Maharashtra State Board HSC Physics (54)
Date: 16 Feb 2026 | Max Marks: 70
SECTION - A
Q. 1. Multiple Choice Questions [10 Marks]
(i) When a number of droplets coalesce to form a single drop, the total surface area of the drop:
Explanation: When small droplets coalesce, the total volume remains constant, but the total surface area decreases. This releases energy.
(ii) In an ideal gas, molecules possess:
Explanation: In an ideal gas, there are no intermolecular forces of attraction, hence potential energy is zero. They only possess kinetic energy due to motion.
(iii) If the frequency of incident radiation is increased above threshold frequency, keeping intensity and potential constant then the photoelectric current:
Explanation: Photoelectric current depends on the intensity (number of photons), not the frequency (energy of photons), provided the frequency is above the threshold.
(iv) The process in which heat is neither absorbed nor released by a system is called:
(v) The period of conical pendulum in terms of its length (l), semi vertical angle (\(\theta\)) and acceleration due to gravity (g) is:
Note: Option (a) in the source image has the typo \( \frac{1}{2\pi} \), but based on standard physics derivation, \( T = 2\pi\sqrt{\frac{h}{g}} = 2\pi\sqrt{\frac{l\cos\theta}{g}} \).
(vi) A conducting rod of length l, rotates about one of its ends in a uniform magnetic field B, with a constant angular velocity \(\omega\). If the plane of rotation is perpendicular to B, the e.m.f. induced between the ends of rod is:
(vii) A metal surface is illuminated by photons of energy 5 eV and 2.5 eV respectively. The ratio of their wavelengths of emitted radiation is:
Solution: \( E = \frac{hc}{\lambda} \Rightarrow E \propto \frac{1}{\lambda} \).
\( \frac{\lambda_1}{\lambda_2} = \frac{E_2}{E_1} = \frac{2.5}{5} = \frac{1}{2} \).
\( \frac{\lambda_1}{\lambda_2} = \frac{E_2}{E_1} = \frac{2.5}{5} = \frac{1}{2} \).
(viii) A particle is subjected to two parallel S.H.M.s such that \( x=2 \sin \omega t \) and \( y=2 \sin(\omega t+\frac{\pi}{3}) \). The amplitude of resultant S.H.M. will be:
Solution: \( R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi} \).
\( R = \sqrt{2^2 + 2^2 + 2(2)(2)\cos(60^\circ)} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3} \).
\( R = \sqrt{2^2 + 2^2 + 2(2)(2)\cos(60^\circ)} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3} \).
(ix) A bar magnet of magnetic moment \( 10~Am^{2} \) has a cross sectional area of \( 2.5\times10^{-4}m^{2} \). If the intensity of magnetisation of magnet is \( 10^{6}A/m \), the length of the bar magnet is:
Solution: \( M_z = \frac{m_{net}}{V} = \frac{m_{net}}{A \cdot L} \).
\( L = \frac{m_{net}}{M_z \cdot A} = \frac{10}{10^6 \cdot 2.5 \times 10^{-4}} = \frac{10}{2.5 \times 10^2} = \frac{10}{250} = 0.04m = 4cm \).
\( L = \frac{m_{net}}{M_z \cdot A} = \frac{10}{10^6 \cdot 2.5 \times 10^{-4}} = \frac{10}{2.5 \times 10^2} = \frac{10}{250} = 0.04m = 4cm \).
(x) In series LCR circuit for \( X_{L}>X_{C} \), \(\tan \phi\) will be:
Explanation: \( \tan \phi = \frac{X_L - X_C}{R} \). Since \( X_L > X_C \), the numerator is positive.
Q. 2. Answer the following questions [8 Marks]
(i) State the formula for electric field intensity due to uniformly charged spherical shell.
Answer: For a point outside the shell (r > R):
\( E = \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \)
For a point inside the shell (r < R): \( E = 0 \)
For a point inside the shell (r < R): \( E = 0 \)
(ii) Name an instrument for measurement of e.m.f. of a cell.
Answer: Potentiometer.
(iii) Calculate the magnitude of force experienced by a stationary charge exposed to uniform magnetic field.
Answer: The magnetic force is given by \( F = qvB \sin\theta \). Since the charge is stationary, \( v = 0 \). Therefore, the force \( F = 0 \).
(iv) Which property of bar magnet is used in navigation?
Answer: The directive property (a freely suspended magnet always aligns itself in the North-South direction).
(v) In Young's double slit experiment, width of the two slits are in the ratio 25:1. Calculate the ratio of amplitudes.
Answer:
\( \frac{W_1}{W_2} = \frac{I_1}{I_2} = \frac{25}{1} \)
Since \( I \propto A^2 \), \( \frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{25}{1}} = \frac{5}{1} \).
Ratio of amplitudes is 5:1.
Since \( I \propto A^2 \), \( \frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{25}{1}} = \frac{5}{1} \).
Ratio of amplitudes is 5:1.
(vi) What is beta plus decay?
Answer: \(\beta^+\) decay is a type of radioactive decay in which a proton inside the nucleus converts into a neutron, releasing a positron (\(e^+\)) and a neutrino (\(\nu\)).
\( p \rightarrow n + e^+ + \nu \)
\( p \rightarrow n + e^+ + \nu \)
(vii) If the tension in sonometer wire is increased by 21%, compare the initial frequency with the later.
Answer:
Frequency \( n \propto \sqrt{T} \).
Let \( T_1 = T \). Then \( T_2 = T + 0.21T = 1.21T \).
\( \frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{T}{1.21T}} = \frac{1}{1.1} = \frac{10}{11} \).
Ratio \( n_1:n_2 = 10:11 \).
Let \( T_1 = T \). Then \( T_2 = T + 0.21T = 1.21T \).
\( \frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{T}{1.21T}} = \frac{1}{1.1} = \frac{10}{11} \).
Ratio \( n_1:n_2 = 10:11 \).
(viii) Define second's pendulum.
Answer: A simple pendulum whose time period is exactly 2 seconds is called a second's pendulum.
SECTION - B
Attempt any EIGHT questions [16 Marks]
Q. 3. What are Eddy currents? State its two applications.
Answer:
Eddy Currents: Circulating currents induced in a bulk piece of conductor when the magnetic flux linked with it changes are called Eddy currents (or Foucault currents).
Applications:
Applications:
- Dead beat galvanometer: To stop the oscillation of the coil quickly.
- Induction Furnace: Used to melt metals using heat produced by eddy currents.
- Electric Brakes: Used in trains.
Q. 4. State any two sources of error in meter bridge experiment. Explain how they can be minimised.
Answer:
Sources of Error:
- Contact resistance at the points where wire is connected to copper strips.
- Non-uniformity of the bridge wire radius.
- Ends of the wire may not coincide exactly with the 0 and 100 cm marks of the scale (End error).
- Errors are minimized by obtaining the null point near the center of the wire (between 34cm and 66cm).
- By interchanging the positions of the unknown resistance and resistance box and taking the average.
Q. 5. Draw a ray diagram showing position of virtual sources and region of interference in biprism experiment.
Answer:
(Note: In an exam, draw a diagram showing a slit S, the biprism, two virtual sources S1 and S2 created by refraction, and the overlapping region on the screen/eyepiece forming interference bands.)
Q. 6. Derive an expression for radius of nth Bohr orbit.
Derivation:
1. Centripetal force = Electrostatic force: \( \frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0}\frac{Ze^2}{r^2} \Rightarrow mv^2r = \frac{Ze^2}{4\pi\epsilon_0} \) ...(i)
2. Bohr's quantization condition: \( mvr = \frac{nh}{2\pi} \Rightarrow v = \frac{nh}{2\pi mr} \) ...(ii)
Substitute (ii) into (i): \( m(\frac{nh}{2\pi mr})^2 r = \frac{Ze^2}{4\pi\epsilon_0} \)
\( \frac{n^2 h^2}{4\pi^2 m r} = \frac{Ze^2}{4\pi\epsilon_0} \)
\( r = \frac{\epsilon_0 n^2 h^2}{\pi m Z e^2} \)
1. Centripetal force = Electrostatic force: \( \frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0}\frac{Ze^2}{r^2} \Rightarrow mv^2r = \frac{Ze^2}{4\pi\epsilon_0} \) ...(i)
2. Bohr's quantization condition: \( mvr = \frac{nh}{2\pi} \Rightarrow v = \frac{nh}{2\pi mr} \) ...(ii)
Substitute (ii) into (i): \( m(\frac{nh}{2\pi mr})^2 r = \frac{Ze^2}{4\pi\epsilon_0} \)
\( \frac{n^2 h^2}{4\pi^2 m r} = \frac{Ze^2}{4\pi\epsilon_0} \)
\( r = \frac{\epsilon_0 n^2 h^2}{\pi m Z e^2} \)
Q. 7. A ceiling fan has moment of inertia of 2 kg \(m^{2}\). It attains maximum frequency of 60 r.p.m. in \(2\pi\) seconds. Calculate its power rating.
Solution:
\( I = 2 kg m^2 \)
\( n = 60 rpm = 1 rps \Rightarrow \omega_f = 2\pi n = 2\pi rad/s \)
\( \omega_i = 0 \)
\( t = 2\pi s \)
Angular acceleration \( \alpha = \frac{\omega_f - \omega_i}{t} = \frac{2\pi - 0}{2\pi} = 1 rad/s^2 \)
Torque \( \tau = I\alpha = 2 \times 1 = 2 Nm \)
Power \( P = \tau \omega_f = 2 \times 2\pi = 4\pi \) Watts (approx 12.56 W).
\( I = 2 kg m^2 \)
\( n = 60 rpm = 1 rps \Rightarrow \omega_f = 2\pi n = 2\pi rad/s \)
\( \omega_i = 0 \)
\( t = 2\pi s \)
Angular acceleration \( \alpha = \frac{\omega_f - \omega_i}{t} = \frac{2\pi - 0}{2\pi} = 1 rad/s^2 \)
Torque \( \tau = I\alpha = 2 \times 1 = 2 Nm \)
Power \( P = \tau \omega_f = 2 \times 2\pi = 4\pi \) Watts (approx 12.56 W).
Q. 8. An electric dipole consists of two unlike charges of magnitude \(2\times10^{-6}C\) each and separated by 4 cm. The dipole is placed in an external electric field of \(10^{5}\) N/C. Calculate the work done by an external agent to turn the dipole through 180°.
Solution:
\( q = 2\times 10^{-6} C \), \( 2l = 4 cm = 0.04 m \), \( E = 10^5 N/C \)
Dipole moment \( p = q \times 2l = 2\times 10^{-6} \times 0.04 = 8 \times 10^{-8} Cm \)
Work done \( W = pE(\cos\theta_1 - \cos\theta_2) \)
Assuming initial position is stable equilibrium (\(0^\circ\)) and turned to \(180^\circ\).
\( W = 8 \times 10^{-8} \times 10^5 (\cos 0^\circ - \cos 180^\circ) \)
\( W = 8 \times 10^{-3} (1 - (-1)) = 8 \times 10^{-3} (2) = 16 \times 10^{-3} J = 0.016 J \).
\( q = 2\times 10^{-6} C \), \( 2l = 4 cm = 0.04 m \), \( E = 10^5 N/C \)
Dipole moment \( p = q \times 2l = 2\times 10^{-6} \times 0.04 = 8 \times 10^{-8} Cm \)
Work done \( W = pE(\cos\theta_1 - \cos\theta_2) \)
Assuming initial position is stable equilibrium (\(0^\circ\)) and turned to \(180^\circ\).
\( W = 8 \times 10^{-8} \times 10^5 (\cos 0^\circ - \cos 180^\circ) \)
\( W = 8 \times 10^{-3} (1 - (-1)) = 8 \times 10^{-3} (2) = 16 \times 10^{-3} J = 0.016 J \).
Q. 9. Derive an expression for the magnetic field produced by a current in a circular arc of a wire using Biot-Savart law.
Answer:
Using \( dB = \frac{\mu_0}{4\pi} \frac{I dl \sin\theta}{r^2} \).
For a circular arc, the angle between current element \(dl\) and radius vector \(r\) is always \(90^\circ\) (\(\sin 90 = 1\)).
\( B = \int dB = \frac{\mu_0 I}{4\pi r^2} \int dl \).
\( \int dl \) is the length of the arc \( s = r\theta \).
\( B = \frac{\mu_0 I}{4\pi r^2} (r\theta) = \frac{\mu_0 I}{4\pi r} \theta \).
For a circular arc, the angle between current element \(dl\) and radius vector \(r\) is always \(90^\circ\) (\(\sin 90 = 1\)).
\( B = \int dB = \frac{\mu_0 I}{4\pi r^2} \int dl \).
\( \int dl \) is the length of the arc \( s = r\theta \).
\( B = \frac{\mu_0 I}{4\pi r^2} (r\theta) = \frac{\mu_0 I}{4\pi r} \theta \).
Q. 10. State advantages and disadvantages of photodiode.
Answer:
Advantages:
Advantages:
- Quick response (very fast switching speed).
- Linear response (Photocurrent is directly proportional to incident light intensity).
- Compact size and low cost.
- Its properties are temperature dependent (dark current increases with temperature).
- Active area is small, so it requires optical lenses to focus light.
- Requires external reverse bias voltage.
Q. 11. Distinguish between harmonics and overtones. [Any Two points]
Answer:
| Harmonics | Overtones |
|---|---|
| Harmonics are integral multiples of the fundamental frequency (n, 2n, 3n...). | Overtones are the actual frequencies present in the vibration above the fundamental frequency. |
| All harmonics may or may not be present in a given sound note. | Overtones are only those frequencies that are actually generated by the instrument. |
| The fundamental frequency is called the first harmonic. | The first frequency higher than the fundamental is called the first overtone. |
Q. 12. A steel ball with radius 0.3 mm is falling with velocity of \(2~m/s\) through a tube filled with glycerine. Calculate viscous force acting on the steel ball. [Given: \(\eta_{glycerine}=0.833~Ns/m^{2}\)]
Solution:
Given:
\( r = 0.3 \text{ mm} = 0.3 \times 10^{-3} \text{ m} = 3 \times 10^{-4} \text{ m} \)
\( v = 2 \text{ m/s} \)
\( \eta = 0.833 \text{ Ns/m}^2 \)
Formula: Stokes' Law
\( F = 6\pi \eta r v \)
Calculation:
\( F = 6 \times 3.142 \times 0.833 \times 3 \times 10^{-4} \times 2 \)
\( F = (6 \times 2 \times 3) \times 3.142 \times 0.833 \times 10^{-4} \)
\( F = 36 \times 3.142 \times 0.833 \times 10^{-4} \)
\( F \approx 94.22 \times 10^{-4} \text{ N} \)
Answer: The viscous force is \( 9.42 \times 10^{-3} \text{ N} \).
Given:
\( r = 0.3 \text{ mm} = 0.3 \times 10^{-3} \text{ m} = 3 \times 10^{-4} \text{ m} \)
\( v = 2 \text{ m/s} \)
\( \eta = 0.833 \text{ Ns/m}^2 \)
Formula: Stokes' Law
\( F = 6\pi \eta r v \)
Calculation:
\( F = 6 \times 3.142 \times 0.833 \times 3 \times 10^{-4} \times 2 \)
\( F = (6 \times 2 \times 3) \times 3.142 \times 0.833 \times 10^{-4} \)
\( F = 36 \times 3.142 \times 0.833 \times 10^{-4} \)
\( F \approx 94.22 \times 10^{-4} \text{ N} \)
Answer: The viscous force is \( 9.42 \times 10^{-3} \text{ N} \).
Q. 13. Calculate the temperature at which the average kinetic energy of a molecule of a gas will be same as that of an electron accelerated through 1 volt. [Given: \(k_{B}=1.4\times10^{-23}J/K\), \(e=1.6\times10^{-19}C\)]
Solution:
Condition: KE of gas molecule = Energy of electron
\( \frac{3}{2} k_B T = eV \)
\( T = \frac{2eV}{3k_B} \)
Calculation:
\( T = \frac{2 \times 1.6 \times 10^{-19} \times 1}{3 \times 1.4 \times 10^{-23}} \)
\( T = \frac{3.2}{4.2} \times 10^{4} \)
\( T = 0.7619 \times 10000 \)
\( T = 7619 \text{ K} \)
Condition: KE of gas molecule = Energy of electron
\( \frac{3}{2} k_B T = eV \)
\( T = \frac{2eV}{3k_B} \)
Calculation:
\( T = \frac{2 \times 1.6 \times 10^{-19} \times 1}{3 \times 1.4 \times 10^{-23}} \)
\( T = \frac{3.2}{4.2} \times 10^{4} \)
\( T = 0.7619 \times 10000 \)
\( T = 7619 \text{ K} \)
Q. 14. An inductor of inductance 200 mH is connected to an A.C. source of peak e.m.f. 220 V and frequency 50 Hz. Calculate the peak current in the circuit.
Given:
Inductance (\(L\)) = \( 200 \text{ mH} = 200 \times 10^{-3} \text{ H} = 0.2 \text{ H} \)
Peak e.m.f. (\(E_0\)) = \( 220 \text{ V} \)
Frequency (\(f\)) = \( 50 \text{ Hz} \)
To Find:
Peak current (\(I_0\)) = ?
Formulae:
1. Inductive Reactance: \( X_L = 2\pi f L \)
2. Peak Current: \( I_0 = \frac{E_0}{X_L} \)
Calculation:
First, calculate the Inductive Reactance (\(X_L\)):
\( X_L = 2 \times 3.142 \times 50 \times 0.2 \)
\( X_L = 3.142 \times 100 \times 0.2 \)
\( X_L = 3.142 \times 20 \)
\( X_L = 62.84 \, \Omega \)
Now, calculate the Peak Current (\(I_0\)):
\( I_0 = \frac{220}{62.84} \)
Using log tables (as per exam instructions):
\( \log(220) = 2.3424 \)
\( \log(62.84) = 1.7982 \)
Subtracting logs: \( 2.3424 - 1.7982 = 0.5442 \)
Antilog(0.5442) \( \approx 3.501 \)
Alternatively, by direct division:
\( I_0 \approx 3.501 \text{ A} \)
Answer:
The peak current in the circuit is 3.501 A.
Inductance (\(L\)) = \( 200 \text{ mH} = 200 \times 10^{-3} \text{ H} = 0.2 \text{ H} \)
Peak e.m.f. (\(E_0\)) = \( 220 \text{ V} \)
Frequency (\(f\)) = \( 50 \text{ Hz} \)
To Find:
Peak current (\(I_0\)) = ?
Formulae:
1. Inductive Reactance: \( X_L = 2\pi f L \)
2. Peak Current: \( I_0 = \frac{E_0}{X_L} \)
Calculation:
First, calculate the Inductive Reactance (\(X_L\)):
\( X_L = 2 \times 3.142 \times 50 \times 0.2 \)
\( X_L = 3.142 \times 100 \times 0.2 \)
\( X_L = 3.142 \times 20 \)
\( X_L = 62.84 \, \Omega \)
Now, calculate the Peak Current (\(I_0\)):
\( I_0 = \frac{220}{62.84} \)
Using log tables (as per exam instructions):
\( \log(220) = 2.3424 \)
\( \log(62.84) = 1.7982 \)
Subtracting logs: \( 2.3424 - 1.7982 = 0.5442 \)
Antilog(0.5442) \( \approx 3.501 \)
Alternatively, by direct division:
\( I_0 \approx 3.501 \text{ A} \)
Answer:
The peak current in the circuit is 3.501 A.
SECTION - C
Attempt any EIGHT questions [24 Marks]
Q. 15. In thermodynamics, define: (a) Mechanical equilibrium (b) Chemical equilibrium (c) Thermal equilibrium
Answer:
(a) Mechanical Equilibrium: When there are no unbalanced forces within the system and between the system and its surroundings (Pressure is constant).
(b) Chemical Equilibrium: When the chemical composition of the system does not change with time (No chemical reactions).
(c) Thermal Equilibrium: When the temperature of the system is uniform throughout and does not change with time.
(a) Mechanical Equilibrium: When there are no unbalanced forces within the system and between the system and its surroundings (Pressure is constant).
(b) Chemical Equilibrium: When the chemical composition of the system does not change with time (No chemical reactions).
(c) Thermal Equilibrium: When the temperature of the system is uniform throughout and does not change with time.
Q. 16. Derive an expression for resonant frequency of series resonant circuit.
Answer:
At resonance, current is maximum, impedance (Z) is minimum.
\( Z = \sqrt{R^2 + (X_L - X_C)^2} \). For Z to be minimum, \( X_L = X_C \).
\( \omega L = \frac{1}{\omega C} \Rightarrow \omega^2 = \frac{1}{LC} \)
\( \omega = \frac{1}{\sqrt{LC}} \)
Since \( \omega = 2\pi f_r \), \( 2\pi f_r = \frac{1}{\sqrt{LC}} \)
\( f_r = \frac{1}{2\pi\sqrt{LC}} \)
\( Z = \sqrt{R^2 + (X_L - X_C)^2} \). For Z to be minimum, \( X_L = X_C \).
\( \omega L = \frac{1}{\omega C} \Rightarrow \omega^2 = \frac{1}{LC} \)
\( \omega = \frac{1}{\sqrt{LC}} \)
Since \( \omega = 2\pi f_r \), \( 2\pi f_r = \frac{1}{\sqrt{LC}} \)
\( f_r = \frac{1}{2\pi\sqrt{LC}} \)
Q. 17. Obtain an expression for period of a bar magnet vibrating in a uniform magnetic field and performing angular S.H.M.
Result: \( T = 2\pi\sqrt{\frac{I}{\mu B}} \) where I is moment of inertia, \(\mu\) is magnetic dipole moment, B is magnetic field.
Q. 18. Define magnetization. State its S.I. unit and dimensions. What is the relation between permeability and magnetic susceptibility?
Answer:
Magnetization (Mz): The net magnetic dipole moment per unit volume. \( M_z = \frac{m_{net}}{V} \).
SI Unit: Ampere/meter (A/m).
Dimensions: \( [L^{-1} M^0 T^0 I^1] \).
Relation: \( \mu = \mu_0 (1 + \chi) \) where \(\chi\) is susceptibility.
Magnetization (Mz): The net magnetic dipole moment per unit volume. \( M_z = \frac{m_{net}}{V} \).
SI Unit: Ampere/meter (A/m).
Dimensions: \( [L^{-1} M^0 T^0 I^1] \).
Relation: \( \mu = \mu_0 (1 + \chi) \) where \(\chi\) is susceptibility.
Q. 19. Derive an expression for electric potential due to a point charge.
Derivation:
Consider a point charge \( +q \) placed at origin \( O \). We want to determine the electric potential at a point \( P \) at a distance \( r \) from \( O \).
1. Definition: Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point against the electrostatic force.
2. Force at intermediate point: Consider an intermediate point \( M \) at a distance \( x \) from \( O \) on the path from infinity to \( P \). The electrostatic force on a unit positive charge at \( M \) is:
$$ F = \frac{1}{4\pi\epsilon_0} \frac{q \times 1}{x^2} $$ (Directed away from the charge).
3. Work done for small displacement: The work done \( dW \) to move the unit charge against this force through a small distance \( dx \) (towards \( O \)) is:
$$ dW = -F dx $$ (Negative sign indicates work is done against the repulsive force).
4. Total Work Done: Total work done in moving the unit charge from \( \infty \) to \( r \) is obtained by integrating \( dW \):
$$ W = \int_{\infty}^{r} - \left( \frac{1}{4\pi\epsilon_0} \frac{q}{x^2} \right) dx $$
$$ W = - \frac{q}{4\pi\epsilon_0} \int_{\infty}^{r} x^{-2} dx $$
Using \( \int x^n dx = \frac{x^{n+1}}{n+1} \):
$$ W = - \frac{q}{4\pi\epsilon_0} \left[ \frac{x^{-1}}{-1} \right]_{\infty}^{r} $$
$$ W = \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{x} \right]_{\infty}^{r} $$
$$ W = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{r} - \frac{1}{\infty} \right) $$
$$ W = \frac{1}{4\pi\epsilon_0} \frac{q}{r} $$
5. Conclusion: By definition, this work done is the electrostatic potential \( V \).
$$ V = \frac{1}{4\pi\epsilon_0} \frac{q}{r} $$
Consider a point charge \( +q \) placed at origin \( O \). We want to determine the electric potential at a point \( P \) at a distance \( r \) from \( O \).
1. Definition: Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point against the electrostatic force.
2. Force at intermediate point: Consider an intermediate point \( M \) at a distance \( x \) from \( O \) on the path from infinity to \( P \). The electrostatic force on a unit positive charge at \( M \) is:
$$ F = \frac{1}{4\pi\epsilon_0} \frac{q \times 1}{x^2} $$ (Directed away from the charge).
3. Work done for small displacement: The work done \( dW \) to move the unit charge against this force through a small distance \( dx \) (towards \( O \)) is:
$$ dW = -F dx $$ (Negative sign indicates work is done against the repulsive force).
4. Total Work Done: Total work done in moving the unit charge from \( \infty \) to \( r \) is obtained by integrating \( dW \):
$$ W = \int_{\infty}^{r} - \left( \frac{1}{4\pi\epsilon_0} \frac{q}{x^2} \right) dx $$
$$ W = - \frac{q}{4\pi\epsilon_0} \int_{\infty}^{r} x^{-2} dx $$
Using \( \int x^n dx = \frac{x^{n+1}}{n+1} \):
$$ W = - \frac{q}{4\pi\epsilon_0} \left[ \frac{x^{-1}}{-1} \right]_{\infty}^{r} $$
$$ W = \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{x} \right]_{\infty}^{r} $$
$$ W = \frac{q}{4\pi\epsilon_0} \left( \frac{1}{r} - \frac{1}{\infty} \right) $$
$$ W = \frac{1}{4\pi\epsilon_0} \frac{q}{r} $$
5. Conclusion: By definition, this work done is the electrostatic potential \( V \).
$$ V = \frac{1}{4\pi\epsilon_0} \frac{q}{r} $$
Q. 20. Obtain an expression for the de-Broglie wavelength associated with an electron accelerated from rest through a potential difference of V volts.
Derivation:
Consider an electron with mass \( m \) and charge \( e \) accelerated from rest through a potential difference \( V \).
1. Kinetic Energy: The work done on the electron by the electric field appears as its kinetic energy (\( E_k \)).
$$ E_k = eV $$ ...(i)
2. Momentum relation: If \( v \) is the velocity of the electron, then \( E_k = \frac{1}{2}mv^2 \). Multiplying and dividing by \( m \):
$$ E_k = \frac{m^2v^2}{2m} = \frac{p^2}{2m} $$
Where \( p = mv \) is the momentum. Thus:
$$ p = \sqrt{2mE_k} $$ ...(ii)
3. de-Broglie Wavelength: According to de-Broglie's hypothesis, the wavelength \( \lambda \) associated with a material particle of momentum \( p \) is:
$$ \lambda = \frac{h}{p} $$
Substituting value of \( p \) from (ii):
$$ \lambda = \frac{h}{\sqrt{2mE_k}} $$
Substituting \( E_k = eV \) from (i):
$$ \lambda = \frac{h}{\sqrt{2meV}} $$
4. Standard Calculation (Optional but recommended): Substituting standard values: \( h = 6.63 \times 10^{-34} Js \) \( m = 9.1 \times 10^{-31} kg \) \( e = 1.6 \times 10^{-19} C \)
$$ \lambda = \frac{1.228}{\sqrt{V}} \text{ nm} $$
Consider an electron with mass \( m \) and charge \( e \) accelerated from rest through a potential difference \( V \).
1. Kinetic Energy: The work done on the electron by the electric field appears as its kinetic energy (\( E_k \)).
$$ E_k = eV $$ ...(i)
2. Momentum relation: If \( v \) is the velocity of the electron, then \( E_k = \frac{1}{2}mv^2 \). Multiplying and dividing by \( m \):
$$ E_k = \frac{m^2v^2}{2m} = \frac{p^2}{2m} $$
Where \( p = mv \) is the momentum. Thus:
$$ p = \sqrt{2mE_k} $$ ...(ii)
3. de-Broglie Wavelength: According to de-Broglie's hypothesis, the wavelength \( \lambda \) associated with a material particle of momentum \( p \) is:
$$ \lambda = \frac{h}{p} $$
Substituting value of \( p \) from (ii):
$$ \lambda = \frac{h}{\sqrt{2mE_k}} $$
Substituting \( E_k = eV \) from (i):
$$ \lambda = \frac{h}{\sqrt{2meV}} $$
4. Standard Calculation (Optional but recommended): Substituting standard values: \( h = 6.63 \times 10^{-34} Js \) \( m = 9.1 \times 10^{-31} kg \) \( e = 1.6 \times 10^{-19} C \)
$$ \lambda = \frac{1.228}{\sqrt{V}} \text{ nm} $$
Q. 21. With a neat circuit diagram, explain the working of a full wave rectifier. Draw input-output waveforms.
1. Circuit Diagram:
The circuit consists of a center-tapped transformer, two diodes (\(D_1\) and \(D_2\)), and a load resistor (\(R_L\)).
2. Working:
3. Conclusion: Since current flows through the load resistor in the same direction during both half cycles of the input AC voltage, the output is unidirectional (DC). This process is called full wave rectification.
4. Input-Output Waveforms:
[Image of input and output waveforms of full wave rectifier]
The output waveform shows pulsating DC voltage with a frequency twice that of the input AC frequency (Ripple frequency = \(2f\)).
The circuit consists of a center-tapped transformer, two diodes (\(D_1\) and \(D_2\)), and a load resistor (\(R_L\)).
2. Working:
- Positive Half Cycle: During the positive half cycle of the AC input, terminal A of the secondary coil becomes positive with respect to the center tap (C), and terminal B becomes negative.
- Diode \(D_1\) is forward biased and conducts current.
- Diode \(D_2\) is reverse biased and does not conduct.
- Current flows through \(R_L\) from X to Y.
- Negative Half Cycle: During the negative half cycle of the AC input, terminal A becomes negative with respect to C, and terminal B becomes positive.
- Diode \(D_1\) is reverse biased and does not conduct.
- Diode \(D_2\) is forward biased and conducts current.
- Current again flows through \(R_L\) from X to Y (same direction).
3. Conclusion: Since current flows through the load resistor in the same direction during both half cycles of the input AC voltage, the output is unidirectional (DC). This process is called full wave rectification.
4. Input-Output Waveforms:
[Image of input and output waveforms of full wave rectifier]
The output waveform shows pulsating DC voltage with a frequency twice that of the input AC frequency (Ripple frequency = \(2f\)).
Q. 22. The string of a guitar is 80 cm long and has a fundamental frequency of 112 Hz. If a guitarist wishes to produce a frequency of 160 Hz, where should he press the string?
Solution:
According to the law of length, frequency is inversely proportional to vibrating length (\(n \propto \frac{1}{l}\)).
\( n_1 l_1 = n_2 l_2 \)
\( 112 \times 80 = 160 \times l_2 \)
\( l_2 = \frac{112 \times 80}{160} = \frac{112}{2} = 56 cm \).
The string should be pressed at 56 cm from the bridge (or the vibrating part should be 56 cm).
According to the law of length, frequency is inversely proportional to vibrating length (\(n \propto \frac{1}{l}\)).
\( n_1 l_1 = n_2 l_2 \)
\( 112 \times 80 = 160 \times l_2 \)
\( l_2 = \frac{112 \times 80}{160} = \frac{112}{2} = 56 cm \).
The string should be pressed at 56 cm from the bridge (or the vibrating part should be 56 cm).
Q. 23. 0.5 mole of an ideal gas at 300 K, expands isothermally from an initial volume of 2 L to a final volume of 6 L. Calculate: (a) work done by the gas (b) heat supplied to the gas.
Solution:
Isothermal Process (T constant).
\( W = nRT \ln(\frac{V_f}{V_i}) = 2.303 nRT \log_{10}(\frac{V_f}{V_i}) \)
\( W = 2.303 \times 0.5 \times 8.31 \times 300 \times \log(\frac{6}{2}) \)
\( W = 2.303 \times 0.5 \times 8.31 \times 300 \times 0.4771 \)
\( W \approx 1369.5 J \)
(b) For isothermal, \( \Delta U = 0 \), so \( Q = W = 1369.5 J \).
Isothermal Process (T constant).
\( W = nRT \ln(\frac{V_f}{V_i}) = 2.303 nRT \log_{10}(\frac{V_f}{V_i}) \)
\( W = 2.303 \times 0.5 \times 8.31 \times 300 \times \log(\frac{6}{2}) \)
\( W = 2.303 \times 0.5 \times 8.31 \times 300 \times 0.4771 \)
\( W \approx 1369.5 J \)
(b) For isothermal, \( \Delta U = 0 \), so \( Q = W = 1369.5 J \).
Q. 24. A galvanometer has a resistance of 40\(\Omega\) and a current of 4 mA is needed for full scale deflection. What is the resistance and how is it to be connected to convert the galvanometer (a) into an ammeter of 0.4 A range and (b) into a voltmeter of 5 V range?
Solution:
Given: \( G = 40\Omega, I_g = 4mA = 0.004 A \).
(a) Ammeter (0.4A): Connect Shunt (S) in parallel.
\( S = \frac{I_g G}{I - I_g} = \frac{0.004 \times 40}{0.4 - 0.004} = \frac{0.16}{0.396} \approx 0.404 \Omega \).
(b) Voltmeter (5V): Connect Resistance (X) in series.
\( X = \frac{V}{I_g} - G = \frac{5}{0.004} - 40 = 1250 - 40 = 1210 \Omega \).
Given: \( G = 40\Omega, I_g = 4mA = 0.004 A \).
(a) Ammeter (0.4A): Connect Shunt (S) in parallel.
\( S = \frac{I_g G}{I - I_g} = \frac{0.004 \times 40}{0.4 - 0.004} = \frac{0.16}{0.396} \approx 0.404 \Omega \).
(b) Voltmeter (5V): Connect Resistance (X) in series.
\( X = \frac{V}{I_g} - G = \frac{5}{0.004} - 40 = 1250 - 40 = 1210 \Omega \).
Q. 25. A coaxial cable consists of a central conducting core wire of radius 'a' and a coaxial cylindrical outer conductor of radius 'b'. The two conductors carry equal current in opposite directions, in and out of the plane of the paper. What will be the magnitude of magnetic induction B for (i) \(a < r < b\) and (ii) \(b < r\)? What will be its direction? where 'r' is the radius of the Ampere's circular loop.
Solution using Ampere's Circuital Law:
Ampere's Law states: \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed} \)
Case (i): For \( a < r < b \) (Inside the cable, between conductors)
Consider an Amperian loop of radius \( r \) such that \( a < r < b \).
The loop encloses only the current flowing through the inner conductor (radius \( a \)). Let this current be \( I \).
\( \oint B dl = B (2\pi r) \)
\( I_{enclosed} = I \)
Therefore, \( B (2\pi r) = \mu_0 I \)
Magnitude: \( B = \frac{\mu_0 I}{2\pi r} \)
Direction: Tangential to the circular loop (determined by Right Hand Thumb Rule).
Case (ii): For \( r > b \) (Outside the cable)
Consider an Amperian loop of radius \( r \) such that \( r > b \).
The loop encloses currents from both conductors:
Using Ampere's Law:
\( B (2\pi r) = \mu_0 (0) \)
\( B (2\pi r) = 0 \)
Magnitude: \( B = 0 \)
Direction: Not applicable (as field is zero).
Ampere's Law states: \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed} \)
Case (i): For \( a < r < b \) (Inside the cable, between conductors)
Consider an Amperian loop of radius \( r \) such that \( a < r < b \).
The loop encloses only the current flowing through the inner conductor (radius \( a \)). Let this current be \( I \).
\( \oint B dl = B (2\pi r) \)
\( I_{enclosed} = I \)
Therefore, \( B (2\pi r) = \mu_0 I \)
Magnitude: \( B = \frac{\mu_0 I}{2\pi r} \)
Direction: Tangential to the circular loop (determined by Right Hand Thumb Rule).
Case (ii): For \( r > b \) (Outside the cable)
Consider an Amperian loop of radius \( r \) such that \( r > b \).
The loop encloses currents from both conductors:
- Inner conductor carries current \( +I \) (e.g., out of page).
- Outer conductor carries current \( -I \) (equal magnitude, opposite direction, e.g., into page).
Using Ampere's Law:
\( B (2\pi r) = \mu_0 (0) \)
\( B (2\pi r) = 0 \)
Magnitude: \( B = 0 \)
Direction: Not applicable (as field is zero).
Q. 26. Energy of an electron in second Bohr orbit is -3.4 eV. Calculate its kinetic energy and potential energy in third Bohr orbit.
Solution:
\( E_n \propto \frac{1}{n^2} \).
\( E_2 = -3.4 eV \). Also \( E_2 = \frac{E_1}{2^2} \Rightarrow E_1 = 4 \times (-3.4) = -13.6 eV \).
Energy in 3rd orbit: \( E_3 = \frac{E_1}{3^2} = \frac{-13.6}{9} = -1.51 eV \).
Kinetic Energy (3rd): \( K.E. = |E_3| = 1.51 eV \).
Potential Energy (3rd): \( P.E. = 2 \times E_3 = 2 \times (-1.51) = -3.02 eV \).
\( E_n \propto \frac{1}{n^2} \).
\( E_2 = -3.4 eV \). Also \( E_2 = \frac{E_1}{2^2} \Rightarrow E_1 = 4 \times (-3.4) = -13.6 eV \).
Energy in 3rd orbit: \( E_3 = \frac{E_1}{3^2} = \frac{-13.6}{9} = -1.51 eV \).
Kinetic Energy (3rd): \( K.E. = |E_3| = 1.51 eV \).
Potential Energy (3rd): \( P.E. = 2 \times E_3 = 2 \times (-1.51) = -3.02 eV \).
SECTION - D
Attempt any THREE questions [12 Marks]
Q. 27. Derive Laplace's law for spherical membrane of bubble due to surface tension.
Answer:
For a soap bubble (2 surfaces):
Work done by excess pressure = Increase in Surface Energy
\( (P_i - P_o) \cdot 4\pi r^2 \cdot \Delta r = T \cdot 2 \cdot (8\pi r \Delta r) \)
\( P_i - P_o = \frac{4T}{r} \).
Work done by excess pressure = Increase in Surface Energy
\( (P_i - P_o) \cdot 4\pi r^2 \cdot \Delta r = T \cdot 2 \cdot (8\pi r \Delta r) \)
\( P_i - P_o = \frac{4T}{r} \).
Q. 28. Derive the relation between coefficient of absorption, coefficient of reflection and coefficient of transmission.
Derivation:
Let \( Q \) be the total amount of radiant energy incident on the surface of a body.
When this radiation falls on the body, it is partly absorbed, partly reflected, and partly transmitted.
Let:
$$Q_a + Q_r + Q_t = Q$$
Dividing both sides by \( Q \):
$$\frac{Q_a}{Q} + \frac{Q_r}{Q} + \frac{Q_t}{Q} = \frac{Q}{Q}$$
$$\frac{Q_a}{Q} + \frac{Q_r}{Q} + \frac{Q_t}{Q} = 1$$
By definition:
$$a + r + t_r = 1$$
Conclusion: The sum of the coefficients of absorption, reflection, and transmission is always equal to unity (1).
Let \( Q \) be the total amount of radiant energy incident on the surface of a body.
When this radiation falls on the body, it is partly absorbed, partly reflected, and partly transmitted.
Let:
- \( Q_a \) = Amount of radiant energy absorbed.
- \( Q_r \) = Amount of radiant energy reflected.
- \( Q_t \) = Amount of radiant energy transmitted.
$$Q_a + Q_r + Q_t = Q$$
Dividing both sides by \( Q \):
$$\frac{Q_a}{Q} + \frac{Q_r}{Q} + \frac{Q_t}{Q} = \frac{Q}{Q}$$
$$\frac{Q_a}{Q} + \frac{Q_r}{Q} + \frac{Q_t}{Q} = 1$$
By definition:
- Coefficient of absorption \( a = \frac{Q_a}{Q} \)
- Coefficient of reflection \( r = \frac{Q_r}{Q} \)
- Coefficient of transmission \( t_r \) (or \( t \)) \( = \frac{Q_t}{Q} \)
$$a + r + t_r = 1$$
Conclusion: The sum of the coefficients of absorption, reflection, and transmission is always equal to unity (1).
Q. 29. Compare the r.m.s. speed of hydrogen molecule at 127°C with r.m.s. speed of oxygen molecule at 27°C, given that molecular masses of hydrogen and oxygen are 2 and 32 respectively.
Given:
Hydrogen (\(H_2\)):
Temperature \( T_1 = 127^\circ C = 127 + 273 = 400 K \)
Molecular Mass \( M_1 = 2 \)
Oxygen (\(O_2\)):
Temperature \( T_2 = 27^\circ C = 27 + 273 = 300 K \)
Molecular Mass \( M_2 = 32 \)
Formula:
Root Mean Square speed \( v_{rms} = \sqrt{\frac{3RT}{M}} \)
Since \( R \) is constant, \( v_{rms} \propto \sqrt{\frac{T}{M}} \)
Calculation:
Let \( v_1 \) be the r.m.s speed of Hydrogen and \( v_2 \) be the r.m.s speed of Oxygen.
$$\frac{v_1}{v_2} = \sqrt{\frac{T_1}{M_1} \times \frac{M_2}{T_2}}$$
$$\frac{v_1}{v_2} = \sqrt{\frac{400}{2} \times \frac{32}{300}}$$
$$\frac{v_1}{v_2} = \sqrt{200 \times \frac{32}{300}}$$
$$\frac{v_1}{v_2} = \sqrt{\frac{2 \times 32}{3}}$$
$$\frac{v_1}{v_2} = \sqrt{\frac{64}{3}}$$
$$\frac{v_1}{v_2} = \frac{8}{\sqrt{3}}$$
Answer:
The ratio of r.m.s speed of Hydrogen to Oxygen is \( 8 : \sqrt{3} \) (or approx \( 4.62 : 1 \)).
Hydrogen (\(H_2\)):
Temperature \( T_1 = 127^\circ C = 127 + 273 = 400 K \)
Molecular Mass \( M_1 = 2 \)
Oxygen (\(O_2\)):
Temperature \( T_2 = 27^\circ C = 27 + 273 = 300 K \)
Molecular Mass \( M_2 = 32 \)
Formula:
Root Mean Square speed \( v_{rms} = \sqrt{\frac{3RT}{M}} \)
Since \( R \) is constant, \( v_{rms} \propto \sqrt{\frac{T}{M}} \)
Calculation:
Let \( v_1 \) be the r.m.s speed of Hydrogen and \( v_2 \) be the r.m.s speed of Oxygen.
$$\frac{v_1}{v_2} = \sqrt{\frac{T_1}{M_1} \times \frac{M_2}{T_2}}$$
$$\frac{v_1}{v_2} = \sqrt{\frac{400}{2} \times \frac{32}{300}}$$
$$\frac{v_1}{v_2} = \sqrt{200 \times \frac{32}{300}}$$
$$\frac{v_1}{v_2} = \sqrt{\frac{2 \times 32}{3}}$$
$$\frac{v_1}{v_2} = \sqrt{\frac{64}{3}}$$
$$\frac{v_1}{v_2} = \frac{8}{\sqrt{3}}$$
Answer:
The ratio of r.m.s speed of Hydrogen to Oxygen is \( 8 : \sqrt{3} \) (or approx \( 4.62 : 1 \)).
Q. 30. A conducting loop of area \(1m^{2}\) is placed normal to a uniform magnetic field of \(3~Wb/m^{2}\) If the magnetic field is uniformly reduced to \(1~Wb/m^{2}\) in 0.5 second, calculate the induced e.m.f. produced in the coil.
Given:
Area of loop (\(A\)) = \( 1 m^2 \)
Initial Magnetic Field (\(B_1\)) = \( 3 Wb/m^2 \)
Final Magnetic Field (\(B_2\)) = \( 1 Wb/m^2 \)
Time interval (\(dt\)) = \( 0.5 s \)
Formula:
According to Faraday's Law of Electromagnetic Induction:
$$ |e| = \left| \frac{d\phi}{dt} \right| = \left| \frac{d(BA)}{dt} \right| = A \left| \frac{dB}{dt} \right| $$
Calculation:
Change in Magnetic Field (\(dB\)) = \( B_2 - B_1 \)
\( dB = 1 - 3 = -2 Wb/m^2 \)
Magnitude of change \( |dB| = 2 Wb/m^2 \)
Substituting in the formula:
$$ |e| = 1 \times \frac{2}{0.5} $$
$$ |e| = \frac{2}{0.5} = 4 V $$
Answer:
The induced e.m.f. produced in the coil is 4 Volts.
Area of loop (\(A\)) = \( 1 m^2 \)
Initial Magnetic Field (\(B_1\)) = \( 3 Wb/m^2 \)
Final Magnetic Field (\(B_2\)) = \( 1 Wb/m^2 \)
Time interval (\(dt\)) = \( 0.5 s \)
Formula:
According to Faraday's Law of Electromagnetic Induction:
$$ |e| = \left| \frac{d\phi}{dt} \right| = \left| \frac{d(BA)}{dt} \right| = A \left| \frac{dB}{dt} \right| $$
Calculation:
Change in Magnetic Field (\(dB\)) = \( B_2 - B_1 \)
\( dB = 1 - 3 = -2 Wb/m^2 \)
Magnitude of change \( |dB| = 2 Wb/m^2 \)
Substituting in the formula:
$$ |e| = 1 \times \frac{2}{0.5} $$
$$ |e| = \frac{2}{0.5} = 4 V $$
Answer:
The induced e.m.f. produced in the coil is 4 Volts.
Q. 31. Using analytical method, obtain an expression for the fringe width of two interfering waves.
Derivation:
Consider Young's double slit experiment setup:
1. Path Difference:
The path difference between the waves reaching \( P \) from \( S_1 \) and \( S_2 \) is:
$$ \Delta x = S_2P - S_1P $$
From geometry, for \( D >> d \), the path difference is approximated as:
$$ \Delta x = \frac{y d}{D} $$
2. Condition for Bright Fringes (Constructive Interference):
For a bright fringe at \( P \), the path difference must be an integral multiple of wavelength (\( n\lambda \)).
$$ \frac{y_n d}{D} = n\lambda $$
Where \( n = 0, 1, 2, ... \)
Therefore, the distance of the \( n^{th} \) bright fringe from the center is:
$$ y_n = \frac{n \lambda D}{d} $$
3. Expression for Fringe Width (\( X \)):
Fringe width is defined as the distance between two consecutive bright (or dark) fringes.
Let's find the distance between the \( n^{th} \) and \( (n+1)^{th} \) bright fringe.
Distance of \( (n+1)^{th} \) bright fringe:
$$ y_{n+1} = \frac{(n+1) \lambda D}{d} $$
Fringe Width \( X = y_{n+1} - y_n \)
$$ X = \frac{(n+1) \lambda D}{d} - \frac{n \lambda D}{d} $$
$$ X = \frac{\lambda D}{d} (n + 1 - n) $$
$$ X = \frac{\lambda D}{d} $$
Conclusion:
The expression for fringe width is \( X = \frac{\lambda D}{d} \).
It shows that fringe width is directly proportional to wavelength (\( \lambda \)) and distance of screen (\( D \)), and inversely proportional to slit separation (\( d \)).
Consider Young's double slit experiment setup:
- Let \( S_1 \) and \( S_2 \) be two coherent monochromatic sources separated by distance \( d \).
- Let \( D \) be the distance between the sources and the screen.
- Let \( \lambda \) be the wavelength of light.
- Consider a point \( P \) on the screen at a distance \( y \) (or \( x \)) from the central bright point \( O \).
1. Path Difference:
The path difference between the waves reaching \( P \) from \( S_1 \) and \( S_2 \) is:
$$ \Delta x = S_2P - S_1P $$
From geometry, for \( D >> d \), the path difference is approximated as:
$$ \Delta x = \frac{y d}{D} $$
2. Condition for Bright Fringes (Constructive Interference):
For a bright fringe at \( P \), the path difference must be an integral multiple of wavelength (\( n\lambda \)).
$$ \frac{y_n d}{D} = n\lambda $$
Where \( n = 0, 1, 2, ... \)
Therefore, the distance of the \( n^{th} \) bright fringe from the center is:
$$ y_n = \frac{n \lambda D}{d} $$
3. Expression for Fringe Width (\( X \)):
Fringe width is defined as the distance between two consecutive bright (or dark) fringes.
Let's find the distance between the \( n^{th} \) and \( (n+1)^{th} \) bright fringe.
Distance of \( (n+1)^{th} \) bright fringe:
$$ y_{n+1} = \frac{(n+1) \lambda D}{d} $$
Fringe Width \( X = y_{n+1} - y_n \)
$$ X = \frac{(n+1) \lambda D}{d} - \frac{n \lambda D}{d} $$
$$ X = \frac{\lambda D}{d} (n + 1 - n) $$
$$ X = \frac{\lambda D}{d} $$
Conclusion:
The expression for fringe width is \( X = \frac{\lambda D}{d} \).
It shows that fringe width is directly proportional to wavelength (\( \lambda \)) and distance of screen (\( D \)), and inversely proportional to slit separation (\( d \)).
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