Coulomb's law states that the electric force (\( F \)) between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them:
\[ F = k \frac{|q_1 q_2|}{r^2} \]

where:

\( F \) is the magnitude of the Coulomb force,

\( q_1 \) and \( q_2 \) are the charges,

\( r \) is the distance between the charges, and

\( k \) is Coulomb's constant.


From the problem, we are given that:

- The force variation is shown with respect to \( 1/r^2 \).

- The charge \( q_2 \) is positive and has the least magnitude.


Step 1: Analyzing the Force Variation

The force between any two charges is proportional to the product of their magnitudes, i.e.,
\[ F \propto |q_1 q_2| \]

Thus, the force acting between two charges is greater when the product of their magnitudes is larger.

Step 2: Considering the Relative Magnitudes

Since \( q_2 \) is given as positive and the least in magnitude, we know that:

- The force between charges \( q_1 \) and \( q_2 \) will be higher if \( |q_1| \) is larger than \( |q_3| \) because \( F \propto |q_1 q_2| \).

- The force between charges \( q_2 \) and \( q_3 \) will be weaker if \( |q_3| \) is smaller than \( |q_1| \), assuming \( r^2 \) is the same for both pairs.

Step 3: Conclusion on the Magnitudes of the Charges

From the above analysis, we can conclude that:


- \( |q_1| \) must be greater than \( |q_3| \) to produce a higher force with \( q_2 \).

- \( |q_2| \) is the least, as stated in the problem, meaning \( q_2 \) is smaller than both \( q_1 \) and \( q_3 \) in magnitude.


Therefore, the relationship between the magnitudes of the charges is:
\[ |q_2| < |q_3| < |q_1| \]

Thus, the correct answer is:
\[ q_2 < q_3 < q_1 \] Quick Tip: In Coulomb's law, the force increases with the product of the magnitudes of the charges and decreases with the square of the distance between them. Always remember that a higher charge magnitude results in a stronger Coulomb force.

The resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]

where:

\( \rho \) is the resistivity of the material,

\( L \) is the length of the wire,

\( A \) is the cross-sectional area of the wire.


For wire P, the resistance is \( R = \rho \frac{L}{A_P} \), where \( A_P \) is the cross-sectional area of wire P.


For wire Q:

- The length of wire Q is half that of wire P: \( L_Q = \frac{L_P}{2} \).

- The diameter of wire Q is twice that of wire P: \( d_Q = 2d_P \), so the cross-sectional area of wire Q is four times that of wire P (since area \( A \propto d^2 \)).


Thus, the cross-sectional area of wire Q is \( A_Q = 4A_P \).

Now, the resistance of wire Q becomes:
\[ R_Q = \rho \frac{L_Q}{A_Q} = \rho \frac{\frac{L_P}{2}}{4A_P} = \frac{R_P}{8} \]

Therefore, the resistance of wire Q is \( \frac{R}{8} \). Quick Tip: The resistance is inversely proportional to the area and directly proportional to the length of the wire. So, doubling the diameter of a wire reduces its resistance significantly.

The magnetic force on a current-carrying wire is given by:
\[ \vec{F} = I \ell \hat{L} \times \vec{B} \]

where:

\( I = 0.5 \, A \) is the current,

\( \ell = 1 \, cm = 0.01 \, m \) is the length of the wire,

\( \hat{L} \) is the unit vector along the direction of current (in the +x direction: \( \hat{i} \)),

\( \vec{B} = (0.4 \times 10^{-3}) \hat{j} + (0.6 \times 10^{-3}) \hat{k} \) is the magnetic field.


First, calculate the cross product \( \hat{L} \times \vec{B} \):
\[ \hat{i} \times (0.4 \times 10^{-3} \hat{j} + 0.6 \times 10^{-3} \hat{k}) = 0.4 \times 10^{-3} \hat{k} - 0.6 \times 10^{-3} \hat{j} \]

Now, the force is:
\[ \vec{F} = (0.5) (0.01) \left( 0.4 \times 10^{-3} \hat{k} - 0.6 \times 10^{-3} \hat{j} \right) \]
\[ \vec{F} = (-3\hat{i} + 2\hat{k}) \, \muN \] Quick Tip: Remember that the force on a current-carrying wire in a magnetic field depends on the direction of both the current and the magnetic field. The right-hand rule helps you determine the direction of the force.

The induced charge \( Q \) in a coil due to a changing magnetic field is given by:
\[ Q = N \cdot \frac{d\Phi}{dt} \]

where:
- \( N = 100 \) is the number of turns,
- \( \Phi = B \cdot A \) is the magnetic flux,
- \( B = 90 \, mT = 90 \times 10^{-3} \, T \),
- \( A = 0.05 \, m^2 \) is the area of each turn.

The rate of change of magnetic flux is:
\[ \frac{d\Phi}{dt} = B \cdot A \cdot \frac{1}{R} \]

Substituting the given values:
\[ Q = 100 \cdot \frac{(90 \times 10^{-3}) \cdot 0.05}{1.5} = 0.30 \, C \]

Thus, the induced charge is 0.30 C . Quick Tip: When calculating induced charge, remember that the change in magnetic flux depends on both the magnetic field and the area of the coil. The higher the rate of change of flux, the greater the induced charge.

The inductance of an air-core solenoid is given by the formula: \[ L = \frac{\mu_0 N^2 A}{l} \]
Where:
\( L = 0.016\,H \),
\( l = 0.81\,m \),
\( r = 0.02\,m \Rightarrow A = \pi r^2 = \pi (0.02)^2 = 1.2566 \times 10^{-3} \, m^2 \),
\( \mu_0 = 4\pi \times 10^{-7} \, H/m \)


Substitute values: \[ 0.016 = \frac{4\pi \times 10^{-7} \times N^2 \times 1.2566 \times 10^{-3}}{0.81} \] \[ N^2 = \frac{0.016 \times 0.81}{4\pi \times 10^{-7} \times 1.2566 \times 10^{-3}} \approx 8214057.59 \] \[ N = \sqrt{8214057.59} \approx 2866 \] Quick Tip: Use the formula \( L = \frac{\mu_0 N^2 A}{l} \) for solenoids and always check units. Square root at the end gives the number of turns.

The average power consumed in an AC circuit over a complete cycle is given by: \[ P_{avg} = \frac{1}{T} \int_0^T v(t) \cdot i(t)\, dt \]
Given, \( v(t) = v_0 \sin(\omega t), \quad i(t) = i_0 \sin(\omega t + \phi) \)


Using identity: \[ \sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] \] \[ P_{avg} = \frac{1}{T} \int_0^T \frac{v_0 i_0}{2} [\cos(\phi) - \cos(2\omega t + \phi)] dt \]
The integral of \(\cos(2\omega t + \phi)\) over a full cycle is zero. So, \[ P_{avg} = \frac{v_0 i_0}{2} \cos \phi \] Quick Tip: For average power in AC circuits with phase difference \( \phi \), use: \[ P_{avg} = \frac{V_0 I_0}{2} \cos \phi \] It accounts for phase shift between voltage and current.

From the graph, we know:

Energy is inversely proportional to wavelength \( \left( E = \frac{hc}{\lambda} \right) \).

P is at highest energy (shortest wavelength) → corresponds to X-rays

Q is at lowest energy (longest wavelength) → corresponds to Microwaves

R lies in between → corresponds to UV radiation


So the correct assignment is:

P → X-rays, Q → Microwaves, R → UV radiation
Quick Tip: Recall the order of electromagnetic waves based on energy: X-rays \> UV radiation \> Microwaves. Use the relation \( E \propto \frac{1}{\lambda} \) to map them properly.

When light passes from a denser medium (water) to a rarer medium (air), the object appears to be at a shallower depth than it actually is.
The apparent depth \( h' \) is given by: \[ h' = \frac{h}{\mu} \]
Where:

\( h \) is the real depth = \( H \)

\( \mu \) is the refractive index = \( \frac{4}{3} \)


Substituting the values: \[ h' = \frac{H}{\frac{4}{3}} = \frac{3H}{4} \] Quick Tip: To find apparent depth in optics, divide real depth by refractive index: \[ Apparent depth = \frac{Real depth}{\mu} \] Use this when viewing objects in water or glass containers.

Step 1: Use Einstein's photoelectric equation: \[ eV_0 = h\nu - \phi \]
Where:

\( e \) = charge of electron

\( V_0 \) = stopping potential

\( h \) = Planck's constant

\( \nu \) = frequency of incident light

\( \phi \) = work function


Step 2: Rearranging the equation: \[ V_0 = \frac{h}{e} \nu - \frac{\phi}{e} \]
This is a straight line of the form \( y = mx + c \), where:

\( y = V_0 \)

\( x = \nu \)

\( m = \frac{h}{e} \) (slope of the line)


Step 3: Solving for charge of the electron \( e \): \[ m = \frac{h}{e} \Rightarrow e = \frac{h}{m} \] Quick Tip: Always compare equations with standard straight-line form. For photoelectric graphs, slope \( m = \frac{h}{e} \), so to find \( e \), rearrange to get \( e = \frac{h}{m} \).

Step 1: Use de Broglie wavelength formula: \[ \lambda = \frac{h}{mv} \]
Where:

\( \lambda \) = de Broglie wavelength

\( h \) = Planck's constant

\( m \) = mass of particle

\( v \) = speed of particle


Step 2: Given that all particles have the same speed, \[ \lambda \propto \frac{1}{m} \]
So, the smaller the mass, the greater the wavelength.

Step 3: Compare masses:

Electron mass \( m_e \) is smallest

Proton mass \( m_p \approx 1836 \times m_e \)

Deuteron mass \( m_d \approx 2 \times m_p \)


Step 4: Therefore, \[ (B) \quad \lambda_e > \lambda_p > \lambda_d \] Quick Tip: For particles at equal speed, lighter particles have larger de Broglie wavelength. So compare masses to deduce the correct order.

The binding energy per nucleon increases rapidly for light nuclei, reaches a maximum around iron (mass number 56), and then gradually decreases for heavier nuclei. This curve explains nuclear stability and why energy is released during both fusion (of light nuclei) and fission (of heavy nuclei).

Option (A) correctly depicts this: a steep rise followed by a gentle decline, with a peak at A = 56.
Quick Tip: Binding energy per nucleon peaks near iron (A = 56). Use this to identify the correct curve in such questions.

In a forward-biased p-n junction diode, an external voltage is applied such that it reduces the built-in potential barrier. This causes more charge carriers to move across the junction, reducing the width of the depletion layer.

Hence, both the potential barrier and depletion width decrease.
Quick Tip: In forward bias, think “easier flow” — the depletion region shrinks and the barrier becomes lower to allow current.

Step 1: Understanding Assertion (A):

When a magnet is moved into a coil with many turns and the circuit is closed, a large amount of current is induced due to electromagnetic induction. This current opposes the motion of the magnet (Lenz's Law), making it harder to move the magnet. So, Assertion (A) is true.


Step 2: Understanding Reason (R):

Lenz’s Law states that the direction of induced current is such that it opposes the cause producing it — in this case, the motion of the magnet. Hence, Reason (R) is also true.


Step 3: Checking the relationship between A and R:

The opposition to motion described in the assertion is explained by Lenz’s Law stated in the reason. Therefore, Reason (R) correctly explains Assertion (A).
Quick Tip: Lenz’s Law is key to explaining resistance to magnet motion in closed loops.

Step 1: Verifying Assertion (A):

Galvanometers are designed such that their deflection is proportional to the current. The torque on the coil causes angular deflection proportional to current. Hence, Assertion (A) is true.


Step 2: Verifying Reason (R):

In a galvanometer, a radial magnetic field is used so that the plane of the coil is always perpendicular to the magnetic field. This ensures torque is always proportional to current. So, Reason (R) is true.


Step 3: Explanation linkage:

Since the radial magnetic field ensures consistent torque-current relation, Reason (R) correctly explains Assertion (A).
Quick Tip: Radial magnetic fields ensure uniform behavior in measuring instruments like galvanometers.

Step 1: Evaluate Assertion (A):

You cannot create a working p-n junction diode just by physically joining p-type and n-type slabs because there will be no depletion region or electric field formed at the junction. Therefore, Assertion (A) is true.


Step 2: Evaluate Reason (R):

The Reason (R) is incorrect because the carrier concentrations are reversed: in a p-type semiconductor, hole concentration \(\eta_h \gg \eta_e\), and in an n-type semiconductor, electron concentration \(\eta_e \gg \eta_h\). So Reason (R) is false.


Step 3: Relationship between A and R:

Assertion is true, but Reason is false — hence, option (C) is correct.
Quick Tip: Always remember: p-n junctions must be chemically fabricated to allow diffusion and depletion region formation.

Step 1: Understanding Assertion (A):

In Bohr’s model, the potential energy of an electron in an orbit is negative and equal to \(-2\) times the kinetic energy. Hence, Assertion (A) is false.


Step 2: Understanding Reason (R):

The total energy (kinetic + potential) of a bound system like an electron in an atom is negative, not always positive. So Reason (R) is also false.


Step 3: Final evaluation:

Both statements are incorrect, so option (D) is the right choice.
Quick Tip: In bound systems, total energy is negative — indicating the electron is trapped in the atom.

Let the emf of the battery be \( E \) volts and internal resistance be \( r \) ohms.

Using Ohm’s Law: \[ Terminal voltage = E - Ir \]

Case 1: When current \( I = 2\,A \), terminal voltage = 5 V \[ E - 2r = 5 \quad \cdots (1) \]

Case 2: When current \( I = 4\,A \), terminal voltage = 4 V \[ E - 4r = 4 \quad \cdots (2) \]

Now subtract equation (2) from equation (1): \[ (E - 2r) - (E - 4r) = 5 - 4 \Rightarrow 2r = 1 \Rightarrow r = 0.5\,\Omega \]

From (1): \( E = 5 + 2r \)
Substitute into (2): \[ 5 + 2r - 4r = 4 \Rightarrow 5 - 2r = 4 \Rightarrow 2r = 1 \Rightarrow r = 0.5\,\Omega \]
Then, \[ E = 5 + 2(0.5) = 6\,V \]

But this leads to inconsistency with the second condition: \[ E - 4r = 6 - 4(0.5) = 6 - 2 = 4 \]

So it is consistent. The earlier error was in labeling.

Thus, final answer: \[ \boxed{E = 6\,V, \quad r = 0.5\,\Omega} \] Quick Tip: Use the terminal voltage formula \( V = E - Ir \) directly when internal resistance is involved. Setting up simultaneous equations is the key.

N/A Quick Tip: In single-slit diffraction, the position of the first minimum is given by \( a \sin \theta = m \lambda \), with \( m = 1 \) for the first minimum.

N/A Quick Tip: To calculate intensity in interference, convert path difference to phase using \( \phi = \frac{2\pi}{\lambda} \Delta x \), then apply \( I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \).

Given:


Refractive index of the rod (\( n_2 \)): \( \frac{2}{\sqrt{3}} \)
Refractive index of air (\( n_1 \)): \( 1 \)
The light ray travels along the surface of the rod, indicating Total Internal Reflection (TIR).


Step 1: Critical Angle for TIR

For Total Internal Reflection (TIR) to occur:

The light must travel from a denser medium (rod) to a rarer medium (air).
The angle of incidence inside the rod must be greater than or equal to the critical angle (\( C \)).


The critical angle (\( C \)) is given by: \[ \sin C = \frac{n_1}{n_2} = \frac{1}{\frac{2}{\sqrt{3}}} = \frac{\sqrt{3}}{2} \] \[ C = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60^\circ \]

Step 2: Relating Critical Angle to \(\theta\)

When the light enters the rod:

The angle of refraction (\( r \)) inside the rod is complementary to the critical angle (\( C \)) because the light travels along the surface: \[ r = 90^\circ - C = 90^\circ - 60^\circ = 30^\circ \]


Using Snell's Law at the point of entry: \[ n_1 \sin \theta = n_2 \sin r \]
Substituting the known values: \[ 1 \cdot \sin \theta = \frac{2}{\sqrt{3}} \cdot \sin 30^\circ \] \[ \sin \theta = \frac{2}{\sqrt{3}} \cdot \frac{1}{2} = \frac{1}{\sqrt{3}} \] \[ \theta = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

Final Answer: \[ \boxed{\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)} \]


\begin{quicktipbox
When a light ray grazes along a surface, total internal reflection occurs at the critical angle. Always apply Snell’s Law at the correct interface.
\end{quicktipbox Quick Tip: When a light ray grazes along a surface, total internal reflection occurs at the critical angle. Always apply Snell’s Law at the correct interface.

We begin with the Bohr model for the hydrogen atom. The electron revolves in a circular orbit under electrostatic attraction from the nucleus.

Step 1: Centripetal force = Electrostatic force
\[ \frac{mv^2}{r} = \frac{ke^2}{r^2} \Rightarrow mv^2 = \frac{ke^2}{r} \quad \cdots (1) \]

Where:

\( m \) is the mass of the electron

\( v \) is its orbital speed

\( r \) is the orbit radius

\( k = \frac{1}{4\pi\varepsilon_0} \)


Step 2: Bohr's quantization condition
\[ mvr = n\hbar \quad \cdots (2) \Rightarrow v = \frac{n\hbar}{mr} \quad \cdots (3) \]

Step 3: Use (3) in (1)
\[ m \left( \frac{n\hbar}{mr} \right)^2 = \frac{ke^2}{r} \Rightarrow \frac{n^2\hbar^2}{mr^2} = \frac{ke^2}{r} \Rightarrow r = \frac{n^2\hbar^2}{mke^2} \quad \cdots (4) \]

So, \( r \propto n^2 \)

Step 4: Time period of revolution \( T = \frac{2\pi r}{v} \)

From (3), \( v = \frac{n\hbar}{mr} \)
\[ T = \frac{2\pi r}{v} = \frac{2\pi r \cdot mr}{n\hbar} = \frac{2\pi m r^2}{n\hbar} \]

Now use \( r \propto n^2 \Rightarrow r^2 \propto n^4 \)
\[ T \propto \frac{r^2}{n} \propto \frac{n^4}{n} = n^3 \]
\[ \boxed{T_n \propto n^3} \]

\begin{quicktipbox
In Bohr's model, derive the radius and speed expressions first using quantization and Coulomb force balance before finding time period.
\end{quicktipbox Quick Tip: In Bohr's model, derive the radius and speed expressions first using quantization and Coulomb force balance before finding time period.

Step 1: Understand the doping ratio

1 dopant atom is added for every \( 5 \times 10^7 \) silicon atoms.


Step 2: Use the given silicon atom density

Number density of silicon atoms is: \[ 5 \times 10^{28} \, atoms/m^3 \]

Step 3: Calculate the number of dopant atoms per m\(^3\)
\[ No. of holes created per m^3 = \frac{5 \times 10^{28}}{5 \times 10^7} = 10^{21} \]

Step 4: Convert to holes per cm\(^3\)
\[ No. of holes per cm^3 = \frac{10^{21}}{10^6} = 10^{15} \]

Step 5: Example of a dopant

One example of such a trivalent dopant is: Aluminium (Al), Indium (In), or Gallium (Ga).
Quick Tip: To find charge carriers from doping, divide atom density by the number of host atoms per dopant, then convert units properly.

Let:

\( E_1 = 3\,V, \quad r_1 = 0.2\,\Omega \)

\( E_2 = 6\,V, \quad r_2 = 0.4\,\Omega \)


Step 1: Equivalent EMF of the parallel combination: \[ E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{\frac{3}{0.2} + \frac{6}{0.4}}{\frac{1}{0.2} + \frac{1}{0.4}} = \frac{15 + 15}{5 + 2.5} = \frac{30}{7.5} = 4\,V \]

Step 2: Equivalent internal resistance of the combination: \[ r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{0.2 \times 0.4}{0.2 + 0.4} = \frac{0.08}{0.6} = \frac{2}{15}\,\Omega \approx 0.133\,\Omega \]

Step 3: Total resistance in the circuit: \[ R_{total} = r_{eq} + R = 0.133 + 4 = 4.133\,\Omega \]

Step 4: Current drawn from the combination: \[ I = \frac{E_{eq}}{R_{total}} = \frac{4}{4.133} \approx 0.968\,A \] Quick Tip: To combine batteries in parallel, use the formula: \[ E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2}, \quad r_{eq} = \frac{r_1 r_2}{r_1 + r_2} \] Then apply Ohm’s law with total external resistance.

Part (i): Relation between \( R' \) and \( R \)

We know resistance of a wire: \[ R = \rho \frac{l}{A} \]
When length becomes \( 2l \), assuming volume is constant: \[ A \propto \frac{1}{l} \Rightarrow A' = \frac{A}{2} \] \[ R' = \rho \frac{2l}{A/2} = 4 \cdot \frac{\rho l}{A} = 4R \]
So, \( R' = 4R \)

Part (ii): Relation between \( v_d' \) and \( v_d \)

Drift velocity is given by: \[ v_d = \frac{eE\tau}{m} \]
For a given potential \( V \), \[ E = \frac{V}{l} \Rightarrow v_d \propto \frac{1}{l} \]
When length becomes \( 2l \): \[ v_d' = \frac{v_d}{2} \] Quick Tip: When a wire is stretched to double its length, its resistance becomes 4 times and drift velocity becomes half due to the inverse dependence on length.

Step 1: Understanding electron motion in a conductor

In a conductor, free electrons are always in random thermal motion, even without an electric field. These motions are in all directions and are very fast (random velocities).

Step 2: Effect of applying an electric field

When a potential difference is applied, an electric field is established. This field causes a small net velocity (called drift velocity) superimposed on the random motion.

Step 3: Direction of drift velocity

Though the electrons still move randomly, the average motion of all electrons is in a direction opposite to the electric field (from lower to higher potential).

Step 4: Conclusion

Therefore, not all electrons move in the same direction. They still move randomly, but with a small average (drift) velocity in a particular direction. Quick Tip: Drift velocity is a small net movement superimposed on random thermal motion. Electrons do not all move in a straight line; instead, they exhibit a slow drift on top of rapid, random motion.

Step 1: Apply Biot–Savart law:

Biot–Savart law: \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I \, d\vec{l} \times \hat{r}}{r^2} \]

Step 2: Geometry of the loop:

Let a circular loop of radius \( R \) carry current \( I \), and we need to find magnetic field at point \( P \) on its axis at distance \( x \).


Step 3: Component analysis:

Due to symmetry, only axial components of \( d\vec{B} \) add up, tangential components cancel.


So, total magnetic field at axial point is: \[ B = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \]

Step 4: At the center of the loop (i.e., \( x = 0 \)):
\[ B = \frac{\mu_0 I R^2}{2 R^3} = \frac{\mu_0 I}{2R} \] Quick Tip: Always use symmetry when applying Biot–Savart law to circular loops — it simplifies vector components greatly.

Step 1: Recall the emf induced in an inductor

When current changes through an inductor, a back emf is induced: \[ \varepsilon = L \frac{dI}{dt} \]

Step 2: Use power expression \( P = \varepsilon \cdot I \)
\[ P = L \frac{dI}{dt} \cdot I = L I \frac{dI}{dt} \]

Step 3: Energy is the integral of power over time
\[ U = \int P \, dt = \int L I \frac{dI}{dt} \, dt \]

Step 4: Simplify using substitution
\[ U = L \int I \, dI = L \left[ \frac{I^2}{2} \right]_0^I = \frac{1}{2} L I^2 \] Quick Tip: Energy stored in an inductor increases with square of the current — similar to energy in a capacitor \( \frac{1}{2} C V^2 \).

Step 1: Energy stored in an inductor is \( U = \frac{1}{2} L I^2 \)


For a solenoid of length \( l \), cross-section \( A \), and \( n \) turns per unit length: \[ L = \mu_0 n^2 A l \Rightarrow U = \frac{1}{2} \mu_0 n^2 A l I^2 \]

Step 2: Volume of solenoid = \( A l \)


Step 3: Magnetic field inside solenoid is \( B = \mu_0 n I \)
\[ \Rightarrow n I = \frac{B}{\mu_0} \Rightarrow (n I)^2 = \frac{B^2}{\mu_0^2} \]

Step 4: Substitute in energy expression:
\[ U = \frac{1}{2} \mu_0 \cdot A l \cdot \frac{B^2}{\mu_0^2} = \frac{1}{2} \cdot \frac{A l B^2}{\mu_0} \]

Step 5: Energy density \( u = \frac{U}{A l} \)
\[ u = \frac{1}{2} \cdot \frac{B^2}{\mu_0} \] Quick Tip: Energy density in magnetic fields is proportional to \( B^2 \), just as electric energy density is proportional to \( E^2 \).

Step 1: Understanding the two types of current

\( I_c \): The conduction current in the wires and resistor parts of the circuit.

\( I_d \): The displacement current between the plates of the capacitor (where no conduction current flows).

Step 2: Maxwell’s correction to Ampere’s Law

To maintain continuity of current in the entire AC circuit, Maxwell introduced displacement current: \[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
This makes the modified Ampere’s law: \[ \oint \vec{B} \cdot d\vec{l} = \mu_0 (I_c + I_d) \]

Step 3: Applying to the capacitor circuit

As the current \( I_c \) flows into one plate of the capacitor, it changes the electric field between the plates. This changing electric field gives rise to \( I_d \).

Step 4: Conclusion

The net current (conduction + displacement) is the same throughout the circuit at every point: \[ I_{net} = I_c + I_d = constant \] Quick Tip: In an AC circuit with a capacitor, conduction current in the wires is matched by displacement current in the gap between plates. This keeps total current continuous.

Step 1: Kirchhoff’s first rule (junction rule):

It states that the algebraic sum of currents meeting at a junction is zero: \[ \sum I = 0 \]

Step 2: Problem at capacitor plates

At the capacitor plates, conduction current flows into the plate but no conduction current flows through the gap. This seems to violate the junction rule.

Step 3: Inclusion of displacement current

Maxwell resolved this by defining displacement current \( I_d \), which flows through the dielectric gap and acts like a current to maintain continuity.

Step 4: Conclusion

With the displacement current considered, the current entering a plate equals the current (displacement) leaving the plate. So Kirchhoff’s junction rule still holds true. Quick Tip: Junction rule remains valid at the capacitor plates if we include displacement current as part of the current flow. This is crucial in time-varying electric fields.

Step 1: Understand the photoelectric equation:

Einstein's photoelectric equation: \[ K.E. = h\nu - \phi \]
where \( h\nu \) is the photon energy, and \( \phi \) is the work function of the metal.


Step 2: Electrons are not all at the same energy level in the metal:

Electrons inside the metal have different binding energies depending on their position (some are more tightly bound than others).


Step 3: Resulting kinetic energies vary:

Even though the incident light is monochromatic (i.e., all photons have same energy), emitted electrons start from different energy levels → they gain different kinetic energies.
Quick Tip: Photoelectrons vary in kinetic energy due to their varying initial energy states inside the metal.

Step 1: Definition of saturation current:

Saturation current is the maximum photoelectric current when all emitted electrons are collected.


Step 2: Effect of light intensity:

Higher light intensity → more photons striking the surface → more electrons emitted.


Step 3: Result:

Greater intensity increases number of emitted electrons (not their energy), thus increasing saturation current.
Quick Tip: Saturation current depends on the number of emitted electrons, which increases with intensity.

Step 1: Photon energy decreases with increasing wavelength:
\[ E = \frac{hc}{\lambda} \Rightarrow As \lambda \uparrow, E \downarrow \]

Step 2: Threshold condition:

To emit photoelectrons, photon energy must be at least equal to work function \( \phi \).


Step 3: Cut-off wavelength:

There exists a maximum wavelength \( \lambda_0 \) (called threshold wavelength) beyond which photons don't have enough energy to liberate electrons.


Step 4: Final conclusion:

If \( \lambda > \lambda_0 \), no photoelectrons are emitted regardless of intensity.
Quick Tip: No emission occurs beyond the threshold wavelength, even if intensity is high — energy per photon is insufficient.

Step 1: Mass Defect

The mass defect of a nucleus is the difference between the sum of the masses of its individual protons and neutrons and the actual mass of the nucleus. It arises due to the conversion of mass into binding energy.
\[ \Delta m = Z m_p + (A - Z) m_n - M_nucleus \]

Step 2: Binding Energy

Binding energy is the energy required to disassemble a nucleus into its constituent protons and neutrons. It is equivalent to the mass defect by Einstein’s relation: \[ E_b = \Delta m \cdot c^2 \]

Step 3: Fission Process and Binding Energy Per Nucleon

In fission, a heavy nucleus (like uranium) splits into smaller nuclei with higher binding energy per nucleon. Since products have higher binding energy per nucleon, the total binding energy increases, and the difference is released as energy. This explains the huge energy release in nuclear fission.
Quick Tip: Higher binding energy per nucleon means greater stability. Fission releases energy because smaller fragments are more stable.

Step 1: Use known values
\[ m_p = 1.007277\, u, \quad m_n = 1.008665\, u, \quad m_d = 2.013553\, u \]

Step 2: Calculate expected mass without binding
\[ m_p + m_n = 1.007277 + 1.008665 = 2.015942\, u \]

Step 3: Find mass defect
\[ \Delta m = 2.015942 - 2.013553 = 0.002389\, u \]

Step 4: Convert mass defect into energy using \( 1\, u = 931.5\, MeV/c^2 \)
\[ E_b = 0.002389 \times 931.5 = 2.224\, MeV \] Quick Tip: Use accurate atomic mass values and multiply the mass defect by 931.5 to get binding energy in MeV.

Step 1: Components of the setup:

Power supply (variable DC source)

Diode under test

Resistor (current limiting)

Voltmeter across diode

Ammeter in series to measure current


Step 2: Forward bias arrangement:

Positive terminal of battery connected to p-side and negative terminal to n-side through resistor. Voltmeter is parallel to diode and ammeter in series.




Step 3: Reverse bias arrangement:

Battery polarity reversed — positive to n-side and negative to p-side. Rest of the arrangement remains the same.



Quick Tip: Forward bias: p to +, n to –. Reverse bias: p to –, n to +. Use voltmeter and ammeter correctly.

Step 1: Forward bias region:

Very small current until threshold (knee) voltage.

After this, current rises exponentially with small increase in voltage.

Step 2: Reverse bias region:

Very small leakage current (reverse saturation current).

Sharp increase only if breakdown occurs.


Graph shape:

Forward bias: Exponential rise after knee voltage.

Reverse bias: Almost flat line near zero (reverse current), till breakdown.


Quick Tip: Remember: Diode conducts significantly only after crossing threshold voltage in forward bias.

Step 1: Understand the circuit configuration and the condition for final charge.

The problem describes an RC series circuit. When key \(S_1\) is closed and \(S_2\) is open, the capacitor C is connected in series with the resistor R and the battery of emf V. The capacitor will charge through the resistor until it is fully charged.



Step 2: Identify the given numerical values.

From the last line of the preceding paragraph, we are given:


Resistance, \(R = 20 k\Omega = 20 \times 10^3 \Omega\)
Capacitance, \(C = 500 \textmu F = 500 \times 10^{-6} F\)
Voltage of the battery, \(V = 10 V\)




Step 3: Calculate the final charge (Q) on the capacitor.

When the capacitor is fully charged, the potential difference across it equals the battery voltage. The formula for the final charge (Q) on a capacitor is given by:
\(Q = C \times V\)


Substitute the given values into the formula:
\(Q = (500 \times 10^{-6} F) \times (10 V)\)
\(Q = 5000 \times 10^{-6} C\)
\(Q = 5 \times 10^3 \times 10^{-6} C\)
\(Q = 5 \times 10^{-3} C\)




Step 4: Convert the final charge to appropriate units and evaluate the options.

The calculated final charge is \(5 \times 10^{-3} C\).

We know that 1 millicoulomb (mC) = \(10^{-3}\) C.

So, \(Q = 5 mC\).


Now, let's compare this with the given options:


(A) 5 \(\muC\) = \(5 \times 10^{-6}\) C (Incorrect)
(B) 5 mC = \(5 \times 10^{-3}\) C (Correct)
(C) 25 mC = \(25 \times 10^{-3}\) C (Incorrect)
(D) 0.1 C (Incorrect)




Step 5: Conclusion.

The final charge on the capacitor when key \(S_1\) is closed and \(S_2\) is open is 5 mC.
\[ \boxed{5 mC} \] Quick Tip: Always use \( Q = CV \) to find the maximum charge stored on a capacitor in a DC circuit.

Step 1: Initial condition — key \( S_1 \) closed, \( S_2 \) open:

In this condition, the capacitor gets charged fully by the battery. \[ Q = C \cdot V = 500 \times 10^{-6} \cdot 10 = 5 \times 10^{-3} = 5 mC \]

Step 2: Now \( S_2 \) is closed and \( S_1 \) is opened:

The capacitor is now connected across just the resistor (no battery). This leads to discharging of the capacitor through the resistor.

Step 3: Final condition after long time:

As time \( t \to \infty \), the charge on the capacitor: \[ q(t) = Q \cdot e^{-t/RC} \to 0 \]

So, the final charge on the capacitor is: \[ \boxed{0} \] Quick Tip: When a charged capacitor is connected across a resistor without any battery, it discharges completely over time. Final charge becomes zero.

Resistance: \( [R] = [M L^2 T^{-3} A^{-2}] \)

Capacitance: \( [C] = [M^{-1} L^{-2} T^4 A^2] \)

\[ [RC] = [R] \cdot [C] = [T] \] Quick Tip: Time constant \( RC \) always has the dimension of time \([T]\). Multiply units directly to confirm.

\[ I(t) = \frac{V}{R} e^{-t/RC} \] \[ RC = 10\,s,\quad t = 5\,s \Rightarrow \frac{t}{RC} = \frac{1}{2} \] \[ I = \frac{10}{20000} \cdot e^{-1/2} = \frac{1}{2000} \cdot \frac{1}{\sqrt{e}} = \frac{1}{2\sqrt{e}} mA \] Quick Tip: Use \( I(t) = \frac{V}{R} e^{-t/RC} \) to find current at any time \( t \).

Step 1: Use formula for initial charging current: \[ I(0) = \frac{V}{R} \]

Step 2: Given values:

\( V = 10\,V \)

\( R = 20\,k\Omega = 2 \times 10^4\,\Omega \)

\[ I(0) = \frac{10}{2 \times 10^4} = \frac{1}{2000} = 0.0005\,A = 0.5\,mA \] Quick Tip: The initial charging current in an RC circuit is \( I = \frac{V}{R} \). Be sure to convert resistance to ohms and express current in milliamperes.

Step 1: Understanding the setup:

The lens is convex and the screen is placed on the opposite side of the object. A real image is formed when object is beyond the focal length.


Step 2: Nature of real image:

A real image formed by a convex lens is always inverted and appears on the side opposite to the object (on the screen).


Step 3: Image matching:

Option (C) shows an inverted image on the correct side — the screen. Hence, it is correct.
Quick Tip: Convex lenses form real, inverted images on the opposite side of the screen when the object is beyond focus.

Step 1: Understanding convex mirrors:

Convex mirrors always form virtual, erect, and diminished images — so their magnification is always positive, not negative.


Step 2: Review other statements:

(B) is correct — virtual images have positive magnification.

(C) is correct — concave lenses also always form virtual, erect, diminished images (positive magnification).

(D) is correct — real and inverted implies negative magnification.


Conclusion:

Only (A) is incorrect.
Quick Tip: Virtual images always have positive magnification; convex mirrors form virtual images only.

Step 1: Understanding the type of cut

Here, the lens is cut parallel to the principal axis, i.e., it is sliced horizontally into two thinner lenses.


Step 2: Impact on power and focal length

The power of a lens is given by: \[ P = \frac{1}{f} \]
Cutting along the principal axis results in lenses with half the power: \[ P' = \frac{P}{2} \Rightarrow f' = \frac{1}{P'} = \frac{1}{P/2} = 2f \]

Step 3: Conclusion

Each half has a focal length of \( 2f \), hence the correct answer is (B).
Quick Tip: Cutting a lens \textbf{parallel} to the principal axis reduces its power by half, doubling the focal length.

Step 1: Optical Geometry

In biconvex lens geometry, using Newton’s form of the lens formula: \[ X_1 \cdot X_2 = f^2 \]

Step 2: Solve for \( f \)
\[ f = \sqrt{X_1 X_2} \]

Step 3: Final answer

So, focal length is the geometric mean of object and image distances from their respective focal points.
Quick Tip: Use Newton’s lens formula: \( X_1 X_2 = f^2 \) to find focal length in such cases.

Step 1: Electrostatic Potential Energy Without External Field
\[ U_{initial} = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_1 q_2}{r} \]
where \( r = 6\,cm = 0.06\,m \). Substituting: \[ U_{initial} = \frac{1}{4\pi\epsilon_0} \cdot \frac{(5 \times 10^{-6})(-1 \times 10^{-6})}{0.06} = -\frac{5 \times 10^{-12}}{4\pi\epsilon_0 \times 0.06} \] \[ = -\frac{5 \times 10^{-12} \cdot 9 \times 10^9}{0.06} = -\frac{45 \times 10^{-3}}{0.06} = -0.75\,J \]

Step 2: Electrostatic Potential Energy Due to External Field

The electric potential due to the field is: \[ V(r) = -\int_\infty^r \vec{E} \cdot d\vec{r} = -\int_\infty^r \frac{3 \times 10^5}{r^2} \, dr = \frac{3 \times 10^5}{r} \]
\[ U_{ext} = q_1 V(r_1) + q_2 V(r_2) \]
with \( r_1 = r_2 = 3\,cm = 0.03\,m \):
\[ U_{ext} = \left( 5 \times 10^{-6} \cdot \frac{3 \times 10^5}{0.03} \right) + \left( -1 \times 10^{-6} \cdot \frac{3 \times 10^5}{0.03} \right) \]
\[ = (5 - 1) \times 10^{-6} \cdot 10^7 = 4 \times 10^{-6} \cdot 10^7 = 40\,J \]

Step 3: Total Potential Energy With External Field
\[ U_{total} = U_{initial} + U_{ext} = -0.75 + 40 = 39.25\,J \] Quick Tip: To calculate the change in electrostatic energy due to an external field, use: \[ \Delta U = \sum q_i V(\vec{r}_i) \] especially when mutual interaction remains constant.

\[ C = \frac{Q}{V} = \frac{80 \times 10^{-6}}{16} = 5 \times 10^{-6}\,F \] Quick Tip: Capacitance is defined by \( C = \frac{Q}{V} \), where \( Q \) is the magnitude of charge on either conductor.

For a spherical shell with a charge distributed over its surface, the electric field at a point outside the shell (at distance \( r \) from the center, where \( r > R \)) is given by Coulomb's Law: \[ E = \frac{KQ}{r^2} \]
where \( Q \) is the total charge on the shell, and \( r \) is the distance from the center.

At a distance \( 3R \) (which is outside all the shells), the electric field depends only on the total charge on the shell, and not on the distribution of charge inside. Therefore, for each shell at a distance \( 3R \), the electric field magnitude is given by: \[ E = \frac{KQ}{(3R)^2} = \frac{KQ}{9R^2} \]

So, for each shell at \( 3R \), the electric field magnitudes are:

For shell A: \( E_A = \frac{K \cdot 6q}{9R^2} \)

For shell B: \( E_B = \frac{K \cdot (-4q)}{9R^2} \)

For shell C: \( E_C = \frac{K \cdot 14q}{9R^2} \)


Since the magnitude depends only on the charge, the electric field magnitudes are proportional to the absolute values of the charges: \[ |E_A| = |E_B| = |E_C| \]
Thus, the electric field magnitudes at a distance of \( 3R \) are the same for all shells. Quick Tip: For a spherical shell with charge \( Q \), the electric field outside the shell depends only on the total charge and not on its internal charge distribution.

We are calculating the work done on the charge \( +5\,\mu C \) as it is moved from point \( C \) to point \( A \) along the semicircular path. The work done is given by: \[ W = q \cdot [V_A - V_C] \]

Step 1: Calculate the potential at point \( C \) (due to both charges at \( B \) and \( D \))
\[ V_C = \left[ \frac{k \cdot 6 \times 10^{-6}}{5 \times 10^{-2}} \right] - \left[ \frac{k \cdot 6 \times 10^{-6}}{5 \times 10^{-2}} \right] \] \[ V_C = 0 \]

Step 2: Calculate the potential at point \( A \):
\[ V_A = \left[ \frac{k \cdot 6 \times 10^{-6}}{15 \times 10^{-2}} \right] - \left[ \frac{k \cdot 6 \times 10^{-6}}{5 \times 10^{-2}} \right] \] \[ V_A = \frac{k \cdot 6 \times 10^{-6}}{15 \times 10^{-2}} \left[ 1 - \frac{3}{15} \right] \] \[ V_A = \frac{9 \times 10^9 \cdot 6 \times 10^{-6} \cdot 2}{15 \times 10^{-2}} \] \[ V_A = -7.2 \times 10^5\, V \]

Step 3: Calculate the work done: \[ W = q \cdot [V_A - V_C] \] \[ W = 5 \times 10^{-6} \cdot \left[ -7.2 \times 10^5 - 0 \right] \] \[ W = -3.6\, J \]

Thus, the work done is \( W = -3.6\, J \). Quick Tip: When moving a charge in an electric field, the work done is the charge times the potential difference between the initial and final positions. The potential at a point due to multiple charges is the sum of the potentials due to each charge.

Step 1: Understanding proton motion in a magnetic field:

The proton is positively charged, and the force acting on it is given by the Lorentz force: \[ \vec{F} = q (\vec{v} \times \vec{B}) \]
where \( q \) is the charge of the proton and \( \vec{B} \) is the magnetic field.


Step 2: Determining the direction of magnetic field:

The proton moves in the plane of the paper, and the force is perpendicular to the velocity, which means the magnetic field is directed perpendicular to the paper.

At point P, the proton is moving along the curved path, and the force must be directed towards the center of the curve. Using the right-hand rule, the direction of \( \vec{B} \) at point P is out of the paper (towards the viewer).

At point Q, since the proton is turning in the opposite direction, the magnetic field direction is into the paper.

At point R, as the proton is turning in the same direction again, the magnetic field direction is again out of the paper.


Step 3: Relative magnitude of magnetic fields:

The magnitude of the magnetic field depends on the curvature of the proton's path. Since the proton moves in a non-uniform magnetic field, the magnetic field strength is greater where the curvature is higher.
Therefore, the magnetic field magnitude at point P (with sharp curvature) is greater than at point R (less curvature). The field at point Q, where the proton changes direction, could be comparable to point P.
Quick Tip: For charged particles in a magnetic field, the direction of the field is determined using the right-hand rule, and the field strength correlates with the curvature of the path.

Step 1: Magnetic field at the center of a current-carrying loop:

The magnetic field \( B \) at the center of a circular loop of radius \( r \) carrying current \( I \) is given by: \[ B = \frac{\mu_0 I}{2r} \]
where \( \mu_0 \) is the permeability of free space.

Step 2: Relating current to magnetic moment:

The magnetic moment \( \vec{\mu} \) of a loop is given by: \[ \vec{\mu} = I A \hat{n} \]
where \( A \) is the area of the loop and \( \hat{n} \) is the unit vector perpendicular to the plane of the loop.

The magnetic moment is also related to the magnetic field at the center by: \[ B = \frac{\mu_0 \mu}{2\pi r^3} \]

Step 3: Substituting the expression for \( B \):

We substitute \( r = \sqrt{\frac{A}{\pi}} \) (from the area of the loop) into the equation for magnetic field: \[ B = \frac{2\mu_0 \mu}{\mu_0 r} \] \[ \Rightarrow \mu = \frac{B A}{2 \mu_0} \]

Thus, the magnetic moment is: \[ \mu = \frac{2BA}{\mu_0} \sqrt{\frac{A}{\pi}} \] Quick Tip: Magnetic moment of a current-carrying loop is proportional to both the current and the area of the loop. The magnetic field at the center of the loop is inversely proportional to the radius.

Let the current loop be a rectangle with dimensions \( l \) and \( w \) (length and width) carrying a current \( I \). The loop is placed in a uniform magnetic field \( \vec{B} \) with the plane of the loop perpendicular to the field.

The torque \( \tau \) acting on the loop is given by: \[ \tau = \vec{\mu} \times \vec{B} \]
where \( \mu \) is the magnetic dipole moment of the current loop, and \( \vec{B} \) is the external magnetic field.

The magnetic dipole moment is defined as: \[ \mu = I \cdot A \]
where \( A \) is the area of the loop. For a rectangular loop, the area is: \[ A = l \cdot w \]

Thus, the magnetic moment becomes: \[ \mu = I \cdot l \cdot w \]

Now, the torque is given by: \[ \tau = \mu B \sin \theta \]
where \( \theta \) is the angle between the magnetic moment and the magnetic field.

For the case where the magnetic moment is perpendicular to the field (\( \theta = 90^\circ \)), the torque simplifies to: \[ \tau = I \cdot l \cdot w \cdot B \]

Thus, the expression for the torque acting on a rectangular current loop suspended in a uniform magnetic field is: \[ \tau = I \cdot A \cdot B \cdot \sin \theta \] Quick Tip: The torque on a current loop in a magnetic field depends on the magnetic moment of the loop and the external field. When the loop is perpendicular to the field, the torque is maximized.

Let the mass of the particle be \( m \), the charge be \( q \), and its velocity be \( v \). The radius \( r \) of the circular path is related to the velocity by: \[ r = \frac{mv}{qB} \]
where \( B \) is the magnetic field strength.

Part (1) Change in Radius:

The particle loses half of its kinetic energy. The initial kinetic energy is: \[ K_i = \frac{1}{2} m v^2 \]
After passing through the sheet of lead, the final kinetic energy is: \[ K_f = \frac{1}{2} K_i = \frac{1}{4} m v^2 \]

The new velocity \( v' \) after the energy loss can be found by equating the final kinetic energy: \[ K_f = \frac{1}{2} m v'^2 \quad \Rightarrow \quad \frac{1}{4} m v^2 = \frac{1}{2} m v'^2 \] \[ v' = \frac{v}{\sqrt{2}} \]

The new radius \( r' \) is given by: \[ r' = \frac{m v'}{qB} = \frac{m \cdot \frac{v}{\sqrt{2}}}{qB} = \frac{r}{\sqrt{2}} \]

Thus, the radius decreases by a factor of \( \sqrt{2} \).

Part (2) Change in Time Period:

The time period \( T \) of revolution is given by: \[ T = \frac{2 \pi r}{v} \]
Substituting the expression for \( r' \), the new time period \( T' \) is: \[ T' = \frac{2 \pi r'}{v'} = \frac{2 \pi \cdot \frac{r}{\sqrt{2}}}{\frac{v}{\sqrt{2}}} = T \]

Thus, the time period remains unchanged. Quick Tip: When a charged particle moves through a material that reduces its kinetic energy, the radius of the path decreases because of the reduced velocity. However, the time period of revolution remains unchanged if only the velocity magnitude changes without altering the charge or magnetic field.

Coherent sources are sources of light that emit waves with a constant phase relationship. In other words, the phase difference between the waves from coherent sources remains constant over time. For sustained interference patterns, the waves need to maintain a fixed phase difference, which results in a stable constructive or destructive interference.

This constant phase relationship is necessary to observe sustained and stable interference patterns. Without coherence, the interference pattern would fade or change over time, making it impossible to observe clear and stable fringes.

(2) Lights from two independent sources are not coherent. Explain.


% Solution
Solution:

When lights come from two independent sources, the phase difference between the waves from each source is random. These sources do not have a fixed phase relationship and therefore are not coherent. As a result, the interference pattern formed by two such sources would be irregular and would not sustain itself, as the random phase differences lead to fluctuating constructive and destructive interference. Quick Tip: For sustained interference, coherence is key, ensuring that the phase difference between the waves remains constant.

The distance between adjacent bright fringes (the fringe separation \( \Delta y \)) on the screen is given by the formula: \[ \Delta y = \frac{\lambda \cdot L}{d} \]
where:

\( \lambda \) is the wavelength of the light,

\( L \) is the distance from the slits to the screen,

\( d \) is the distance between the two slits.


Given:

\( \lambda = 600\,nm = 600 \times 10^{-9}\,m \)

\( L = 1.20\,m \)

\( d = 0.1\,mm = 0.1 \times 10^{-3}\,m \)


Substituting the values: \[ \Delta y = \frac{600 \times 10^{-9} \cdot 1.20}{0.1 \times 10^{-3}} = 7.2 \times 10^{-3}\,m = 7.2\,mm \]

Thus, the adjacent bright fringes are \( 7.2\,mm \) apart.

(2) Find the angular width (in degrees) of the first bright fringe.


% Solution
Solution:

The angular width of the first bright fringe \( \theta \) is given by: \[ \theta = \frac{\lambda}{d} \]
where:
\( \lambda \) is the wavelength of the light,

\( d \) is the distance between the slits.


Substituting the values: \[ \theta = \frac{600 \times 10^{-9}}{0.1 \times 10^{-3}} = 6^\circ \]

Thus, the angular width of the first bright fringe is \( 6^\circ \). Quick Tip: The fringe separation is dependent on the wavelength of the light, the distance between the slits, and the distance to the screen. The angular width is related to the slit separation and wavelength.

Step 1: Definition of a wavefront:

A wavefront is the locus of points in a medium that are all in phase, i.e., points where the waves have the same phase. For a plane wave, all points on the wavefront are at the same distance from the source and oscillate in unison.


Step 2: Incident and refracted wavefronts:

When a plane wave hits a convex lens, the wavefront bends as it passes through the lens.
The part of the wavefront passing through the lens is refracted and converges towards the focal point of the lens, changing the curvature.

Step 3: Diagram for incident and refracted wavefronts:



Quick Tip: For a converging lens, the incident plane wavefront gets refracted and converges at the focal point, changing the curvature.

Ray Diagram:

A parallel beam of light strikes the spherical glass ball.

The rays get refracted as they pass through the spherical surface.

The rays converge to form an image inside the sphere, depending on the radius and refractive index.





Step 1: Refraction at the first surface (from rarer to denser medium)

For refraction from the first surface, we use the formula: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
Where:

\( n_1 = 1 \) (refractive index of air),

\( n_2 = 1.5 \) (refractive index of the glass),

\( R = 15 \, cm \) (radius of the sphere),

\( u = \infty \) (object at infinity).


Substituting the values into the equation: \[ \frac{1.5}{v} - \frac{1}{\infty} = \frac{1.5 - 1}{15} \] \[ \frac{1.5}{v} = \frac{0.5}{15} \] \[ v = 45 \, cm \]

Thus, the image formed by the first surface is at a distance of \( v = 45 \, cm \) inside the sphere.

Step 2: Refraction at the second surface (from denser to rarer medium)

Now, for the refraction at the second surface, we use the formula: \[ \frac{n_1}{v'} - \frac{n_2}{u'} = \frac{n_1 - n_2}{R} \]
Where:

\( n_1 = 1 \) (refractive index of air),

\( n_2 = 1.5 \) (refractive index of the glass),

\( R = -15 \, cm \) (radius of curvature of the second surface),

\( u' = 15 \, cm \) (object distance from the second surface).

Substituting the values into the equation: \[ \frac{1}{v'} - \frac{1.5}{15} = \frac{1 - 1.5}{-15} \] \[ \frac{1}{v'} - \frac{0.1}{1} = \frac{-0.5}{-15} \] \[ \frac{1}{v'} - 0.1 = 0.0333 \] \[ \frac{1}{v'} = 0.1333 \] \[ v' = 7.5 \, cm \]

Thus, the final image position after refraction through the second surface is at \( 7.5 \, cm \) inside the sphere.

Step 3: Final Image:

The final image is formed at a distance of \( 7.5 \, cm \) from the second surface.
The image is real and formed inside the spherical glass ball. Quick Tip: For spherical surfaces, the image formed depends on the object distance and the refractive index of the material.