{10^{-1}} \] \[ B = 2 \times 10^{-6 - (-1)} = 2 \times 10^{-5} \, T \]

Step 4: Final Answer:

The intensity of the magnetic field at the point (10 cm, 0, 0) is \(2 \times 10^{-5}\) Tesla.

(Note: To find the direction, use the right-hand thumb rule. If the thumb points along the current (+Z direction), the curled fingers at the point (+X axis) will point in the +Y direction. So, \(\vec{B} = 2 \times 10^{-5} \hat{j} \, T\)).
Quick Tip: Always ensure all units are in the SI system (meters, Amperes, etc.) before performing calculations. For finding the direction of the magnetic field from a straight wire, the Right-Hand Thumb Rule is the quickest method.

Step 1: Understanding the Concept:

Lenz's law is a fundamental principle in electromagnetic induction, which provides the direction of the induced current. It is a consequence of the law of conservation of energy.


Step 2: Detailed Explanation:

The law can be broken down into key ideas:

Cause: A change in magnetic flux (\(\Delta\Phi_B\)) through a conducting loop.

Effect: An induced current is generated in the loop.

Opposition: This induced current creates its own magnetic field. The direction of this induced magnetic field is such that it opposes the original change in flux.


For example:

If the magnetic flux through a loop is \textit{increasing, the induced current will create a magnetic field in the opposite direction to counteract the increase.

If the magnetic flux through a loop is \textit{decreasing, the induced current will create a magnetic field in the same direction to try and maintain the flux.


Mathematically, Lenz's law is represented by the negative sign in Faraday's law of induction: \[ \mathcal{E = - \frac{d\Phi_B}{dt} \]
where \(\mathcal{E}\) is the induced EMF. The negative sign signifies the opposition.


Step 3: Final Answer:

Lenz's law states that the direction of the induced current is such that it will oppose the change in magnetic flux that is causing it.
Quick Tip: Think of Lenz's law as "electromagnetic inertia." The system (the conducting loop) resists any change to the magnetic flux passing through it, just as a massive object resists changes to its state of motion.

Step 1: Understanding the Concept:

Total Internal Reflection (TIR) is a special case of reflection that occurs under specific conditions at the boundary between two transparent media. It's the principle behind optical fibers.


Step 2: Detailed Explanation:

For TIR to occur, two necessary conditions must be met:

Medium of Travel: The light ray must be traveling from a medium with a higher refractive index (optically denser) to a medium with a lower refractive index (optically rarer). For example, from water to air, or from glass to water.

Angle of Incidence: The angle of incidence (\(i\)) in the denser medium must be greater than a specific angle called the critical angle (\(i_c\)). The critical angle is the specific angle of incidence for which the angle of refraction is 90 degrees.


When these two conditions are satisfied, no light is refracted into the rarer medium; instead, the entire ray is reflected back into the denser medium, following the laws of reflection (angle of incidence = angle of reflection).


Step 3: Final Answer:

Total internal reflection is an optical phenomenon in which a light ray traveling from an optically denser medium to an optically rarer medium is completely reflected back into the denser medium. This happens when the angle of incidence at the interface is greater than the critical angle for the pair of media.
Quick Tip: Remember the two key conditions for TIR: 1. Denser to Rarer medium, and 2. Angle of incidence \(>\) Critical angle (\(i > i_c\)). Applications like optical fibers, mirages, and the sparkling of diamonds are all based on this principle.

Step 1: Understanding the Concept:

The work function is a key concept in the study of the photoelectric effect. It represents the binding energy of the outermost (valence) electrons to the metal. To liberate an electron, this energy barrier must be overcome.


Step 2: Detailed Explanation:


Symbol and Units: The work function is typically denoted by \(\phi_0\) (phi-naught) or \(W\). It is a form of energy and is commonly measured in electron-volts (eV), although the SI unit is the joule (J).

Material Property: It is an intrinsic property of a specific material. Different metals have different work functions. For example, alkali metals like cesium have very low work functions, making them good materials for photoelectric devices.

Photoelectric Effect Context: In the photoelectric effect, if a photon with energy \(E = hf\) strikes the metal, an electron is emitted only if the photon's energy is greater than or equal to the work function (\(hf \geq \phi_0\)). The minimum frequency required is called the threshold frequency (\(f_0\)), where \(\phi_0 = hf_0\).



Step 3: Final Answer:

The work-function (\(\phi_0\)) is defined as the minimum energy that must be supplied to an electron to remove it from the surface of a given metal to a point just outside the metal with zero kinetic energy.
Quick Tip: Think of the work function as an "exit fee" for an electron leaving a metal. A photon's energy is the "payment." If the payment is less than the fee (\(E < \phi_0\)), the electron cannot leave. If it's exactly the fee (\(E = \phi_0\)), the electron leaves with no leftover energy (zero kinetic energy).

Step 1: Understanding the Concept:

In quantum mechanics, light is described as being composed of discrete packets of energy called photons. The energy of a single photon is related to the frequency and wavelength of the corresponding electromagnetic wave.


Step 2: Key Formula or Approach:

The derivation involves two fundamental equations:

The Planck-Einstein relation, which gives the energy of a photon (\(E\)) in terms of its frequency (\(f\)):
\[ E = hf \]
where \(h\) is Planck's constant.

The wave equation, which relates the speed of light (\(c\)), its frequency (\(f\)), and its wavelength (\(\lambda\)):
\[ c = f \lambda \]


Step 3: Detailed Explanation:

Our goal is to express the energy \(E\) in terms of \(h\) and \(\lambda\). The first equation has \(h\) but uses \(f\), not \(\lambda\). We need to replace \(f\) using the second equation.

From the wave equation, we can express frequency as: \[ f = \frac{c}{\lambda} \]
Now, substitute this expression for \(f\) into the Planck-Einstein relation: \[ E = h \left( \frac{c}{\lambda} \right) \] \[ E = \frac{hc}{\lambda} \]
This is the required equation.


Step 4: Final Answer:

The equation for the energy of a photon (\(E\)) in terms of Planck's constant (\(h\)), the speed of light (\(c\)), and wavelength (\(\lambda\)) is: \[ E = \frac{hc}{\lambda} \] Quick Tip: This equation shows an inverse relationship between energy and wavelength: shorter wavelengths (like blue or UV light) correspond to higher-energy photons, while longer wavelengths (like red or infrared light) correspond to lower-energy photons. This is a cornerstone of modern physics.

Step 1: Understanding the Concept:

Kirchhoff's laws are two rules that deal with the conservation of charge and energy within electrical circuits. They provide a systematic way to analyze complex circuits that cannot be simplified into simple series or parallel combinations.


Step 2: Detailed Explanation:

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


Statement: The algebraic sum of all electric currents meeting at any junction (or node) in a circuit is zero. In other words, the total current entering a junction must equal the total current leaving that junction.

Principle: This law is based on the law of conservation of charge. Charge cannot accumulate at a junction.

Equation: \(\sum I = 0\)

Circuit Diagram:

Consider a junction 'A' where currents \(I_1\) and \(I_2\) are entering, and currents \(I_3\) and \(I_4\) are leaving.


\begin{circuitikz[american currents]
\draw (0,0) node[circ] (A) { node[above left] {A;
\draw (A) -- ++(-1,1) to[short, i<=\(I_1\)] ++(-1,0);
\draw (A) -- ++(-1,-1) to[short, i<=\(I_2\)] ++(-1,0);
\draw (A) -- ++(1,1) to[short, i>=\(I_3\)] ++(1,0);
\draw (A) -- ++(1,-1) to[short, i>=\(I_4\)] ++(1,0);
\end{circuitikz

According to KCL, if we consider incoming currents as positive and outgoing currents as negative:

\(I_1 + I_2 - I_3 - I_4 = 0\)

or,

\(I_{in} = I_{out} \implies I_1 + I_2 = I_3 + I_4\)



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


Statement: The algebraic sum of the changes in electric potential (voltage) encountered in a complete traversal of any closed loop in a circuit is zero.

Principle: This law is based on the law of conservation of energy. The net energy gained by a charge after moving around a closed loop must be zero.

Equation: \(\sum \Delta V = 0\)

Circuit Diagram:

Consider a simple closed loop containing a voltage source (EMF \(\mathcal{E}\)) and two resistors \(R_1\) and \(R_2\), with a current \(I\) flowing through them.


\begin{circuitikz[american voltages]
\draw (0,0) to[battery, l=\(\mathcal{E}\)] (0,3)
to[R, l=\(R_1\)] (3,3)
to[R, l=\(R_2\)] (3,0)
to[short] (0,0);
\node at (1.5,1.5) [rotate=-90] {\huge\(\circlearrowright\);
\node at (2.1,1.5) {\(I\);
\end{circuitikz

To apply KVL, we choose a direction (e.g., clockwise) and sum the potential changes:

- Traversing the battery from negative to positive terminal: Potential gain of \(+\mathcal{E}\).

- Traversing resistor \(R_1\) in the direction of current: Potential drop of \(-IR_1\).

- Traversing resistor \(R_2\) in the direction of current: Potential drop of \(-IR_2\).

The sum of these potential changes is zero:

\(\mathcal{E} - IR_1 - IR_2 = 0\)



Step 3: Final Answer:

Kirchhoff's two laws are the Junction Rule, which states that the sum of currents at a junction is zero (\(\sum I = 0\)), and the Loop Rule, which states that the sum of potential differences around any closed loop is zero (\(\sum \Delta V = 0\)).
Quick Tip: When applying KVL, establish a consistent sign convention. A common convention is: EMF is positive if traversed from the negative to the positive terminal. Potential drop across a resistor (\(-IR\)) is negative if traversed in the same direction as the current.

Step 1: Understanding the Concept:

When a ray of light passes through a prism, it deviates from its original path. The angle between the incident ray and the emergent ray is called the angle of deviation (\(\delta\)). This angle depends on the angle of incidence (\(i\)). The (\(i - \delta\)) curve is a graphical representation of this relationship.


Step 2: Detailed Explanation and Diagram:

The graph of the angle of deviation (\(\delta\)) as a function of the angle of incidence (\(i\)) has the following characteristics:


Axes: The angle of incidence (\(i\)) is plotted on the X-axis, and the angle of deviation (\(\delta\)) is plotted on the Y-axis.

Shape of the Curve: As the angle of incidence \(i\) is increased from a small value, the angle of deviation \(\delta\) first decreases, reaches a minimum value, and then starts to increase again. This results in a U-shaped curve that is not symmetric.

Angle of Minimum Deviation (\(\delta_m\)): The lowest point on the curve corresponds to the minimum possible value of the angle of deviation, denoted by \(\delta_m\). At this specific point, the angle of incidence is equal to the angle of emergence (\(i = e\)). For every other value of \(\delta\), there are two corresponding values of the angle of incidence.


Graphical Representation:


\begin{tikzpicture
\begin{axis[
xlabel={Angle of Incidence (\(i\)),
ylabel={Angle of Deviation (\(\delta\)),
xmin=20, xmax=100,
ymin=30, ymax=60,
axis lines=left,
ticks=none,
clip=false,
width=0.7\textwidth,
height=0.5\textwidth,
]
\addplot[smooth, thick, blue, domain=30:90] {37 + 0.01*(x-50)^2;
\node[label={below:\(i_m\), circle, fill, inner sep=1.5pt] at (axis cs:50,37) {;
\draw[dashed, gray] (axis cs:20,37) -- (axis cs:50,37) node[left, black] {\(\delta_m\);
\draw[dashed, gray] (axis cs:50,30) -- (axis cs:50,37);
\end{axis
\end{tikzpicture


Step 3: Final Answer:

The (\(i - \delta\)) curve for a prism shows that the angle of deviation initially decreases with an increase in the angle of incidence, reaches a minimum value (\(\delta_m\)), and then increases. The point where the deviation is minimum is called the angle of minimum deviation.
Quick Tip: The condition for minimum deviation is crucial for many prism-related problems. At \(\delta = \delta_m\), we have \(i = e\) and \(r_1 = r_2\), where \(r_1\) and \(r_2\) are the angles of refraction inside the prism. This symmetry simplifies the derivation of the prism formula.

Step 1: Understanding the Concept:

Nuclear fusion is the process that powers stars, including our Sun. It involves the merging of light nuclei, which results in a release of energy due to a phenomenon called mass defect.


Step 2: Detailed Explanation:


Process: In a fusion reaction, two light nuclei, such as isotopes of hydrogen (like deuterium and tritium), are forced together with enough energy to overcome their mutual electrostatic repulsion (the Coulomb barrier). Once they are close enough, the attractive strong nuclear force takes over and fuses them into a single, heavier nucleus (like helium).

Energy Release: The key to energy release is that the mass of the resulting heavier nucleus is slightly less than the sum of the masses of the original light nuclei. This difference in mass (\(\Delta m\)), known as the mass defect, is converted into a tremendous amount of energy (\(E\)) according to Albert Einstein's mass-energy equivalence principle, \(E = \Delta m c^2\), where \(c\) is the speed of light.

Conditions Required: Such reactions require extreme conditions of temperature (on the order of millions of degrees Celsius) and pressure. These conditions, found in the cores of stars, create a state of matter called plasma where nuclei are stripped of their electrons and have enough kinetic energy to overcome their repulsion and fuse.

Example (D-T Fusion): A common example studied for terrestrial fusion reactors is the fusion of deuterium (\(^2_1H\) or D) and tritium (\(^3_1H\) or T):

\[ ^2_1H + ^3_1H \rightarrow ^4_2He + ^1_0n + 17.6 \, MeV (Energy) \]
Here, a deuterium nucleus and a tritium nucleus fuse to form a helium nucleus and a neutron, releasing 17.6 million electron-volts of energy.



Step 3: Final Answer:

Nuclear fusion is a process where light atomic nuclei merge to form a heavier nucleus. This process releases a significant amount of energy because the final product has less mass than the initial reactants, with the lost mass being converted directly into energy. The process requires extremely high temperatures and pressures.
Quick Tip: Contrast nuclear fusion with nuclear fission. Fusion combines light nuclei (e.g., H + H \(\to\) He), while fission splits a heavy nucleus (e.g., Uranium \(\to\) Barium + Krypton). Both release enormous energy, but fusion generally releases more energy per nucleon and produces less long-lived radioactive waste.

Step 1: Understanding the Concept:

This problem requires analyzing a circuit containing diodes and resistors in parallel. The key is to first determine the biasing of each diode (forward or reverse) to see which branches will conduct current. Then, the circuit can be simplified to calculate the total current using Ohm's law.


Step 2: Detailed Explanation:

1. Analyze the Diode Biasing:

The circuit has a 10 V battery. The positive terminal of the battery is connected to the p-side (anode, the triangle part) of all three diodes. The negative terminal is connected (through the resistors) to the n-side (cathode, the line part) of all three diodes.

Since the p-side of each diode is at a higher potential than its n-side, all three diodes are forward-biased.


2. Simplify the Circuit:

Assuming the diodes are ideal, a forward-biased diode acts as a closed switch, meaning it behaves like a wire with zero resistance. Therefore, we can replace all three diodes with connecting wires.

After this simplification, the circuit consists of three resistors connected in parallel across the 10 V battery:

\(R_1 = 6 \, \Omega\)

\(R_2 = 2 \, \Omega\)

\(R_3 = 3 \, \Omega\)



3. Calculate the Equivalent Resistance (\(R_{eq}\)):

For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances: \[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \] \[ \frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{2} + \frac{1}{3} \]
To add these fractions, we find a common denominator, which is 6: \[ \frac{1}{R_{eq}} = \frac{1}{6} + \frac{3}{6} + \frac{2}{6} = \frac{1+3+2}{6} = \frac{6}{6} = 1 \, \Omega^{-1} \]
Therefore, the equivalent resistance is: \[ R_{eq} = 1 \, \Omega \]

4. Calculate the Total Current:

The current through the battery is the total current flowing through the circuit. Using Ohm's Law, \(V = I_{total} \cdot R_{eq}\): \[ I_{total} = \frac{V}{R_{eq}} \] \[ I_{total} = \frac{10 \, V}{1 \, \Omega} = 10 \, A \]

Step 3: Final Answer:

The current through the battery is 10 A.
Quick Tip: When you see diodes in a DC circuit, the very first step is to check their biasing. A forward-biased ideal diode is a short circuit (wire), and a reverse-biased ideal diode is an open circuit (a break in the wire). This simplification is the key to solving such problems.

Step 1: Statement of Gauss's Law:

Gauss's law in electrostatics states that the total electric flux (\(\Phi_E\)) through any closed hypothetical surface (called a Gaussian surface) is equal to \(\frac{1}{\epsilon_0}\) times the net electric charge (\(q_{enc}\)) enclosed within that surface.

Mathematically, it is expressed as: \[ \Phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0} \]
where \(\vec{E}\) is the electric field, \(d\vec{A}\) is the differential area vector on the closed surface \(S\), and \(\epsilon_0\) is the permittivity of free space.


Step 2: Derivation for a Linear Charge Distribution:

Consider an infinitely long, thin, straight wire with a uniform positive linear charge density \(\lambda\) (charge per unit length). We want to find the electric field \(E\) at a point P at a perpendicular distance \(r\) from the wire.


Symmetry and Gaussian Surface: By symmetry, the electric field \(\vec{E}\) must be directed radially outward from the wire, and its magnitude must be the same at all points equidistant from the wire. We choose a cylindrical Gaussian surface of radius \(r\) and length \(L\), coaxial with the wire, such that point P lies on its curved surface.


\begin{tikzpicture[scale=1.5]
% The charged wire
\draw[thick, red] (0,-1.5) -- (0,1.5) node[above] {+ + +;
\node at (0.2, 0) [right] {\(\lambda\);

% The Gaussian cylinder
\draw[blue, dashed] (1,1) arc (90:270:1 and 0.3);
\draw[blue, dashed] (1,-1) -- (-1,-1);
\draw[blue] (-1,-1) arc (270:360:1 and 0.3) arc (0:90:1 and 0.3);
\draw[blue] (-1,1) -- (1,1);
\draw[blue] (-1,-1) -- (-1,1);
\draw[blue] (1,-1) -- (1,1);

% Labels
\draw[<->] (0,0.5) -- (1,0.5) node[midway, above] {\(r\);
\draw[<->] (-1.2,-1) -- (-1.2,1) node[midway, left] {\(L\);

% E-field vector
\draw[->, thick, magenta] (1,0) -- (1.5,0) node[right] {\(\vec{E}\);
% dA vector on curved surface
\draw[->, thick, green] (1,0) -- (1.5,0) node[below left, xshift=15pt, yshift=5pt] {\(d\vec{A}\);
% dA vector on top surface
\draw[->, thick, green] (0,1) -- (0,1.5) node[above] {\(d\vec{A}\);
\node at (0.7, 1.2) [right, magenta] {\(\vec{E}\);
\draw[->, thick, magenta] (0.7, 1) -- (1.2, 1);
\end{tikzpicture


Calculate Electric Flux (\(\Phi_E\)): The flux integral is taken over the entire closed surface, which consists of two flat circular caps (top and bottom) and the curved cylindrical surface.
\[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \int_{top} \vec{E} \cdot d\vec{A} + \int_{bottom} \vec{E} \cdot d\vec{A} + \int_{curved} \vec{E} \cdot d\vec{A} \]

For the top and bottom caps, the area vector \(d\vec{A}\) is perpendicular to the electric field \(\vec{E}\) (i.e., angle is \(90^\circ\)). So, \(\vec{E} \cdot d\vec{A} = E \, dA \cos(90^\circ) = 0\). The flux through the caps is zero.
For the curved surface, the electric field \(\vec{E}\) is parallel to the area vector \(d\vec{A}\) at every point (angle is \(0^\circ\)). So, \(\vec{E} \cdot d\vec{A} = E \, dA \cos(0^\circ) = E \, dA\).

The total flux is therefore only through the curved surface:
\[ \Phi_E = \int_{curved} E \, dA \]
Since \(E\) is constant in magnitude on this surface, we can take it out of the integral:
\[ \Phi_E = E \int_{curved} dA = E \times (Area of curved surface) = E(2\pi r L) \]

Calculate Enclosed Charge (\(q_{enc}\)): The charge enclosed by the Gaussian surface of length \(L\) is the linear charge density multiplied by the length:
\[ q_{enc} = \lambda L \]

Apply Gauss's Law: Now we equate the flux and the enclosed charge according to Gauss's law:
\[ \Phi_E = \frac{q_{enc}}{\epsilon_0} \implies E(2\pi r L) = \frac{\lambda L}{\epsilon_0} \]

Solve for E: Canceling \(L\) from both sides, we get the expression for the electric field:
\[ E = \frac{\lambda}{2\pi \epsilon_0 r} \] Quick Tip: The key to using Gauss's law is choosing the right Gaussian surface. The surface should be chosen such that the electric field is either parallel or perpendicular to the surface vector \(d\vec{A}\), and the magnitude of E is constant on the parts of the surface where the flux is non-zero.

Step 1: Understanding the Concept (Steady State in DC Circuit):

This is a DC circuit containing a capacitor. When a DC voltage is applied, the capacitor charges up. After a long time, it becomes fully charged and its acts as an open circuit, meaning no current can flow through the branch containing the capacitor. This is called the steady state.


Step 2: Circuit Analysis at Steady State:


Current in the 10 \(\Omega\) resistor: At steady state, the capacitor \(C = 10 \, \muF\) is fully charged and blocks the flow of DC current. Therefore, the current in the middle branch, which contains the \(10 \, \Omega\) resistor, becomes zero.
\[ I_{10\Omega} = 0 A \]

Current in the 2 \(\Omega\) resistor: Since no current flows through the middle branch, the circuit simplifies to a single series loop consisting of the \(2\) V battery, the \(2 \, \Omega\) resistor, and the \(8 \, \Omega\) resistor.

The total resistance in this loop is:
\[ R_{total} = 2 \, \Omega + 8 \, \Omega = 10 \, \Omega \]
Using Ohm's law, the current flowing through this loop is:
\[ I_{2\Omega} = \frac{V}{R_{total}} = \frac{2 V}{10 \, \Omega} = 0.2 A \]
This is the current flowing through both the \(2 \, \Omega\) and \(8 \, \Omega\) resistors.


Charge on the Capacitor: To find the charge \(Q\) on the capacitor, we first need to find the potential difference (voltage) \(V_C\) across it. The capacitor and the 10 \(\Omega\) resistor are connected in parallel with the \(8 \, \Omega\) resistor (based on the common interpretation of this circuit's drawing). Therefore, the voltage across the capacitor branch is the same as the voltage across the \(8 \, \Omega\) resistor.
\[ V_C = V_{8\Omega} \]
The voltage across the \(8 \, \Omega\) resistor can be calculated using Ohm's law, with the current we found in the previous step:
\[ V_{8\Omega} = I_{8\Omega} \times R_{8\Omega} = (0.2 A) \times (8 \, \Omega) = 1.6 V \]
So, the voltage across the capacitor is \(V_C = 1.6\) V. Note that because the current through the 10 \(\Omega\) resistor is zero, there is no voltage drop across it, so the full 1.6 V appears across the capacitor plates.

The charge \(Q\) on the capacitor is given by \(Q = C V_C\):
\[ Q = (10 \, \muF) \times (1.6 V) = 16 \, \muC \]


Step 3: Final Answer:


Current in the \(2 \, \Omega\) resistor = 0.2 A.

Current in the \(10 \, \Omega\) resistor = 0 A.

Charge on the capacitor = 16 \(\mu\)C.
Quick Tip: In any DC circuit problem with capacitors, always analyze the circuit at \(t=0\) (capacitor is a short circuit) and \(t \to \infty\) (capacitor is an open circuit). The question usually implies the steady-state condition unless it asks about the transient phase.

Step 1: Statement of Ampere's Circuital Law:

Ampere's law states that the line integral of the magnetic field \(\vec{B}\) around any closed loop (called an Amperian loop) is equal to \(\mu_0\) times the total current \(I_{enc}\) passing through the area enclosed by the loop. \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} \]

Step 2: Derivation for a Long Solenoid:

Consider a long solenoid with \(n\) turns per unit length, carrying a current \(I\).


Symmetry and Magnetic Field: For an ideal long solenoid, the magnetic field inside is strong, uniform, and directed along the axis of the solenoid. The magnetic field outside is negligibly weak (approximately zero).

Amperian Loop: To apply Ampere's law, we choose a rectangular Amperian loop `pqrs` of length \(L\), as shown in the cross-sectional diagram. The side `pq` is inside the solenoid, parallel to the axis, while the side `rs` is outside.


\begin{tikzpicture
% Solenoid turns (cross-section)
\foreach \x in {0,1,2,3,4,5,6 {
\node at (\x, 0.5) {\(\otimes\); % Current into page
\node at (\x, -0.5) {\(\odot\); % Current out of page

\draw[gray] (-0.5, 0.8) -- (6.5, 0.8);
\draw[gray] (-0.5, -0.8) -- (6.5, -0.8);

% Magnetic field lines
\foreach \y in {-0.3,0,0.3 {
\draw[->, blue, thick] (-0.5, \y) -- (6.5, \y);

\node at (3, 0) [above, blue] {\(\vec{B}\);
\node at (3, 1.2) [above, blue] {\(\vec{B} \approx 0\);

% Amperian loop
\draw[red, thick, ->] (1, -0.2) -- (5, -0.2) node[midway, below] {\(p \to q\);
\draw[red, thick, ->] (5, -0.2) -- (5, 1.2) node[midway, right] {\(q \to r\);
\draw[red, thick, ->] (5, 1.2) -- (1, 1.2) node[midway, above] {\(r \to s\);
\draw[red, thick, ->] (1, 1.2) -- (1, -0.2) node[midway, left] {\(s \to p\);
\node at (0.8, -0.2) [below left] {\(p\);
\node at (5.2, -0.2) [below right] {\(q\);
\node at (5.2, 1.2) [above right] {\(r\);
\node at (0.8, 1.2) [above left] {\(s\);
\draw[<->] (1,-1) -- (5,-1) node[midway, below] {\(L\);
\end{tikzpicture


Evaluate the Line Integral: We evaluate \(\oint \vec{B} \cdot d\vec{l}\) for the loop `pqrs`.
\[ \oint_{pqrs} \vec{B} \cdot d\vec{l} = \int_{p}^{q} \vec{B} \cdot d\vec{l} + \int_{q}^{r} \vec{B} \cdot d\vec{l} + \int_{r}^{s} \vec{B} \cdot d\vec{l} + \int_{s}^{p} \vec{B} \cdot d\vec{l} \]

Along pq (inside): \(\vec{B}\) is parallel to \(d\vec{l}\). So, \(\int_{p}^{q} \vec{B} \cdot d\vec{l} = \int BL \cos(0^\circ) dl = B \int_{0}^{L} dl = BL\).
Along qr and sp (perpendicular): \(\vec{B}\) is perpendicular to \(d\vec{l}\). So, \(\int \vec{B} \cdot d\vec{l} = \int B dl \cos(90^\circ) = 0\).
Along rs (outside): The magnetic field outside is zero (\(B=0\)). So, \(\int_{r}^{s} \vec{B} \cdot d\vec{l} = 0\).

The total value of the line integral is: \(\oint \vec{B} \cdot d\vec{l} = BL + 0 + 0 + 0 = BL\).

Calculate Enclosed Current (\(I_{enc}\)): The number of turns within the loop of length \(L\) is \(N = nL\). Since each turn carries current \(I\), the total current enclosed by the loop is:
\[ I_{enc} = N \times I = nLI \]

Apply Ampere's Law:
\[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} \implies BL = \mu_0 (nLI) \]

Solve for B: Canceling \(L\) from both sides gives the magnetic field inside the solenoid:
\[ B = \mu_0 n I \] Quick Tip: Remember that \(n\) in the formula \(B = \mu_0 n I\) is the number of turns per unit length (\(n = N/L\)), not the total number of turns. This is a common point of confusion. The formula shows that the field inside a long solenoid is independent of its radius and the position within the solenoid.

Step 1: Understanding the Concept (Motional EMF):

When a conducting rod rotates in a uniform magnetic field, an electromotive force (EMF) is induced across its ends. This is a type of motional EMF. Each point on the rod moves with a different linear speed, so we must integrate the small EMFs generated across infinitesimal segments of the rod to find the total EMF.


Step 2: Key Formula and Derivation:

Consider a small element of the rod of length \(dr\) at a distance \(r\) from the axis of rotation.

The linear velocity of this element is given by \(v = \omega r\), where \(\omega\) is the angular frequency.

The small EMF induced across this element \(dr\) is: \[ d\mathcal{E} = B v dr = B(\omega r) dr \]
To find the total EMF induced between the axis (\(r=0\)) and the outer end (the ring at \(r=L\)), we integrate this expression from \(0\) to \(L\): \[ \mathcal{E} = \int_{0}^{L} d\mathcal{E} = \int_{0}^{L} B\omega r \, dr \]
Since \(B\) and \(\omega\) are constant: \[ \mathcal{E} = B\omega \int_{0}^{L} r \, dr = B\omega \left[ \frac{r^2}{2} \right]_{0}^{L} = B\omega \left( \frac{L^2}{2} - 0 \right) \] \[ \mathcal{E} = \frac{1}{2} B \omega L^2 \]

Step 3: Calculation:

We are given the following values:

Length of the rod, \(L = 1.0\) m.
Angular frequency, \(\omega = 400\) rad/s.
Magnetic field strength, \(B = 0.5\) T.

Substitute these values into the derived formula: \[ \mathcal{E} = \frac{1}{2} (0.5 T) (400 rad/s) (1.0 m)^2 \] \[ \mathcal{E} = \frac{1}{2} \times 0.5 \times 400 \times 1 \] \[ \mathcal{E} = 0.25 \times 400 \] \[ \mathcal{E} = 100 V \]

Step 4: Final Answer:

The induced emf between the ring and the axis is 100 V.
Quick Tip: A common mistake is to use the linear velocity of the end of the rod (\(v = \omega L\)) and the formula for a linearly moving rod (\(\mathcal{E} = BvL\)). This gives \(\mathcal{E} = B(\omega L)L = B\omega L^2\), which is incorrect by a factor of 2. Always remember the factor of \(\frac{1}{2}\) for a rotating rod, which comes from the integration.

Step 1: Understanding the Concept and Setup:

The Lens Maker's formula relates the focal length (\(f\)) of a thin lens to the refractive index (\(\mu\)) of its material and the radii of curvature (\(R_1\) and \(R_2\)) of its two surfaces. It is derived by applying the formula for refraction at a single spherical surface twice.

Consider a thin convex lens of refractive index \(\mu_2\) placed in a medium of refractive index \(\mu_1\). A point object O is placed on the principal axis.


\begin{tikzpicture
% Principal Axis
\draw[<->] (-4,0) -- (4,0);
% Lens
\draw (0,1.5) .. controls (0.5,0) .. (0,-1.5);
\draw (0,1.5) .. controls (-0.5,0) .. (0,-1.5);
\node at (0, -2) {Lens (\(\mu_2\));
\node at (2.5, -2) {Medium (\(\mu_1\));
% Object
\node[label=below:O] at (-3,0) {\(\bullet\);
% Image 1
\node[label=below:\(I'\)] at (3.5,0) {\(\circ\);
% Image 2
\node[label=below:I] at (2,0) {\(\bullet\);
% Rays
\draw[->, thick, red] (-3,0) -- (0,0.8);
\draw[->, thick, red, dashed] (0,0.8) -- (3.5,0);
\draw[->, thick, red] (0,0.8) -- (2,0);
% Distances
\draw[<->] (-3,-0.5) -- (0,-0.5) node[midway, below] {\(u\);
\draw[<->] (0,-0.5) -- (2,-0.5) node[midway, below] {\(v\);
\draw[<->] (0,-1) -- (3.5,-1) node[midway, below] {\(v'\);
\end{tikzpicture


Step 2: Refraction at the First Surface (Radius \(R_1\)):

Light travels from the medium (\(\mu_1\)) to the lens (\(\mu_2\)). For the first surface, the object is at O (distance \(u\)). An intermediate image \(I'\) is formed at a distance \(v'\). The formula for refraction at a single spherical surface is: \[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]
Applying this to our case: \[ \frac{\mu_2}{v'} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R_1} \quad \cdots (1) \]

Step 3: Refraction at the Second Surface (Radius \(R_2\)):

The image \(I'\) formed by the first surface acts as a virtual object for the second surface. Light now travels from the lens (\(\mu_2\)) back to the medium (\(\mu_1\)). The final image is formed at I (distance \(v\)).

For this surface, the object distance is \(v'\). Applying the single surface formula (with \(\mu_1\) and \(\mu_2\) interchanged): \[ \frac{\mu_1}{v} - \frac{\mu_2}{v'} = \frac{\mu_1 - \mu_2}{R_2} \] \[ \frac{\mu_1}{v} - \frac{\mu_2}{v'} = -\frac{\mu_2 - \mu_1}{R_2} \quad \cdots (2) \]

Step 4: Combining the Equations:

Adding equation (1) and equation (2): \[ \left( \frac{\mu_2}{v'} - \frac{\mu_1}{u} \right) + \left( \frac{\mu_1}{v} - \frac{\mu_2}{v'} \right) = \left( \frac{\mu_2 - \mu_1}{R_1} \right) - \left( \frac{\mu_2 - \mu_1}{R_2} \right) \]
The term \(\frac{\mu_2}{v'}\) cancels out: \[ \frac{\mu_1}{v} - \frac{\mu_1}{u} = (\mu_2 - \mu_1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Dividing the entire equation by \(\mu_1\): \[ \frac{1}{v} - \frac{1}{u} = \left(\frac{\mu_2}{\mu_1} - 1\right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Step 5: Introducing Focal Length (\(f\)):

By definition, when the object is at infinity (\(u = \infty\)), the image is formed at the focal point (\(v = f\)). Substituting this into the above equation: \[ \frac{1}{f} - \frac{1}{\infty} = \left(\frac{\mu_2}{\mu_1} - 1\right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Since \(\frac{1}{\infty} = 0\), we get: \[ \frac{1}{f} = \left(\frac{\mu_2}{\mu_1} - 1\right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
If the surrounding medium is air or vacuum (\(\mu_1 = 1\)) and the lens material has refractive index \(\mu_2 = \mu\), the formula simplifies to its most common form: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
This is the Lens Maker's formula. Quick Tip: The derivation hinges on the "thin lens approximation," where the thickness of the lens is considered negligible. This allows us to measure all distances from a single point (the optical center) and to use the image distance from the first surface directly as the object distance for the second surface. Remember to use the Cartesian sign convention consistently for \(u, v, R_1,\) and \(R_2\) when solving problems.

1. Radial Magnetic Field:

A radial magnetic field is a magnetic field in which the field lines are always directed along the radii from the center of a circle. In the context of a moving coil galvanometer, it is produced by using cylindrically concave pole pieces of a magnet and placing a soft iron core at the center. This ensures that the plane of the coil is always parallel to the magnetic field lines, and thus the angle between the magnetic field vector (\(\vec{B}\)) and the area vector of the coil (\(\vec{A}\)) is always \(90^\circ\), maximizing the torque.


2. Principle and Working of Moving Coil Galvanometer:


Principle: The working of a moving coil galvanometer is based on the principle that a current-carrying loop placed in a uniform magnetic field experiences a torque.

Working: When a current \(I\) flows through the rectangular coil of \(N\) turns and area \(A\), placed in a magnetic field \(B\), it experiences a deflecting torque (\(\tau_d\)).
\[ \tau_d = NIAB \sin\theta \]
In a radial magnetic field, \(\theta = 90^\circ\), so \(\sin\theta = 1\). The deflecting torque is maximum and constant for any position of the coil:
\[ \tau_d = NIAB \]
This torque causes the coil to rotate. As the coil rotates, the suspension wire gets twisted, producing a restoring torque (\(\tau_r\)) that is proportional to the angle of deflection (\(\phi\)).
\[ \tau_r = k\phi \]
where \(k\) is the torsional constant of the spring. In equilibrium, the deflecting torque equals the restoring torque:
\[ NIAB = k\phi \implies I = \left(\frac{k}{NAB}\right) \phi \]
Since \((k/NAB)\) is a constant, the current is directly proportional to the deflection (\(I \propto \phi\)). This allows for a linear scale for current measurement.



Diagram:

\begin{tikzpicture
% Poles
\node at (-2,0) [left] {N;
\node at (2,0) [right] {S;
\draw[thick] (-1.5,1.5) arc (90:270:1.5);
\draw[thick] (1.5,1.5) arc (90:-90:1.5);
% Iron Core
\draw[fill=gray!30, thick] (0,0) circle (0.8);
\node at (0,0) {Soft Iron Core;
% Radial Field Lines
\foreach \angle in {45,90,135,225,270,315 {
\draw[->, blue] (\angle:0.8) -- (\angle:1.5);

% Coil
\draw[red, thick] (0.8,0.5) -- (0.8,-0.5);
\draw[red, thick] (-0.8,0.5) -- (-0.8,-0.5);
\end{tikzpicture


3. Increasing Sensitivity:

The current sensitivity of a galvanometer is defined as the deflection produced per unit current, \(S_i = \frac{\phi}{I}\). From the equilibrium equation, we have: \[ S_i = \frac{\phi}{I} = \frac{NAB}{k} \]
To increase the sensitivity, we can:

Increase the number of turns in the coil (\(N\)).
Increase the magnetic field strength (\(B\)).
Increase the area of the coil (\(A\)).
Decrease the torsional constant (\(k\)) of the suspension fiber by using a material like phosphor-bronze. Quick Tip: The purpose of the radial field is to make the torque (\(\tau = NIAB\sin\theta\)) maximum (\(\theta=90^\circ\)) and independent of the coil's orientation. This results in a linear relationship between current and deflection (\(I \propto \phi\)), which is essential for a calibrated measuring instrument.

Step 1: Understanding the Concept and Given Data:

This problem deals with a sinusoidal alternating current (AC). We are given the peak value (\(I_0\)) and the frequency (\(f\)) and need to find its RMS value, the time to reach the first peak, and visualize its waveform.

Peak current, \(I_0 = 14.14\) A.
Frequency, \(f = 50\) Hz.


Step 2: Key Formulas and Calculations:

1. R.M.S. Value of Current (\(I_{rms}\)):

The relationship between the peak value and the RMS value of a sinusoidal current is: \[ I_{rms} = \frac{I_0}{\sqrt{2}} \]
We are given \(I_0 = 14.14\) A. Note that \(14.14 \approx 10 \times 1.414 = 10\sqrt{2}\). \[ I_{rms} = \frac{10\sqrt{2}}{\sqrt{2}} = 10 \, A \]

2. Time to Reach the Peak Value:

The time period of one cycle is \(T = \frac{1}{f}\). \[ T = \frac{1}{50 \, Hz} = 0.02 \, s \]
A sinusoidal current starts from zero, reaches its first positive peak at one-quarter of the time period (\(t = T/4\)). \[ t_{peak} = \frac{T}{4} = \frac{0.02 \, s}{4} = 0.005 \, s \]

Step 3: Current-Time Graph for Two Cycles:

We need to draw the graph for a time duration of \(2T = 2 \times 0.02 = 0.04\) s. The graph is a sine wave with an amplitude of 14.14 A.

\begin{tikzpicture
\begin{axis[
xlabel={Time (s),
ylabel={Current (A),
xmin=0, xmax=0.045,
ymin=-16, ymax=16,
axis lines=middle,
grid=both,
xtick={0, 0.005, 0.01, 0.015, 0.02, 0.025, 0.03, 0.035, 0.04,
xticklabels={0, T/4, T/2, 3T/4, T, 5T/4, 3T/2, 7T/4, 2T,
ytick={-14.14, 0, 14.14,
yticklabels={-14.14, 0, 14.14,
width=0.9\textwidth,
height=0.5\textwidth,
]
\addplot[smooth, thick, blue, domain=0:0.04, samples=200] {14.14*sin(2*pi*50*x*180/pi);
\draw[dashed, gray] (axis cs:0.005,0) -- (axis cs:0.005,14.14);
\draw[dashed, gray] (axis cs:0,14.14) -- (axis cs:0.005,14.14);
\end{axis
\end{tikzpicture


Step 4: Final Answer:

The r.m.s. value of the current is 10 A.
The time taken for the current to reach the peak value starting from zero is 0.005 s. Quick Tip: In AC problems, values like 14.14, 1.414, or 0.707 are often hints to use \(\sqrt{2}\). Specifically, \(14.14 \approx 10\sqrt{2}\) and \(0.707 \approx 1/\sqrt{2}\). Recognizing these quickly can save a lot of calculation time.

Step 1: Understanding the Concept:

In a plane electromagnetic (EM) wave propagating in a vacuum, the electric field (\(E\)) and magnetic field (\(B\)) are mutually perpendicular and also perpendicular to the direction of wave propagation. Their amplitudes (\(E_0\) and \(B_0\)) are related by the speed of light (\(c\)). The frequency (\(\nu\)), angular frequency (\(\omega\)), and wavelength (\(\lambda\)) are related through the wave speed equation.


Step 2: Key Formulas:

The formulas needed to solve this problem are:

Relation between electric and magnetic field amplitudes: \(c = \frac{E_0}{B_0}\)
Relation between angular frequency and linear frequency: \(\omega = 2\pi\nu\)
Relation between wave speed, frequency, and wavelength: \(c = \nu\lambda\)

We will use the constant value for the speed of light in vacuum, \(c = 3 \times 10^8\) m/s.


Step 3: Detailed Calculations:

First, convert the given frequency to SI units: \[ \nu = 50 \, MHz = 50 \times 10^6 \, Hz \]
Given \(E_0 = 150\) N/C.

(i) Amplitude of magnetic field (\(B_0\)):

From the formula \(c = E_0 / B_0\), we can rearrange to find \(B_0\): \[ B_0 = \frac{E_0}{c} \]
Substitute the given values: \[ B_0 = \frac{150 \, N/C}{3 \times 10^8 \, m/s} = 50 \times 10^{-8} \, T \] \[ B_0 = 5 \times 10^{-7} \, T \]

(ii) Angular frequency (\(\omega\)):

Using the formula \(\omega = 2\pi\nu\): \[ \omega = 2\pi \times (50 \times 10^6 \, Hz) = 100\pi \times 10^6 \, rad/s \] \[ \omega = \pi \times 10^8 \, rad/s \]
Substituting \(\pi \approx 3.14\): \[ \omega \approx 3.14 \times 10^8 \, rad/s \]

(iii) Wavelength (\(\lambda\)):

From the formula \(c = \nu\lambda\), we can rearrange to find \(\lambda\): \[ \lambda = \frac{c}{\nu} \]
Substitute the values: \[ \lambda = \frac{3 \times 10^8 \, m/s}{50 \times 10^6 \, Hz} = \frac{300 \times 10^6 \, m/s}{50 \times 10^6 \, Hz} \] \[ \lambda = \frac{300}{50} \, m = 6 \, m \]

Step 4: Final Answer:


(i) Amplitude of magnetic field, \(B_0 = 5 \times 10^{-7}\) T.
(ii) Angular frequency, \(\omega = \pi \times 10^8\) rad/s.
(iii) Wavelength, \(\lambda = 6\) m. Quick Tip: A very useful and easy way to remember the relationship between E and B in an EM wave is \(E = cB\). Always ensure that all quantities are in their base SI units (Hz for frequency, m/s for speed, etc.) before starting calculations to avoid errors.

Step 1: Understanding the Concept:

James Clerk Maxwell discovered an inconsistency in Ampere's Circuital Law, particularly when applied to situations with time-varying electric fields, such as the charging or discharging of a capacitor. To resolve this, he introduced the concept of displacement current.


Step 2: Explanation of Displacement Current:

Displacement current (\(I_D\)) is not a current in the traditional sense of flowing charges (which is conduction current, \(I_C\)). Instead, it is a theoretical current that is associated with a changing electric field.

Consider the gap between the plates of a capacitor being charged. While charges flow in the connecting wires (\(I_C\)), no charge actually crosses the gap. However, as charge accumulates on the plates, the electric field (\(E\)) between the plates changes with time. This changing electric field produces a magnetic field in the gap, just as a real current would.

Maxwell defined this displacement current as being proportional to the rate of change of electric flux (\(\Phi_E\)) through a surface.

The displacement current is given by the formula: \[ I_D = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \(\Phi_E = \int \vec{E} \cdot d\vec{A}\) is the electric flux and \(\epsilon_0\) is the permittivity of free space. The existence of displacement current implies that a changing electric field is a source of a magnetic field.


Step 3: Modified Ampere's Law (Ampere-Maxwell Law):

The original Ampere's law was \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_C\). Maxwell modified this by adding the displacement current term to the total current.

The modified law, known as the Ampere-Maxwell Law, states that the line integral of the magnetic field around a closed loop is proportional to the sum of the conduction current (\(I_C\)) and the displacement current (\(I_D\)) passing through the surface enclosed by the loop.

The equation is: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 (I_C + I_D) \]
Substituting the expression for \(I_D\), we get: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 \left( I_C + \epsilon_0 \frac{d\Phi_E}{dt} \right) \] Quick Tip: Remember the symmetry introduced by Maxwell: Faraday's law of induction states that a changing magnetic field produces an electric field. The concept of displacement current completes this symmetry by stating that a changing electric field produces a magnetic field. These two ideas are the foundation of electromagnetic waves.

1. Plane-Polarised Light:

An ordinary light wave (unpolarised light) consists of electric field vectors oscillating in all possible directions perpendicular to the direction of wave propagation. Plane-polarised light is a light wave in which the electric field oscillations are restricted to a single plane containing the direction of propagation. This plane is called the plane of polarisation.


2. Distinguishing between types of light using a Polaroid:

A polaroid (also called an analyser) is a device that only allows light with a specific polarisation orientation to pass through it. By passing the incident light through a polaroid and rotating the polaroid by \(360^\circ\), we can observe the intensity of the transmitted light and distinguish between the three types:



Case 1: Ordinary (Unpolarised) Light:

Process: When unpolarised light is passed through a polaroid, the transmitted intensity is always half of the incident intensity (\(I = I_0/2\)).
Observation: As the polaroid is rotated, the intensity of the transmitted light remains constant. There is no variation in brightness.


Case 2: Totally (Plane) Polarised Light:

Process: When plane-polarised light is passed through a polaroid, the transmitted intensity follows Malus's Law: \(I = I_{max} \cos^2\theta\), where \(\theta\) is the angle between the polarisation plane of the light and the pass-axis of the polaroid.
Observation: As the polaroid is rotated, the intensity of the transmitted light varies from a maximum value to zero. The light will be completely extinguished (intensity becomes zero) at two positions in a full \(360^\circ\) rotation (when \(\theta = 90^\circ\) and \(\theta = 270^\circ\)).


Case 3: Partially Polarised Light:

Process: Partially polarised light is a mixture of unpolarised and plane-polarised light.
Observation: As the polaroid is rotated, the intensity of the transmitted light varies between a maximum value and a non-zero minimum value. Unlike totally polarised light, the intensity never becomes zero. Quick Tip: The key to distinguishing the types of light is to look for the minimum intensity when rotating the analyser. No change in intensity \(\implies\) Unpolarised. Intensity becomes zero \(\implies\) Completely Polarised. Intensity has a non-zero minimum \(\implies\) Partially Polarised.

Step 1: Understanding the Concept (Photoelectric Effect):

This problem uses Einstein's photoelectric equation, which describes the energy balance when a photon strikes a metal surface and ejects an electron (a photoelectron). The energy of the incident photon is used partly to overcome the work function of the metal and the rest is converted into the kinetic energy of the ejected electron.


Step 2: Key Formula or Approach:

Einstein's photoelectric equation is: \[ K.E._{max} = E_{photon} - \phi_0 \]
where:

\(K.E._{max}\) is the maximum kinetic energy of the ejected photoelectron.
\(E_{photon}\) is the energy of the incident photon.
\(\phi_0\) is the work function of the metal.

The energy of a photon can be calculated from its wavelength \(\lambda\). A very useful shortcut for calculations where wavelength is in Angstroms (Å) and energy is needed in electron-volts (eV) is: \[ E_{photon} (in eV) = \frac{hc}{e\lambda} \approx \frac{12400}{\lambda (in Å)} \]

Step 3: Detailed Calculation:


Given Data:

Wavelength of incident light, \(\lambda = 3300\) Å.
Work function of Cesium, \(\phi_0 = 1.9\) eV.


Calculate Photon Energy (\(E_{photon}\)):
Using the shortcut formula:
\[ E_{photon} = \frac{12400}{\lambda (in Å)} \, eV = \frac{12400}{3300} \, eV \]
\[ E_{photon} = \frac{124}{33} \, eV \approx 3.757 \, eV \]
Let's use \(E_{photon} \approx 3.76\) eV for our calculation.

Calculate Maximum Kinetic Energy (\(K.E._{max}\)):
Now, substitute the values into the photoelectric equation:
\[ K.E._{max} = E_{photon} - \phi_0 \]
\[ K.E._{max} = 3.76 \, eV - 1.9 \, eV \]
\[ K.E._{max} = 1.86 \, eV \]


Step 4: Final Answer:

The maximum kinetic energy of the ejected photo-electrons is 1.86 eV. Quick Tip: Memorizing the constant combination \(hc \approx 1240 \, eV \cdot nm\) or \(12400 \, eV \cdot Å\) is extremely helpful for modern physics problems. It allows you to directly convert wavelength in nanometers or Angstroms to photon energy in electron-volts without dealing with Planck's constant, the speed of light, and the charge of an electron individually.

Part 1: Deduction of Potential Energy of a Capacitor


Step 1: Understanding the Concept:

The potential energy stored in a capacitor is equal to the total work done in charging the capacitor. This work is done by an external agent (like a battery) to transfer charge from one plate to another against the electrostatic forces.


Step 2: Derivation:

Let's consider a capacitor of capacitance \(C\). At an intermediate stage during charging, let \(q\) be the charge on the plates and \(V\) be the potential difference between them. By definition, \(V = q/C\).

The work done (\(dW\)) to transfer an additional infinitesimal charge \(dq\) from the negative plate to the positive plate is given by: \[ dW = V \, dq \]
Substituting \(V = q/C\): \[ dW = \frac{q}{C} \, dq \]
To find the total work done (\(W\)) in charging the capacitor from an initial charge of 0 to a final charge of \(Q\), we integrate \(dW\): \[ W = \int dW = \int_{0}^{Q} \frac{q}{C} \, dq \]
Since \(C\) is a constant: \[ W = \frac{1}{C} \int_{0}^{Q} q \, dq = \frac{1}{C} \left[ \frac{q^2}{2} \right]_{0}^{Q} = \frac{1}{C} \left( \frac{Q^2}{2} - 0 \right) = \frac{Q^2}{2C} \]
This work done is stored as electrostatic potential energy (\(U\)) in the capacitor. \[ U = \frac{Q^2}{2C} \]
Using the relation \(Q = CV\), we can express the energy in other forms:

\(U = \frac{(CV)^2}{2C} = \frac{1}{2}CV^2\)
\(U = \frac{Q^2}{2(Q/V)} = \frac{1}{2}QV\)

Thus, the potential energy is \(U = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV\).


Part 2: Derivation of Energy Density


Step 1: Understanding the Concept:

Energy density (\(u_E\)) is the electrostatic potential energy stored per unit volume in the electric field.


Step 2: Derivation:

Consider a parallel plate capacitor with plate area \(A\) and separation distance \(d\). The volume of the space between the plates is \(Volume = A \times d\).

The capacitance of this capacitor is given by: \[ C = \frac{\epsilon_0 A}{d} \]
The electric field \(E\) between the plates is uniform and is related to the potential difference \(V\) by: \[ E = \frac{V}{d} \implies V = Ed \]
Now, let's use the potential energy formula \(U = \frac{1}{2}CV^2\) and substitute the expressions for \(C\) and \(V\): \[ U = \frac{1}{2} \left( \frac{\epsilon_0 A}{d} \right) (Ed)^2 = \frac{1}{2} \frac{\epsilon_0 A}{d} E^2 d^2 \] \[ U = \frac{1}{2} \epsilon_0 E^2 (A \cdot d) \]
Energy density (\(u_E\)) is the energy per unit volume: \[ u_E = \frac{Energy (U)}{Volume} = \frac{U}{A \cdot d} \] \[ u_E = \frac{\frac{1}{2} \epsilon_0 E^2 (A \cdot d)}{A \cdot d} \] \[ u_E = \frac{1}{2} \epsilon_0 E^2 \]
This shows that the energy density in the electric field between the plates of a charged capacitor is \(\frac{1}{2} \epsilon_0 E^2\). Quick Tip: The three forms of capacitor energy (\(U = \frac{1}{2}CV^2\), \(U = \frac{Q^2}{2C}\), \(U = \frac{1}{2}QV\)) are all equivalent. Choose the most convenient one based on the quantities given in a problem. The energy density formula is general and applies to any electric field, not just the one in a capacitor.

Part 1: Unit Derivation


Step 1: Understanding the Concept:

We need to show the equivalence of two sets of units by breaking them down into their fundamental definitions.


Step 2: Derivation:

Let's start with the Left Hand Side (LHS): \(\frac{Farad}{meter}\).

By definition, capacitance \(C = Q/V\), so the unit Farad (F) is equivalent to Coulomb per Volt (C/V). \[ LHS = \frac{C/V}{m} = \frac{C}{V \cdot m} \]
The unit for potential difference, Volt (V), is defined as energy per unit charge, which is Joule per Coulomb (J/C). \[ V = \frac{J}{C} \]
The unit for energy, Joule (J), is defined as work done, which is force times distance, or Newton-meter (\(N \cdot m\)). \[ J = N \cdot m \]
Substituting these back into our expression for the LHS: \[ V = \frac{N \cdot m}{C} \]
Now, substitute this expression for Volt into the LHS: \[ LHS = \frac{C}{\left(\frac{N \cdot m}{C}\right) \cdot m} = \frac{C^2}{N \cdot m^2} \]
In terms of base units, this is \(\frac{coulomb^2}{newton \times meter^2}\), which is exactly the Right Hand Side (RHS). Thus, the relation is proven.


Part 2: Name the Physical Quantity

The unit \(\frac{Farad}{meter}\) is the SI unit for **electric permittivity (\(\epsilon\))**. The permittivity of free space, \(\epsilon_0\), has a value of approximately \(8.854 \times 10^{-12}\) F/m.


Part 3: Circuit Problem


Step 1: Understanding the Circuit:

The circuit shows two capacitors, \(C_1 = 4 \, \muF\) and \(C_2 = 6 \, \muF\), connected in series to a battery of voltage \(V\).


Step 2: Key Principles for Series Capacitors:


The charge stored on each capacitor is the same: \(Q_1 = Q_2 = Q_{total}\).
The total voltage of the battery is the sum of the voltages across each capacitor: \(V = V_1 + V_2\).


Step 3: Calculation:

We are given the potential difference across the \(6 \, \muF\) capacitor: \[ V_2 = 2 \, V \]
First, we find the charge stored on this capacitor using \(Q = CV\): \[ Q_2 = C_2 V_2 = (6 \, \muF)(2 \, V) = 12 \, \muC \]
Since the capacitors are in series, the charge on the \(4 \, \muF\) capacitor is the same: \[ Q_1 = Q_2 = 12 \, \muC \]
Now, we can find the voltage across the \(4 \, \muF\) capacitor: \[ V_1 = \frac{Q_1}{C_1} = \frac{12 \, \muC}{4 \, \muF} = 3 \, V \]
The total potential difference across the battery is the sum of the individual voltages: \[ V = V_1 + V_2 = 3 \, V + 2 \, V = 5 \, V \]

Step 4: Final Answer:

The potential difference across the ends of the battery is 5 V. Quick Tip: For capacitors in series, the charge is the same on all of them. This is the key piece of information. Calculate the charge on the one capacitor where you have enough information (C and V), and then use that charge value for all other capacitors in the series branch.

1. Definition of Wavefront:

A wavefront is defined as the continuous locus of all the particles of a medium which are vibrating in the same phase at any given instant. The direction of propagation of the wave is always perpendicular to the wavefront.


2. Refraction of Waves using Huygen's Principle:

Let us consider a plane wavefront AB incident on a refracting surface XY, separating two media, medium 1 (rarer) and medium 2 (denser). Let the speed of light in the two media be \(v_1\) and \(v_2\) respectively, with \(v_1 > v_2\). The angle of incidence is \(i\).

\begin{tikzpicture[scale=1.5]
% Interface
\draw[thick] (-2,0) -- (3,0) node[right] {Y;
\node at (-2,0) [left] {X;
\node at (0, 1.5) {Medium 1 (speed \(v_1\));
\node at (0, -1) {Medium 2 (speed \(v_2\));

% Incident rays and wavefront
\draw[->, thick, blue] (-1, 2) -- (0,0);
\draw[->, thick, blue] (1.5, 2) -- (2.25, 0);
\draw[thick, red] (-1,2) node[above left] {A -- (1.5,2) node[above right] {B;

% Normals
\draw[dashed] (0,0) -- (0,-2);
\draw[dashed] (2.25,0) -- (2.25,-2);

% Angles of incidence
\draw (0,0.8) arc (90:230:0.8);
\node at (-0.3, 0.5) {\(i\);
\draw (2.25,0.8) arc (90:230:0.8);
\node at (1.95, 0.5) {\(i\);

% Construction
\node at (2.25,0) [above] {C;
\node at (0,0) [above] {A';

% Time calculation
% Let time for B to reach C be t. BC = v1*t.
% BC is hypotenuse of triangle ABC. AC = 2.25 - (-1) = 3.25. Angle is i.
% sin(i) = BC/AC'. A' is (0,2). C is (2.25,0). No, this is complex.
% Let's use the standard way. t = BC/v1. In time t, wavelet from A' travels AD = v2*t.

% Draw the refracted wavefront
% BC = A'C * sin(i). Let A' be origin for a moment. No. Let A' be (0,0).
% B is at (some x, some y). This is too complex. Let's draw it schematically.
\draw[red] (1.5,2) -- (2.25,0); % Ray BC
\node at (1.7, 1.1) {\(v_1 t\);

% Secondary wavelet from A'
% AD = v_2*t = v_2 * (BC/v_1)
\draw[dashed, blue] (0,0) circle (1.2); % AD = v_2*t
\node at (0.9, -0.9) [below right] {D;

% Refracted wavefront
\draw[thick, red] (2.25, 0) -- (0.9, -0.9); % Tangent CD

% Refracted ray
\draw[->, thick, blue] (0,0) -- (0.9, -0.9);

% Angle of refraction
\draw (0, -0.5) arc (270:210:0.5);
\node at (0.2, -0.4) {\(r\);

\end{tikzpicture


According to Huygen's principle, every point on the incident wavefront AB is a source of secondary wavelets.
When the wavefront strikes the surface at point A, A becomes a source of secondary wavelets in the second medium. Meanwhile, point B is still in the first medium.
Let \(t\) be the time taken for the disturbance from point B to reach point C on the surface. Therefore, the distance \(BC = v_1 t\).
In the same time \(t\), the secondary wavelets from point A will travel a distance \(AD = v_2 t\) in the second medium.
To find the new refracted wavefront, we draw a sphere (a circle in 2D) of radius \(AD = v_2 t\) with A as the center.
The tangent drawn from point C to this sphere gives the new refracted wavefront, CD.
Now, in the right-angled triangle ABC:
\[ \sin i = \frac{BC}{AC} = \frac{v_1 t}{AC} \]
And in the right-angled triangle ADC:
\[ \sin r = \frac{AD}{AC} = \frac{v_2 t}{AC} \]
Dividing the two equations, we get:
\[ \frac{\sin i}{\sin r} = \frac{v_1 t / AC}{v_2 t / AC} = \frac{v_1}{v_2} \]
The ratio of the speeds of light is equal to the refractive index of the second medium with respect to the first, \(n_{21}\).
\[ \frac{\sin i}{\sin r} = \frac{v_1}{v_2} = n_{21} \]

This is Snell's law of refraction, thus proving the law using Huygen's principle. Quick Tip: The key to proving laws of reflection or refraction using Huygen's principle is to correctly identify the distances traveled by the wavelets from two different points on the incident wavefront in the same interval of time. The new wavefront is the common tangent to all these secondary wavelets.

Step 1: Understanding the Concept and Given Data:

We are dealing with an astronomical telescope where the object is at a finite distance, not at infinity. For "normal vision" or "normal adjustment", the final image must be formed at infinity. This means the intermediate image formed by the objective lens must be located at the principal focus of the eyepiece.

Focal length of objective, \(f_o = +40\) cm.
Focal length of eyepiece, \(f_e = +4\) cm.
Object distance for objective, \(u_o = -200\) cm.


Step 2: Find the position of the intermediate image (\(v_o\)):

We use the lens formula for the objective lens: \[ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} \] \[ \frac{1}{40} = \frac{1}{v_o} - \frac{1}{-200} \] \[ \frac{1}{40} = \frac{1}{v_o} + \frac{1}{200} \] \[ \frac{1}{v_o} = \frac{1}{40} - \frac{1}{200} = \frac{5 - 1}{200} = \frac{4}{200} = \frac{1}{50} \] \[ v_o = +50 cm \]
The objective lens forms a real, inverted image 50 cm away from it.

Step 3: Find the distance between the lenses (\(L\)):

For the final image to be at infinity (normal vision), the intermediate image formed by the objective must be at the first focal point of the eyepiece. Thus, the distance of the intermediate image from the eyepiece is \(u_e = f_e = 4\) cm.
The total distance between the objective and eyepiece lenses is the sum of their separation: \[ L = |v_o| + |u_e| = v_o + f_e \] \[ L = 50 cm + 4 cm = 54 cm \]

Step 4: Find the magnification (\(M\)):

The angular magnification of a telescope is given by \(M = \frac{\beta}{\alpha}\), where \(\beta\) is the angle subtended by the final image at the eye and \(\alpha\) is the angle subtended by the object at the objective. \[ \alpha \approx \frac{h_o}{|u_o|} \quad and \quad \beta \approx \frac{h_i}{f_e} \]
where \(h_o\) is the object height and \(h_i\) is the height of the intermediate image. \[ M = \frac{\beta}{\alpha} = \frac{h_i/f_e}{h_o/|u_o|} = \frac{h_i}{h_o} \times \frac{|u_o|}{f_e} \]
The linear magnification of the objective is \(m_o = \frac{h_i}{h_o} = \frac{v_o}{u_o}\). The overall image is inverted. \[ M = \frac{v_o}{u_o} \times \frac{|u_o|}{f_e} = \frac{v_o}{f_e} \times \frac{|u_o|}{u_o} = -\frac{v_o}{f_e} \]
Substituting the values: \[ M = -\frac{50}{4} = -12.5 \] Quick Tip: For a telescope with an object at a finite distance, the formula for the length in normal adjustment becomes \(L = v_o + f_e\) and the magnification becomes \(M = -v_o/f_e\). Notice that if the object is at infinity (\(u_o \to \infty\)), then \(v_o \to f_o\), and these formulas reduce to the standard ones: \(L = f_o + f_e\) and \(M = -f_o/f_e\).

1. Bohr's Postulates for the Hydrogen Atom:


Postulate of Stationary Orbits: An electron in an atom can revolve in certain stable, circular orbits called stationary orbits without emitting any radiation. Each orbit has a definite energy associated with it.
Postulate of Quantization of Angular Momentum: The angular momentum (\(L\)) of an electron in a stationary orbit is an integral multiple of \(h/2\pi\), where \(h\) is Planck's constant.
\[ L = m_e v r = n \frac{h}{2\pi} \]
where \(n = 1, 2, 3, \ldots\) is the principal quantum number.
Postulate of Energy Transition: An electron can make a transition from a higher energy stationary orbit to a lower energy one. When it does so, a photon is emitted whose energy is equal to the energy difference between the initial (\(E_i\)) and final (\(E_f\)) orbits.
\[ E_{photon} = h\nu = E_i - E_f \]


2. Proof that \(r \propto n^2\):

Consider an electron of mass \(m_e\) and charge \(-e\) revolving around a nucleus of charge \(+e\) (for hydrogen, Z=1) in a circular orbit of radius \(r\) with a velocity \(v\).


Force Balance: The electrostatic force of attraction between the nucleus and the electron provides the necessary centripetal force for the circular motion.
\[ F_{centripetal} = F_{electrostatic} \]
\[ \frac{m_e v^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} \quad \cdots (1) \]
Bohr's Quantization Condition: From the second postulate:
\[ m_e v r = \frac{nh}{2\pi} \]
From this, we can express the velocity \(v\) as:
\[ v = \frac{nh}{2\pi m_e r} \quad \cdots (2) \]
Substitution: Substitute the expression for \(v\) from equation (2) into equation (1):
\[ \frac{m_e}{r} \left( \frac{nh}{2\pi m_e r} \right)^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} \]
\[ \frac{m_e}{r} \frac{n^2 h^2}{4\pi^2 m_e^2 r^2} = \frac{e^2}{4\pi\epsilon_0 r^2} \]
Simplification and Solving for \(r\):
\[ \frac{n^2 h^2}{4\pi^2 m_e r^3} = \frac{e^2}{4\pi\epsilon_0 r^2} \]
Cancel \(4\pi\) and \(r^2\) from both denominators:
\[ \frac{n^2 h^2}{\pi m_e r} = \frac{e^2}{\epsilon_0} \]
Rearranging the terms to solve for the radius \(r\):
\[ r = \frac{n^2 h^2 \epsilon_0}{\pi m_e e^2} \]
Conclusion: In this expression, \(h, \epsilon_0, \pi, m_e,\) and \(e\) are all fundamental constants. Therefore, we can write the radius as:
\[ r = (constant) \times n^2 \]
This proves that the radius of the orbital is directly proportional to the square of the principal quantum number (\(r \propto n^2\)). Quick Tip: The derivation for Bohr's radius and energy levels relies on two fundamental equations: the force balance equation (\(F_{centripetal} = F_{electric}\)) and the angular momentum quantization condition (\(mvr = nh/2\pi\)). By combining these two, you can derive expressions for \(r\), \(v\), and \(E\) in terms of \(n\).

1. Nuclear Fission:

Nuclear fission is a nuclear reaction in which the nucleus of a heavy, unstable atom (such as Uranium-235 or Plutonium-239) absorbs a neutron and splits into two or more smaller, lighter nuclei (fission fragments). This process also releases a few neutrons and a very large amount of energy. The released neutrons can then go on to cause further fissions, leading to a self-sustaining chain reaction if controlled, as in a nuclear reactor.


2. Calculation of Fissions per Second:


Step 1: List the Given Data and Convert to SI Units:


Energy released per fission, \(E = 200\) MeV.
Power of the reactor, \(P = 4\) MW.

We need to convert these to Joules and Watts (Joules/second) respectively.

\(1 eV = 1.6 \times 10^{-19} J \implies 1 MeV = 1.6 \times 10^{-13} J\)
\[ E = 200 \times 10^6 eV = 200 \times 1.6 \times 10^{-13} J = 3.2 \times 10^{-11} J \]
\(1 MW = 10^6 W\)
\[ P = 4 \times 10^6 W = 4 \times 10^6 J/s \]


Step 2: Relate Power, Energy, and Number of Fissions:

Power is the rate at which energy is produced. If \(N\) is the number of fissions occurring per second, then the total energy produced per second (Power) is the product of \(N\) and the energy per fission (\(E\)). \[ P = N \times E \]
We need to find \(N\), the number of fissions per second. \[ N = \frac{P}{E} \]

Step 3: Calculate the Value of N:

Substitute the values in SI units into the equation: \[ N = \frac{4 \times 10^6 J/s}{3.2 \times 10^{-11} J/fission} \] \[ N = \frac{4}{3.2} \times 10^{6 - (-11)} fissions/s \] \[ N = \frac{40}{32} \times 10^{17} fissions/s = \frac{5}{4} \times 10^{17} fissions/s \] \[ N = 1.25 \times 10^{17} fissions/s \]

Step 4: Final Answer:

The number of nuclei fissioned per second in the reactor is \(1.25 \times 10^{17}\). Quick Tip: In nuclear physics calculations, unit conversion is critical. Always convert energy from MeV to Joules and power from MW (or GW) to Watts before applying the formulas. Remember the relationship: Power = (Number of events per second) \(\times\) (Energy per event).

1. Energy Bands in Solids:

In an isolated atom, electrons occupy discrete energy levels. When atoms are brought close together to form a crystalline solid, the electrons in the outer shells of an atom are influenced by the electrons and nuclei of neighboring atoms. This interaction causes the discrete energy levels of the isolated atoms to split and form a collection of a very large number of closely spaced energy levels, which are effectively continuous. These ranges of allowed energies are called **energy bands**. The two most important bands are:

Valence Band (VB): The energy band containing the valence electrons. It is the highest occupied energy band at 0 K.
Conduction Band (CB): The lowest unoccupied energy band. Electrons in this band are free to move and contribute to electrical conduction.
Forbidden Energy Gap (\(E_g\)): The energy range between the top of the valence band and the bottom of the conduction band, which contains no allowed energy levels for electrons.


2. Differentiation based on Band Theory:

\begin{tikzpicture
% Conductor
\node at (0, 3) {Conductor;
\draw[fill=blue!30] (-1,0) rectangle (1,2);
\draw[fill=red!30] (-1,1.5) rectangle (1,2.5);
\node at (0, 0.75) {Valence Band;
\node at (0, 2) {Conduction Band;
\node at (0, 1.5) [right] {Overlap;

% Insulator
\node at (4, 3) {Insulator;
\draw[fill=blue!30] (3,0) rectangle (5,1);
\draw[fill=red!30] (3,2) rectangle (5,3);
\draw[<->] (5.2, 1) -- (5.2, 2) node[midway, right] {\(E_g > 3\) eV;
\node at (4, 0.5) {Valence Band (Filled);
\node at (4, 2.5) {Conduction Band (Empty);

% Semiconductor
\node at (8, 3) {Semiconductor;
\draw[fill=blue!30] (7,0) rectangle (9,1);
\draw[fill=red!30] (7,1.5) rectangle (9,2.5);
\draw[<->] (9.2, 1) -- (9.2, 1.5) node[midway, right] {\(E_g \approx 1\) eV;
\node at (8, 0.5) {Valence Band;
\node at (8, 2) {Conduction Band;
\end{tikzpicture


Conductors: In conductors (metals), the valence band and the conduction band overlap. There is no forbidden energy gap between them. Due to this overlap, a large number of free electrons are readily available in the conduction band to conduct electricity, even at zero Kelvin.
Insulators: In insulators, the valence band is completely filled, and the conduction band is completely empty. There is a very large forbidden energy gap (\(E_g > 3\) eV) separating them. It is almost impossible for an electron to gain enough energy to jump from the valence band to the conduction band.
Semiconductors: In semiconductors, the band structure is similar to that of insulators, but the forbidden energy gap is much smaller (\(E_g < 3\) eV, e.g., \(\sim 1.1\) eV for Silicon). At 0 K, they behave like insulators. However, at room temperature, some electrons can gain enough thermal energy to jump the gap into the conduction band, allowing for limited conductivity.


3. Effect of Temperature:

Conductors: When the temperature of a conductor is increased, the thermal vibrations of the metal ions (lattice) increase. This leads to more frequent collisions between the free electrons and the ions, which increases the electrical resistance.
Insulators: Due to the large band gap, an increase in temperature has very little effect on the conductivity of insulators, as very few electrons can be thermally excited to the conduction band.
Semiconductors: When the temperature of a semiconductor is increased, more valence electrons gain sufficient thermal energy to jump across the small forbidden gap into the conduction band, creating electron-hole pairs. This increases the number of charge carriers, which significantly increases the conductivity (and decreases the resistance). Quick Tip: The key distinguishing factor is the size of the forbidden energy gap (\(E_g\)). For conductors \(E_g \approx 0\), for semiconductors \(E_g\) is small, and for insulators \(E_g\) is large. This gap size dictates how easily electrons can become charge carriers.

1. Formation of Depletion Layer:

When a p-type semiconductor is joined with an n-type semiconductor to form a p-n junction, two processes begin: diffusion and drift.

Diffusion: Due to the high concentration of holes on the p-side and electrons on the n-side, a concentration gradient exists across the junction. This causes holes to diffuse from the p-side to the n-side, and electrons to diffuse from the n-side to the p-side.
Recombination and Ion Formation: When a hole diffuses to the n-side, it recombines with an electron. When an electron diffuses to the p-side, it recombines with a hole. This process is not instantaneous. The electron leaving the n-side leaves behind a positively charged (ionized) donor atom. Similarly, the hole leaving the p-side leaves behind a negatively charged (ionized) acceptor atom.
Depletion Region: This creates a region near the junction which is depleted of mobile charge carriers (electrons and holes) but contains a layer of fixed positive ions on the n-side and fixed negative ions on the p-side. This region is known as the depletion layer or space-charge region.


2. Potential Barrier:

The accumulation of fixed positive and negative charges in the depletion layer sets up an electric field directed from the positive n-side to the negative p-side. This electric field creates a potential difference across the junction. This potential difference is called the potential barrier or barrier voltage (\(V_B\)).
The potential barrier opposes the further diffusion of majority charge carriers. An equilibrium is reached when the electric force due to the potential barrier on the majority carriers becomes equal and opposite to the force due to the concentration gradient. For silicon, \(V_B \approx 0.7\) V, and for germanium, \(V_B \approx 0.3\) V at room temperature.


3. Avalanche Breakdown:

Avalanche breakdown is a phenomenon that occurs in a p-n junction diode under a high reverse bias voltage.

Reverse Bias: In reverse bias, the applied voltage widens the depletion layer and increases the strength of the electric field across it. The current is very small and is due to minority carriers.
Carrier Acceleration: These minority carriers are accelerated to very high velocities and gain significant kinetic energy as they are swept across the strong electric field in the depletion region.
Impact Ionization: If the reverse voltage is high enough (the breakdown voltage), these energetic minority carriers can collide with the atoms of the semiconductor crystal lattice with enough force to knock valence electrons out of their covalent bonds. This process is called impact ionization, and it creates new electron-hole pairs.
Avalanche Effect: The newly created charge carriers are also accelerated by the strong electric field and cause further impact ionizations, creating even more carriers. This process repeats, leading to a rapid, cumulative multiplication of charge carriers, much like an avalanche.
Breakdown: This rapid multiplication results in a sharp increase in the reverse current, and the diode is said to be in the breakdown region. This process can be destructive to the diode if the current is not limited by an external circuit resistance. Quick Tip: To remember the direction of the internal field in the depletion layer, think about the fixed ions left behind: positive ions are on the N-side and negative ions are on the P-side. The electric field always points from positive to negative, so it points from the N-side to the P-side.