Maharashtra Board HSC Physics 2015
Solved Question Paper
Q. 1. Select and write the most appropriate answer from the given alternatives for each sub-question: [7]
Explanation: The standard formula for the period of a conical pendulum is \(T = 2\pi \sqrt{\frac{l \cos \theta}{g}}\).
Option (c) simplifies to: \(4\pi \sqrt{\frac{1}{4}} \sqrt{\frac{l \cos \theta}{g}} = 4\pi \left(\frac{1}{2}\right) \sqrt{\frac{l \cos \theta}{g}} = 2\pi \sqrt{\frac{l \cos \theta}{g}}\).
Explanation: Rotational Kinetic Energy \(E = \frac{1}{2}I\omega^2\). Here, \(I\) (Moment of Inertia) depends on the distribution of mass, and \(\omega\) is the angular speed.
Explanation: The time period of a simple pendulum \(T = 2\pi \sqrt{\frac{l}{g}}\) is independent of the mass of the bob. Since the size is the same, the length \(l\) (distance to center of mass) remains unchanged.
Explanation: According to Hooke's Law, \(F = kx\) (where \(x\) is extension). This represents a linear relationship with a positive slope.
Explanation: When a longitudinal wave reflects from a rigid (denser) boundary, there is a phase change of \(\pi\) in displacement, but no phase change in pressure. Thus, a high-pressure region (compression) reflects as a high-pressure region (compression).
Explanation: \(G = \frac{Fr^2}{m_1 m_2}\). Dimensions \(= \frac{[MLT^{-2}][L^2]}{[M^2]} = [L^3 M^{-1} T^{-2}]\).
Explanation: Rate of heat loss due to radiation \( \frac{dQ}{dt} \propto A (T^4 - T_0^4) \). Since materials and temperature difference are identical, the rate depends only on surface area \(A = 4\pi r^2\).
Ratio = \(\frac{r_1^2}{r_2^2} = \left(\frac{6}{12}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\).
HSC Physics Board Papers with Solution
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- Physics - March 2024 - Hindi Medium View Answer Key
- Physics - March 2023 - English Medium View Answer Key
- Physics - July 2023 - English Medium View Answer Key
- Physics - March 2022 - English Medium View Answer Key
- Physics - July 2022 - English Medium View Answer Key
- Physics - July 2021 - English Medium View Answer Key
- Physics - February 2020 - English Medium View Answer Key
- Physics - March 2013 View
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- Physics - March 2014 View Answer Key
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- Physics - March 2015 View Answer Key
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- Physics - March 2017 View
- Physics - July 2017 View
Q. 2. Attempt any SIX: [12]
Acceleration \(\vec{a}\) is the rate of change of velocity \(\vec{v}\): $$ \vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(\vec{\omega} \times \vec{r}) $$ Using the product rule for vector differentiation: $$ \vec{a} = \left(\frac{d\vec{\omega}}{dt} \times \vec{r}\right) + \left(\vec{\omega} \times \frac{d\vec{r}}{dt}\right) $$ Substituting \(\frac{d\vec{\omega}}{dt} = \vec{\alpha}\) (angular acceleration) and \(\frac{d\vec{r}}{dt} = \vec{v}\): $$ \vec{a} = (\vec{\alpha} \times \vec{r}) + (\vec{\omega} \times \vec{v}) $$ Here, 1. \(\vec{a}_T = \vec{\alpha} \times \vec{r}\) is the Tangential Acceleration. 2. \(\vec{a}_R = \vec{\omega} \times \vec{v} = \vec{\omega} \times (\vec{\omega} \times \vec{r})\) is the Radial (Centripetal) Acceleration.
The satellite and the astronaut are both in a state of free fall towards the Earth under the influence of gravity. The gravitational force provides the necessary centripetal acceleration for the circular orbit.
If we consider the forces on the astronaut, the gravitational force \(mg\) is acting downwards. Let \(N\) be the normal reaction force from the floor of the satellite.
Equation of motion: \(mg - N = ma\), where \(a\) is the centripetal acceleration.
Since the satellite is in orbit, \(a = g\) (acceleration due to gravity at that height).
Therefore, \(mg - N = mg \implies N = 0\).
Since the feeling of weight comes from the normal reaction force, and \(N = 0\), the astronaut feels weightless.
Theorem of Parallel Axes: The moment of inertia of a body about any axis (\(I_O\)) is equal to the sum of its moment of inertia about a parallel axis passing through its centre of mass (\(I_C\)) and the product of the mass of the body (\(M\)) and the square of the perpendicular distance (\(h\)) between the two parallel axes. $$ I_O = I_C + Mh^2 $$ Theorem of Perpendicular Axes: The moment of inertia of a plane laminar body about an axis perpendicular to its plane (\(I_z\)) is equal to the sum of its moments of inertia about two mutually perpendicular axes (\(I_x\) and \(I_y\)) in its plane and intersecting at the point where the perpendicular axis cuts the lamina. $$ I_z = I_x + I_y $$
(a) Wien's Displacement Law: The wavelength for which the emissive power of a blackbody is maximum (\(\lambda_{max}\)) is inversely proportional to the absolute temperature (\(T\)) of the blackbody. $$ \lambda_{max} \propto \frac{1}{T} \quad \text{or} \quad \lambda_{max}T = b \text{ (constant)} $$ (b) First Law of Thermodynamics: The amount of heat (\(\Delta Q\)) supplied to a system is equal to the sum of the increase in internal energy (\(\Delta U\)) of the system and the external work done (\(\Delta W\)) by the system. $$ \Delta Q = \Delta U + \Delta W $$
Given:
Period \(T = 2\) s
Amplitude \(A = 10\) cm
Position: "4 cm from its positive extreme position".
Since extreme position is at \(x = A = 10\) cm, the position \(x\) is \(10 - 4 = 6\) cm.
(Note: Acceleration depends on displacement from mean position).
Formulae:
Angular frequency \(\omega = \frac{2\pi}{T}\)
Acceleration \(a = \omega^2 x\) (in magnitude)
Calculation:
\(\omega = \frac{2\pi}{2} = \pi\) rad/s.
\(x = 6\) cm.
\(a = \pi^2 (6) = 6\pi^2\) cm/s\(^2\).
Taking \(\pi^2 \approx 9.87\):
\(a \approx 6 \times 9.87 = 59.22\) cm/s\(^2\).
Given:
\(T_0 = 75.5\) dyne/cm (at 0°C)
\(\Delta t = 25 - 0 = 25^\circ\)C
\(\alpha = 2.7 \times 10^{-3} /^\circ\)C
Formula: \(T = T_0(1 - \alpha \Delta t)\)
Calculation:
\(T_{25} = 75.5 [1 - (2.7 \times 10^{-3} \times 25)]\)
\(T_{25} = 75.5 [1 - 0.0675]\)
\(T_{25} = 75.5 [0.9325]\)
\(T_{25} \approx 70.40\) dyne/cm.
Given:
Initial frequency \(n_1 = 15\) r.p.s \(\Rightarrow \omega_1 = 2\pi(15) = 30\pi\) rad/s
Final frequency \(n_2 = 5\) r.p.s \(\Rightarrow \omega_2 = 2\pi(5) = 10\pi\) rad/s
Angular displacement \(\theta = 50\) revs \(= 50 \times 2\pi = 100\pi\) rad
Formula: \(\omega_2^2 = \omega_1^2 + 2\alpha\theta\)
Calculation:
\((10\pi)^2 = (30\pi)^2 + 2\alpha(100\pi)\)
\(100\pi^2 = 900\pi^2 + 200\pi\alpha\)
\(200\pi\alpha = 100\pi^2 - 900\pi^2\)
\(200\pi\alpha = -800\pi^2\)
\(\alpha = \frac{-800\pi^2}{200\pi}\)
\(\alpha = -4\pi\) rad/s\(^2\).
Answer: The angular acceleration is \(-4\pi\) rad/s\(^2\) (or \(-12.56\) rad/s\(^2\)).
Given:
\(R_J / R_E = 5\)
\(T_E = 1\) year
Formula: Kepler's Third Law \(T^2 \propto R^3\) \(\Rightarrow \left(\frac{T_J}{T_E}\right)^2 = \left(\frac{R_J}{R_E}\right)^3\)
Calculation:
\(\left(\frac{T_J}{1}\right)^2 = (5)^3\)
\(T_J^2 = 125\)
\(T_J = \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5}\)
\(T_J \approx 5 \times 2.236 = 11.18\) years.
Q. 3. Attempt any THREE: [9]
Consider a liquid drop of radius \(R\) and surface tension \(T\). Let \(P_i\) be the pressure inside and \(P_o\) be the pressure outside. Excess pressure \(P = P_i - P_o\).
Let the radius increase by a small amount \(dR\) due to this excess pressure.
Work done by excess pressure:
Force = Pressure \(\times\) Surface Area = \(P \times 4\pi R^2\)
Work \(dW = \text{Force} \times \text{displacement} = (P \times 4\pi R^2) dR\) ... (1)
Increase in Surface Potential Energy:
Initial Area \(A_1 = 4\pi R^2\)
Final Area \(A_2 = 4\pi (R+dR)^2 \approx 4\pi (R^2 + 2R dR) = 4\pi R^2 + 8\pi R dR\)
Increase in area \(dA = 8\pi R dR\)
Work done against surface tension \(dW = T \times dA = T(8\pi R dR)\) ... (2)
Equating (1) and (2):
\(P(4\pi R^2)dR = T(8\pi R dR)\)
\(P R = 2T\)
$$ P = \frac{2T}{R} $$
Doppler Effect: The apparent change in the frequency of sound heard by an observer due to the relative motion between the source of sound and the observer is called the Doppler effect.
Applications:
- Radar: To determine the speed of moving vehicles (by police).
- Astronomy: To estimate the speed of stars and galaxies (Red shift/Blue shift).
- Medical Science (Sonography): To study the motion of blood/heart beat in echocardiography.
- Airport Radar: To track the position and speed of aircrafts.
Given:
\(T = 27^\circ\text{C} = 300\) K
\(R = 8320\) J/kmol K
Molar mass of Oxygen (\(O_2\)) \(M = 32\) kg/kmol.
(a) Average KE per kilomole:
\(E = \frac{3}{2} RT\)
\(E = 1.5 \times 8320 \times 300\)
\(E = 3744000\) J \(= 3.744 \times 10^6\) J/kmol.
(b) Average KE per kilogram:
\(E_{kg} = \frac{E_{kmol}}{M}\)
\(E_{kg} = \frac{3.744 \times 10^6}{32}\)
\(E_{kg} = 117000\) J/kg \(= 1.17 \times 10^5\) J/kg.
Given:
Area \(A = 5 \text{ mm}^2 = 5 \times 10^{-6} \text{ m}^2\)
\(\Delta T = 25 - 0 = 25^\circ\)C
\(\alpha = 12 \times 10^{-6} /^\circ\)C
\(Y = 20 \times 10^{10} \text{ N/m}^2\)
Calculations:
1. Thermal Strain:
Strain = \(\alpha \Delta T = (12 \times 10^{-6}) \times 25\)
Strain = \(300 \times 10^{-6} = 3 \times 10^{-4}\).
2. Force:
\(Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\text{Strain}}\)
\(F = Y \cdot A \cdot \text{Strain}\)
\(F = (20 \times 10^{10}) \times (5 \times 10^{-6}) \times (3 \times 10^{-4})\)
\(F = (20 \times 5 \times 3) \times 10^{10-6-4}\)
\(F = 300 \times 10^0 = 300\) N.
Answer: Force = 300 N, Strain = \(3 \times 10^{-4}\).
Q. 4. [7]
Forced Vibrations: Vibrations in which a body vibrates with a frequency other than its natural frequency under the influence of an external periodic force are called forced vibrations.
Resonance: It is a special case of forced vibration where the frequency of the external driving force matches the natural frequency of the body, causing the body to vibrate with maximum amplitude.
Pipe closed at one end (Odd Harmonics):
Let \(L\) be the length of the pipe and \(v\) be the speed of sound.
Fundamental Mode (1st Harmonic):
Node at closed end, Antinode at open end. Length \(L = \lambda/4 \Rightarrow \lambda = 4L\).
Frequency \(n_1 = \frac{v}{\lambda} = \frac{v}{4L}\).
First Overtone (3rd Harmonic):
\(L = \frac{3\lambda_1}{4} \Rightarrow \lambda_1 = \frac{4L}{3}\).
Frequency \(n_2 = \frac{v}{\lambda_1} = \frac{3v}{4L} = 3n_1\).
Second Overtone (5th Harmonic):
\(L = \frac{5\lambda_2}{4} \Rightarrow \lambda_2 = \frac{4L}{5}\).
Frequency \(n_3 = \frac{v}{\lambda_2} = \frac{5v}{4L} = 5n_1\).
Since frequencies are \(n_1, 3n_1, 5n_1, \dots\), only odd harmonics are present.
Given:
\(n_1 = 256\) Hz
\(n_2 = 320\) Hz
\(l_2 = l_1 - 10\) cm
Force \(T\) and mass per unit length \(m\) are constant.
Formula: \(n \propto \frac{1}{l} \implies n_1 l_1 = n_2 l_2\)
Calculation:
\(256 \cdot l_1 = 320 (l_1 - 10)\)
\(256 l_1 = 320 l_1 - 3200\)
\(320 l_1 - 256 l_1 = 3200\)
\(64 l_1 = 3200\)
\(l_1 = \frac{3200}{64} = 50\)
Answer: Original length of wire = 50 cm.
Derivation:
Restoring force \(F = -kx\).
Work done to displace particle by \(dx\) against this force is \(dW = -F dx = kx dx\).
Total work done to displace from 0 to \(x\):
\(W = \int_0^x kx \, dx = \frac{1}{2}kx^2\).
This work is stored as Potential Energy. \(PE = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2\).
Evaluations:
(a) At mean position (\(x=0\)): \(PE = \frac{1}{2}k(0)^2 = 0\).
(b) At extreme position (\(x=\pm A\)): \(PE = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2\).
Given:
Initial freq \(n_1 = 100\) rpm.
Initial MI \(I_1 = 2 \times 10^{-4}\) kg m\(^2\).
Mass of wax \(m = 20 \text{ g} = 0.02\) kg.
Distance \(r = 5 \text{ cm} = 0.05\) m.
Principle: Conservation of Angular Momentum (\(L_1 = L_2 \implies I_1 n_1 = I_2 n_2\)).
Calculation:
MI of wax \(I_{wax} = mr^2 = 0.02 \times (0.05)^2 = 0.02 \times 0.0025 = 0.5 \times 10^{-4}\) kg m\(^2\).
Total Final MI \(I_2 = I_1 + I_{wax} = 2 \times 10^{-4} + 0.5 \times 10^{-4} = 2.5 \times 10^{-4}\) kg m\(^2\).
Using \(I_1 n_1 = I_2 n_2\):
\((2 \times 10^{-4}) \times 100 = (2.5 \times 10^{-4}) \times n_2\)
\(200 = 2.5 n_2\)
\(n_2 = \frac{200}{2.5} = \frac{2000}{25} = 80\) rpm.
Answer: New frequency = 80 r.p.m.
Q. 5. Select and write the most appropriate answer from the given alternatives for each sub-question: [7]
(Since \(K = h\nu - \Phi\), doubling \(\nu\) gives \(K' = 2h\nu - \Phi = 2(K+\Phi) - \Phi = 2K + \Phi\), which is > 2K).
(Velocity \(v \propto 1/n\), so \(p = mv \propto 1/n\)).
(Indium is a Group 13/trivalent element, which acts as an acceptor).
Explanation: \(d = \frac{\lambda}{2NA} = \frac{6 \times 10^{-7}}{2(0.12)} = \frac{6}{0.24} \times 10^{-7} = 25 \times 10^{-7}\) m.
Q. 6. Attempt any SIX: [12]
Polaroid: A large sheet of synthetic material capable of producing plane polarized light from unpolarized light is called a polaroid. It contains long chain molecules aligned in a specific direction.
Uses:
1. Used in sunglasses to reduce glare.
2. Used in photographic cameras to eliminate reflections.
3. Used in 3D movie glasses.
(1) Magnetization (M): The net magnetic dipole moment developed per unit volume of a material when placed in a magnetizing field. (\(M = m_{net}/V\)).
(2) Magnetic Intensity (H): The ability of a magnetizing field to magnetize a material medium. It is defined as the number of ampere-turns per unit length of a solenoid/toroid needed to produce the field. (\(B = \mu H\)).
Block Diagram: Information Source \(\rightarrow\) Transmitter \(\rightarrow\) Channel (with Noise) \(\rightarrow\) Receiver \(\rightarrow\) User of Information.
Given:
Length \(L = 3.142\) m \(\approx \pi\) m.
Turns: 2 layers \(\times\) 500 = 1000 turns.
Current \(I = 5\) A.
Formula: \(B = \mu_0 n I\) where \(n = N/L\).
Calculation:
\(n = \frac{1000}{\pi}\).
\(B = (4\pi \times 10^{-7}) \times \frac{1000}{\pi} \times 5\)
\(B = 4 \times 1000 \times 5 \times 10^{-7}\)
\(B = 20000 \times 10^{-7} = 2 \times 10^{-3}\) Tesla.
Formula: \(M = NIA\)
Calculation:
\(M = 300 \times 15 \times 5 \times 10^{-3}\)
\(M = 4500 \times 0.005\)
\(M = 22.5\) A m\(^2\).
Formula: \(|e| = \frac{d\phi}{dt}\)
Calculation:
\(\phi = 8t^2 + 6t + C\)
\(\frac{d\phi}{dt} = 16t + 6\)
At \(t=2\): \(e = 16(2) + 6 = 38\) mWb/s.
Since unit of \(\phi\) is milliweber, \(e = 38\) mV.
Interpretation:
The problem asks for the ionization energy for "this atom" (Hydrogen, given -13.6eV ground state). Standard Ionization Energy is defined as the energy required to remove an electron from the ground state to infinity.
Ionization Energy of H-atom = \(+13.6\) eV.
However, if the question implies calculating the energy required to remove the electron from the current 5th orbit (often called binding energy of that state):
Energy of \(n\)th orbit \(E_n = \frac{-13.6}{n^2}\) eV.
For \(n=5\): \(E_5 = \frac{-13.6}{25} = -0.544\) eV.
Energy required to ionize from \(n=5\) is \(+0.544\) eV.
Q. 7. Attempt any THREE: [9]
1. Centripetal Force = Electrostatic Force:
\(\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2} \implies mv^2 = \frac{e^2}{4\pi\epsilon_0 r}\) ... (1)
2. Bohr's Quantization Condition:
\(mvr = \frac{nh}{2\pi} \implies v = \frac{nh}{2\pi mr}\) ... (2)
3. Substitute (2) in (1):
\(m \left(\frac{nh}{2\pi mr}\right)^2 = \frac{e^2}{4\pi\epsilon_0 r}\)
\(m \frac{n^2 h^2}{4\pi^2 m^2 r^2} = \frac{e^2}{4\pi\epsilon_0 r}\)
\(\frac{n^2 h^2}{\pi m r} = \frac{e^2}{\epsilon_0}\)
\(r = \frac{\epsilon_0 n^2 h^2}{\pi m e^2}\)
\(\alpha\) (Current Amplification Factor in CB mode): Ratio of collector current to emitter current at constant collector-base voltage. \(\alpha = \frac{I_C}{I_E}\).
\(\beta\) (Current Amplification Factor in CE mode): Ratio of collector current to base current at constant collector-emitter voltage. \(\beta = \frac{I_C}{I_B}\).
Relation:
We know \(I_E = I_B + I_C\).
Divide by \(I_C\): \(\frac{I_E}{I_C} = \frac{I_B}{I_C} + 1\)
Since \(\frac{I_E}{I_C} = \frac{1}{\alpha}\) and \(\frac{I_B}{I_C} = \frac{1}{\beta}\):
\(\frac{1}{\alpha} = \frac{1}{\beta} + 1\)
\(\frac{1}{\beta} = \frac{1}{\alpha} - 1 = \frac{1-\alpha}{\alpha}\)
\(\beta = \frac{\alpha}{1-\alpha}\)
Concept: By Gauss's Law, Total Normal Electric Induction (T.N.E.I) over a closed surface is equal to the algebraic sum of charges enclosed (\(Q_{enclosed}\)).
Calculations:
Sphere 1: \(R_1 = 2\) mm \(= 2\times 10^{-3}\) m, \(\sigma_1 = 5 \mu C/m^2\).
\(q_1 = \sigma_1 \times 4\pi R_1^2 = (5\times 10^{-6}) \times 4\pi (2\times 10^{-3})^2\)
\(q_1 = 20\pi \times 10^{-6} \times 4 \times 10^{-6} = 80\pi \times 10^{-12}\) C.
Sphere 2: \(R_2 = 1\) mm \(= 1\times 10^{-3}\) m, \(\sigma_2 = -2 \mu C/m^2\).
\(q_2 = \sigma_2 \times 4\pi R_2^2 = (-2\times 10^{-6}) \times 4\pi (1\times 10^{-3})^2\)
\(q_2 = -8\pi \times 10^{-12}\) C.
Total Charge \(Q = q_1 + q_2 = (80\pi - 8\pi) \times 10^{-12} = 72\pi \times 10^{-12}\) C.
\(Q = 72 \times 3.142 \times 10^{-12} \approx 226.2 \times 10^{-12}\) C (or 226.2 pC).
Given:
\(\lambda_0 = 3800 \mathring{A}\)
\(\lambda = 2600 \mathring{A}\)
\(hc \approx 12400\) eV \(\mathring{A}\) (Useful approximation, or calculate using constants given).
Using constants: \(hc = 6.63 \times 10^{-34} \times 3 \times 10^8 = 19.89 \times 10^{-26}\) Jm.
Formula: \(K.E._{max} = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}\)
Calculation (in eV):
Energy of photon \(E = \frac{12400}{2600} \approx 4.769\) eV.
Work Function \(\Phi = \frac{12400}{3800} \approx 3.263\) eV.
\(K.E._{max} = 4.769 - 3.263 = 1.506\) eV.
(Using strict calculation: \(K = hc(\frac{1}{2.6 \times 10^{-7}} - \frac{1}{3.8 \times 10^{-7}})\) J, then convert to eV. Result is approx 1.51 eV).
Q. 8. [7]
Derivation:
Consider a coil of \(N\) turns and area \(A\) rotating with angular velocity \(\omega\) in magnetic field \(B\).
Angle \(\theta = \omega t\).
Magnetic Flux \(\phi = N \vec{B} \cdot \vec{A} = NBA \cos(\omega t)\).
Induced EMF \(e = -\frac{d\phi}{dt} = -NBA \frac{d}{dt}(\cos \omega t)\).
\(e = -NBA (-\sin \omega t \cdot \omega)\).
\(e = NBA\omega \sin(\omega t) = e_0 \sin(\omega t)\).
This is the equation for sinusoidal alternating EMF.
Graph: A sine wave starting from 0, reaching \(+e_0\) at \(T/4\), 0 at \(T/2\), \(-e_0\) at \(3T/4\), etc.
Given:
Resistance/length \(\sigma = 0.1 \Omega\)/cm.
Cell 1: \(E_1 = 1.5\) V, \(L_1 = 300\) cm.
Cell 2: \(E_2 = 1.4\) V, \(L_2 = ?\)
1. Calculate Current (I):
Potential drop across \(L_1\) equals \(E_1\).
\(V_1 = I \times R_{300} = I \times (300 \times 0.1) = I \times 30\).
\(1.5 = 30 I \implies I = \frac{1.5}{30} = 0.05\) A.
2. Calculate \(L_2\):
Using Principle of Potentiometer: \(\frac{E_1}{E_2} = \frac{L_1}{L_2}\).
\(\frac{1.5}{1.4} = \frac{300}{L_2}\)
\(L_2 = \frac{1.4 \times 300}{1.5} = 1.4 \times 200 = 280\) cm.
Answer: Current = 0.05 A, Balancing length = 280 cm.
Description:
1. A narrow slit S is illuminated by monochromatic light.
2. Light passes through a biprism, creating two virtual coherent sources \(S_1\) and \(S_2\).
3. Interference fringes are observed on a screen/eyepiece at distance \(D\).
4. Measure fringe width \(X\) (or \(\beta\)). Measure distance \(d\) between virtual sources using conjugate foci method (displacement method with convex lens).
5. Formula: \(\lambda = \frac{X d}{D}\).
Given:
Critical angle \(C = \sin^{-1}(3/5) \implies \sin C = 3/5\).
Formulae:
Refractive index \(\mu = \frac{1}{\sin C}\).
Brewster's Law: \(\tan i_p = \mu\).
Calculation:
\(\mu = \frac{1}{3/5} = \frac{5}{3}\).
\(\tan i_p = \frac{5}{3} \approx 1.6667\).
\(i_p = \tan^{-1}(1.6667)\) or \(i_p \approx 59^\circ\).