Board Question Paper: July 2023
Physics - HSC Maharashtra Board
(i) \( c = 3 \times 10^8 \) m/s
(ii) \( h = 6.63 \times 10^{-34} \) Js
(iii) \( \pi = 3.142 \)
(iv) \( g = 9.8 \) m/s²
(v) \( \epsilon_0 = 8.85 \times 10^{-12} \) C²/Nm²
(vi) \( \mu_0 = 4\pi \times 10^{-7} \) Wb/A-m
(vii) \( R_H = 1.097 \times 10^7 \) m⁻¹
(viii) \( R = 8.31 \) J mol⁻¹ K⁻¹
(i) A body performing uniform circular motion has constant _______.
Reason: In uniform circular motion, the speed is constant. Since kinetic energy depends only on speed (\( K.E. = \frac{1}{2}mv^2 \)), it remains constant. Velocity, displacement, and acceleration change direction continuously.
(ii) When soluble substance such as common salt (i.e. sodium chloride) is dissolved in water, surface tension of water _______.
Reason: Adding highly soluble impurities like common salt (NaCl) increases the intermolecular force of attraction, thereby increasing the surface tension of water.
(iii) Periodic time of angular oscillations of a bar magnet is proportional to the _______.
Formula: \( T = 2\pi \sqrt{\frac{I}{\mu B}} \). Therefore, \( T \propto \sqrt{\frac{I}{B}} \).
(iv) The coefficient of absorption of a body is equal to its coefficient of emission at a given temperature is called _______.
Statement: At a given temperature, the coefficient of absorption of a body is equal to its coefficient of emission (\(a = e\)).
(v) The second law of thermodynamics deals with the transfer of _______.
Reason: The second law imposes restrictions on the direction of heat transfer and the efficiency of heat engines.
(vi) Which of the following phenomenon proves that light is a transverse wave _______.
Reason: Polarization restricts the vibrations of a wave to a single plane, which is only possible for transverse waves.
(vii) Magnitude of induced e.m.f. produced between the ends of a conductor of length ‘\( \frac{L}{2} \)’ moving with a uniform velocity ‘v’ at right angles to a uniform magnetic field of intensity ‘2B’ will be _______.
Calculation: Induced EMF \( e = B_{eff} \cdot L_{eff} \cdot v \).
Here, \( B_{eff} = 2B \) and \( L_{eff} = \frac{L}{2} \).
\( e = (2B) \times (\frac{L}{2}) \times v = BLv \).
(viii) Solar cell operates on the principle of _______.
(ix) A body performing linear S. H. M. experiences a force of 0.2 N when displaced through 4 cm from the mean position. Its force constant will be _______.
Calculation: \( F = kx \Rightarrow k = \frac{F}{x} \).
\( F = 0.2 \) N, \( x = 4 \text{ cm} = 0.04 \) m.
\( k = \frac{0.2}{0.04} = \frac{20}{4} = 5 \) N/m.
(x) If a current of 1A flows through a solenoid of length 25 cm and made up of 250 turns of copper wire then the magnitude of magnetic induction inside the solenoid will be _______.
Calculation: \( B = \mu_0 n I \).
\( n = \frac{N}{L} = \frac{250}{0.25} = 1000 \) turns/m.
\( B = 4\pi \times 10^{-7} \times 1000 \times 1 = 4 \times 3.142 \times 10^{-4} \)
\( B = 12.568 \times 10^{-4} = 1.2568 \times 10^{-3} \) T.
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(i) State the formula for end correction in a resonance tube experiment.
(ii) Categorize the following into polar and non-polar dielectrics: (a) H₂O (b) CO₂
(a) H₂O: Polar dielectric
(b) CO₂: Non-polar dielectric
(iii) Define potential gradient.
(iv) Calculate the period of a particle performing linear S.H.M. with maximum speed of 0.08 m/s and maximum acceleration of 0.32 m/s².
Given: \( v_{max} = A\omega = 0.08 \) m/s, \( a_{max} = A\omega^2 = 0.32 \) m/s².
Dividing \( a_{max} \) by \( v_{max} \): \[ \frac{A\omega^2}{A\omega} = \omega = \frac{0.32}{0.08} = 4 \text{ rad/s} \] Period \( T = \frac{2\pi}{\omega} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ s} \approx 1.57 \text{ s} \).
(v) Define gyromagnetic ratio.
(vi) State the conditions for current and impedance in parallel resonance circuit.
1. The impedance is maximum (ideally infinite).
2. The current drawn from the source is minimum.
(vii) Radius of the third Bohr orbit is 0.477 nm. Calculate the radius of the second Bohr orbit.
Radius of \( n^{th} \) orbit \( r_n \propto n^2 \).
\( \frac{r_2}{r_3} = \frac{2^2}{3^2} = \frac{4}{9} \)
\( r_2 = r_3 \times \frac{4}{9} = 0.477 \times \frac{4}{9} \)
\( r_2 = 0.053 \times 4 = 0.212 \) nm.
(viii) Name the logic gate having single input- single output.
Attempt any EIGHT questions of the following: [16]
Q.3. State the conditions for a steady interference pattern.
- The two sources of light must be coherent (maintain a constant phase difference).
- The two sources must be monochromatic (emit light of a single wavelength).
- The two sources must be of equal amplitude (for better contrast).
- The two sources must be close to each other.
- The screen should be at a sufficient distance from the sources.
- The sources should be narrow.
Q.4. Derive an expression for electric field intensity due to an infinitely long straight charged wire.
Consider a uniformly charged infinite wire with linear charge density \( \lambda \).
Imagine a cylindrical Gaussian surface of radius \( r \) and length \( L \) coaxial with the wire.
According to Gauss's Law: \( \oint \vec{E} \cdot d\vec{s} = \frac{q}{\epsilon_0} \).
The electric field is perpendicular to the wire. The flux through the flat ends is zero.
Flux through curved surface = \( E \times 2\pi r L \).
Charge enclosed \( q = \lambda L \).
\[ E (2\pi r L) = \frac{\lambda L}{\epsilon_0} \] \[ E = \frac{\lambda}{2\pi \epsilon_0 r} \]
Q.5. What are eddy currents? Write two applications of eddy currents.
Definition: Eddy currents are circulating currents induced in a thick conductor (bulk piece of metal) when it is subjected to a changing magnetic flux.
Applications:
- Induction Furnace: Used to melt metals by producing large heat via eddy currents.
- Magnetic Braking: Used in electric trains where strong magnetic fields induce eddy currents in the wheels/rails to oppose motion.
Q.6. Define and state formulae for: (a) Inductive reactance (b) Capacitive reactance
(a) Inductive Reactance (\( X_L \)): The opposition offered by an inductor to the flow of AC.
Formula: \( X_L = \omega L = 2\pi f L \)
(b) Capacitive Reactance (\( X_C \)): The opposition offered by a capacitor to the flow of AC.
Formula: \( X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \)
Q.7. Draw a p –V diagram and explain the concept of positive work done and negative work done.
The p-V diagram shows pressure on the Y-axis and volume on the X-axis.
Positive Work Done: When a gas expands (volume increases from \( V_1 \) to \( V_2 \), \( V_2 > V_1 \)), the work done by the gas is positive. The arrow on the p-V curve points to the right.
Negative Work Done: When a gas is compressed (volume decreases from \( V_1 \) to \( V_2 \), \( V_2 < V_1 \)), the work done on the gas is negative. The arrow on the p-V curve points to the left.
(Note: In an exam, students should draw a graph showing expansion and compression curves).
Q.8. State the law of length and the law of linear density for a vibrating string.
Law of Length: The fundamental frequency of vibration of a string is inversely proportional to its vibrating length, provided tension and linear density are constant. (\( n \propto \frac{1}{l} \))
Law of Linear Density: The fundamental frequency is inversely proportional to the square root of its linear density (mass per unit length), provided tension and length are constant. (\( n \propto \frac{1}{\sqrt{m}} \))
Q.9. For a moving coil galvanometer, show that \( S = \frac{G}{n-1} \), where S is shunt resistance, G is galvanometer resistance, n is ratio of total current to full scale deflection current.
Let \( I \) be total current and \( I_g \) be full scale deflection current.
Shunt \( S \) is connected in parallel with Galvanometer \( G \).
Voltage across \( S \) = Voltage across \( G \).
\( (I - I_g)S = I_g G \)
\( S = \frac{I_g G}{I - I_g} \)
Divide numerator and denominator by \( I_g \):
\( S = \frac{G}{\frac{I}{I_g} - 1} \)
Given \( n = \frac{I}{I_g} \), therefore:
\( S = \frac{G}{n - 1} \)
Q.10. The de - Broglie wavelengths associated with an electron and a proton are same. Calculate the ratio of their kinetic energies. [Given: \( m_p = 1836 m_e \)]
De-Broglie wavelength \( \lambda = \frac{h}{p} \). Since wavelengths are same, momentum \( p \) is same for both.
Kinetic Energy \( K = \frac{p^2}{2m} \).
Ratio: \( \frac{K_e}{K_p} = \frac{p^2/2m_e}{p^2/2m_p} = \frac{m_p}{m_e} \)
Given \( \frac{m_p}{m_e} = 1836 \).
\( \therefore \text{Ratio of Kinetic Energies } (K_e : K_p) = 1836 : 1 \).
Q.11. Derive an expression for 'Half Life Time' of a radioactive material using the ‘Law of Radioactive Decay’.
Law of decay: \( N = N_0 e^{-\lambda t} \)
At half life \( t = T_{1/2} \), the number of nuclei remaining is half the initial number: \( N = \frac{N_0}{2} \).
\( \frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \)
\( \frac{1}{2} = e^{-\lambda T_{1/2}} \)
Inverting: \( 2 = e^{\lambda T_{1/2}} \)
Taking natural log: \( \ln(2) = \lambda T_{1/2} \)
\( T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda} \).
Q.12. A motorcyclist performs stunt along the cylindrical wall of a ‘Well of Death’ of inner radius 4 m. Coefficient of static friction between the tyres and the wall is 0.4. Calculate the maximum period of revolution. [Use \( g = 10 \) m/s²]
To prevent slipping down, friction \( f_s \ge mg \).
\( \mu N \ge mg \), where Normal force \( N = mr\omega^2 \).
\( \mu m r \omega^2 \ge mg \Rightarrow \omega^2 \ge \frac{g}{\mu r} \).
For maximum period, we need minimum frequency/angular velocity. We consider the limiting case:
\( \omega_{min}^2 = \frac{g}{\mu r} \)
Since \( T = \frac{2\pi}{\omega} \), \( T_{max} = 2\pi \sqrt{\frac{\mu r}{g}} \).
Calculation: \( T_{max} = 2\pi \sqrt{\frac{0.4 \times 4}{10}} = 2\pi \sqrt{0.16} \)
\( T_{max} = 2\pi (0.4) = 0.8 \times 3.142 \)
\( T_{max} \approx 2.51 \) s.
Q.13. Calculate the diameter of a water drop, if the excess pressure inside the drop is 80 N/m²: [Surface Tension of water = \( 7.2 \times 10^{-2} \) N/m]
Excess pressure inside a drop \( P = \frac{2T}{r} \).
\( 80 = \frac{2 \times 7.2 \times 10^{-2}}{r} \)
\( r = \frac{14.4 \times 10^{-2}}{80} = 0.18 \times 10^{-2} \text{ m} = 1.8 \times 10^{-3} \text{ m} = 1.8 \text{ mm} \).
Diameter \( d = 2r = 3.6 \) mm.
Q.14. Calculate the current through a long straight wire at a distance of 2.4 cm, where the magnetic field intensity is 16 μT.
Formula: \( B = \frac{\mu_0 I}{2\pi r} \)
Given: \( B = 16 \times 10^{-6} \) T, \( r = 2.4 \times 10^{-2} \) m.
\( 16 \times 10^{-6} = \frac{4\pi \times 10^{-7} \times I}{2\pi \times 2.4 \times 10^{-2}} \)
\( 16 \times 10^{-6} = \frac{2 \times 10^{-7} \times I}{2.4 \times 10^{-2}} \)
\( I = \frac{16 \times 10^{-6} \times 2.4 \times 10^{-2}}{2 \times 10^{-7}} \)
\( I = \frac{38.4 \times 10^{-8}}{2 \times 10^{-7}} = 19.2 \times 10^{-1} = 1.92 \) A.
Attempt any EIGHT questions of the following: [24]
Q.15. Explain construction and working of Ferry’s Black Body.
Construction: It consists of a double-walled hollow sphere. The space between the walls is evacuated to prevent heat loss by conduction and convection. The inner surface is coated with lampblack (absorptivity ~98%). There is a small aperture acting as the inlet for radiation. Opposite the aperture, there is a conical projection.
Working: When radiation enters through the aperture, it strikes the conical projection and is reflected. Due to the geometry, the radiation suffers multiple internal reflections. At each reflection, about 98% is absorbed by lampblack. After multiple reflections, almost 100% of the radiation is absorbed. Thus, the aperture acts as a perfect black body.
Q.16. Show that the deflection produced in a moving coil galvanometer is directly proportional to the current flowing through its coil or vice-versa.
When current \( I \) flows through a coil of \( N \) turns and area \( A \) in a magnetic field \( B \), the deflecting torque is \( \tau_d = NIAB \sin \theta \).
With a radial magnetic field, \( \sin \theta = 1 \), so \( \tau_d = NIAB \).
The suspension fiber provides a restoring torque \( \tau_r = C \phi \), where \( C \) is twist constant and \( \phi \) is deflection angle.
At equilibrium, \( \tau_d = \tau_r \).
\( NIAB = C \phi \)
\( \phi = (\frac{NAB}{C}) I \)
Since \( N, A, B, C \) are constants, \( \phi \propto I \).
Q.17. With the help of a neat circuit diagram, explain the working of a half wave rectifier. Draw input-output waveforms.
Circuit: Consists of a transformer, a single p-n junction diode connected in series with a load resistor \( R_L \).
Working:
- During positive half cycle of AC input, the diode is forward biased and conducts current. Voltage appears across \( R_L \).
- During negative half cycle, the diode is reverse biased and does not conduct (open switch). Output voltage is zero.
Q.18. Obtain the expression for the period of simple pendulum performing S.H.M.
Let a pendulum of length \( L \) and mass \( m \) be displaced by angle \( \theta \).
Restoring force \( F = -mg \sin \theta \).
For small \( \theta \), \( \sin \theta \approx \theta = \frac{x}{L} \).
\( F = -mg \frac{x}{L} \).
Compare with \( F = -kx \), so \( k = \frac{mg}{L} \).
Period \( T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{m}{mg/L}} \).
\( T = 2\pi \sqrt{\frac{L}{g}} \).
Q.19. Show that the beat frequency of two interfering sound waves is the difference between the individual frequencies of the two sound waves.
Let two waves be \( y_1 = a \sin(2\pi n_1 t) \) and \( y_2 = a \sin(2\pi n_2 t) \).
By superposition, \( y = y_1 + y_2 = a [\sin(2\pi n_1 t) + \sin(2\pi n_2 t)] \).
Using trig identity: \( y = 2a \cos[2\pi(\frac{n_1-n_2}{2})t] \sin[2\pi(\frac{n_1+n_2}{2})t] \).
Resultant amplitude \( A = 2a \cos[2\pi(\frac{n_1-n_2}{2})t] \).
Intensity is max when \( \cos \) term is \(\pm 1\). This happens when \( \pi(n_1-n_2)t = k\pi \).
Time interval between successive maxima is \( T_{beat} = \frac{1}{n_1 - n_2} \).
Beat frequency \( N = \frac{1}{T_{beat}} = |n_1 - n_2| \).
Q.20. Obtain an expression for orbital magnetic moment of an electron revolving around the nucleus of an atom.
Electron revolves in orbit of radius \( r \) with velocity \( v \).
Current \( I = \frac{e}{T} = \frac{e}{2\pi r / v} = \frac{ev}{2\pi r} \).
Magnetic moment \( M_{orb} = I \times A = I \times (\pi r^2) \).
\( M_{orb} = (\frac{ev}{2\pi r}) (\pi r^2) = \frac{evr}{2} \).
Since angular momentum \( L = m_e v r \), \( vr = L/m_e \).
\( M_{orb} = \frac{e}{2m_e} L \).
Q.21. State Einstein's photoelectric equation. Hence, explain any two characteristics of photoelectric effect.
Equation: \( h\nu = \phi_0 + K_{max} \)
where \( h\nu \) is photon energy, \( \phi_0 \) is work function, \( K_{max} \) is max kinetic energy.
Characteristics:
- Threshold Frequency: If \( h\nu < \phi_0 \), no emission occurs regardless of intensity. This explains why emission stops below a certain frequency \( \nu_0 \).
- KE vs Frequency: \( K_{max} = h\nu - \phi_0 \). Max kinetic energy depends linearly on frequency and is independent of intensity.
Q.22. Calculate the shortest wavelength of Paschen series and longest wavelength of Balmer series for H – atom.
\( R_H \approx 1.097 \times 10^7 \text{ m}^{-1} \).
1. Shortest Paschen (\( n_1=3, n_2=\infty \)):
\( \frac{1}{\lambda_s} = R_H (\frac{1}{3^2} - \frac{1}{\infty}) = \frac{R_H}{9} \)
\( \lambda_s = \frac{9}{R_H} \approx 8.20 \times 10^{-7} \text{ m} = 820 \text{ nm} \).
2. Longest Balmer (\( n_1=2, n_2=3 \)):
\( \frac{1}{\lambda_l} = R_H (\frac{1}{2^2} - \frac{1}{3^2}) = R_H (\frac{1}{4} - \frac{1}{9}) = R_H (\frac{5}{36}) \)
\( \lambda_l = \frac{36}{5 R_H} \approx 6.56 \times 10^{-7} \text{ m} = 656 \text{ nm} \).
Q.23. A 60 W filament lamp loses all its energy by radiation from its surface. The emissivity of the filament surface is 0.5 and the surface area is \( 5 \times 10^{-5} \) m². Calculate the temperature of the filament.
Stefan-Boltzmann Law: \( P = \sigma \epsilon A T^4 \)
\( 60 = (5.67 \times 10^{-8}) \times 0.5 \times (5 \times 10^{-5}) \times T^4 \)
\( 60 = 14.175 \times 10^{-13} T^4 \)
\( T^4 = \frac{60 \times 10^{13}}{14.175} \approx 4.23 \times 10^{13} = 42.3 \times 10^{12} \)
\( T = (42.3)^{1/4} \times 10^3 \approx 2.55 \times 10^3 \)
\( T \approx 2550 \) K.
Q.24. Two capacitors of capacities \( C_1 \) and \( C_2 \) are connected in parallel and this combination is connected in series with a capacitor of capacity \( C_3 \). Calculate the equivalent capacity.
Parallel combination \( C_p = C_1 + C_2 \).
Series combination of \( C_p \) and \( C_3 \):
\( \frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C_3} \)
\( C_{eq} = \frac{C_p C_3}{C_p + C_3} = \frac{(C_1 + C_2)C_3}{C_1 + C_2 + C_3} \).
Q.25. With an unknown resistance X in the left gap and a resistance of 30 Ω in the right gap of a meter-bridge, the null point is obtained at 40 cm from the left end. Calculate the unknown resistance and shift in the position of the null point, when resistance in each gap is increased by 15Ω.
Case 1: \( \frac{X}{30} = \frac{40}{60} = \frac{2}{3} \Rightarrow X = 20 \Omega \).
Case 2: Resistances become \( X' = 20+15 = 35 \Omega \) and \( R' = 30+15 = 45 \Omega \).
\( \frac{35}{45} = \frac{L_x}{100-L_x} \)
\( \frac{7}{9} = \frac{L_x}{100-L_x} \Rightarrow 700 - 7L_x = 9L_x \Rightarrow 16L_x = 700 \).
\( L_x = 43.75 \) cm.
Shift = \( 43.75 - 40 = 3.75 \) cm (towards right).
Q.26. An inductor of inductance 200 mH is connected to an A.C. source of peak e.m.f. 210 V and frequency 50 Hz. Calculate the peak current and instantaneous voltage of the source when the current is at its peak value.
\( L = 0.2 \) H, \( E_0 = 210 \) V, \( f = 50 \) Hz.
\( X_L = 2\pi f L = 100\pi \times 0.2 = 20\pi \approx 62.83 \Omega \).
Peak Current \( I_0 = \frac{E_0}{X_L} = \frac{210}{62.83} \approx 3.34 \) A.
In a purely inductive circuit, voltage leads current by \( 90^\circ \).
When current is at peak, phase of current is \( 90^\circ \) (sine max). Voltage phase is \( 90^\circ + 90^\circ = 180^\circ \).
Instantaneous voltage \( e = E_0 \sin(\omega t + \pi/2) \). If we reference current as \( I = I_0 \sin \omega t \), then \( e = E_0 \cos \omega t \).
When \( I = I_0 \), \( \sin \omega t = 1 \implies \omega t = \pi/2 \).
Then \( e = E_0 \cos(\pi/2) = 0 \).
Instantaneous voltage is zero.
Attempt any THREE questions of the following: [12]
Q.27. State and prove theorem of parallel axes.
Statement: 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 center 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.
\( I_o = I_c + Mh^2 \).
Proof: Consider a body of mass \( M \). Let \( C \) be center of mass. Consider element \( dm \) at point \( P \). Let distance from CM axis be \( CP = r \) and from parallel axis be \( OP = q \). Extend \( OC \) to \( D \) such that \( PD \perp OC \).
By Pythagoras/Geometry in \( \triangle OPD \), \( OP^2 = OC^2 + CP^2 + 2 OC \cdot CD \).
Integrating \( \int q^2 dm = \int r^2 dm + h^2 \int dm + 2h \int x dm \).
\( I_o = I_c + Mh^2 + 0 \) (Since \( \int x dm = 0 \) about CM).
Q.28. Define: (i) Isothermal process (ii) Adiabatic process.
One mole of an ideal gas is enclosed in an ideal cylinder at 1.0 mPa and 27°C. The gas is allowed to expand till its volume is doubled. Calculate the work done if the expansion is isobaric.
(i) Isothermal process: A thermodynamic process in which the temperature of the system remains constant throughout.
(ii) Adiabatic process: A process in which there is no exchange of heat between the system and its surroundings.
Numerical:
Isobaric expansion (Constant P). Volume doubles (\( V_2 = 2V_1 \)).
Ideal Gas Law: \( PV = nRT \).
If P is constant and V doubles, T must double. \( T_1 = 300 \) K, \( T_2 = 600 \) K.
Work \( W = P(V_2 - V_1) = P V_1 \).
Using gas law: \( P V_1 = n R T_1 \).
\( W = 1 \times 8.31 \times 300 = 2493 \) J.
(Note: "1.0 mPa" (millipascal) seems to be a very low pressure, likely a typo for MPa, but the calculation using \( W = nRT \) is independent of the specific pressure value as long as \( n \) and \( T \) are known).
Q.29. Distinguish between streamline flow and turbulent flow. (Any Two points). Calculate the terminal velocity...
Distinction:
1. Streamline: Smooth, orderly flow. Turbulent: Irregular, chaotic flow.
2. Streamline: Velocity at a point is constant. Turbulent: Velocity changes randomly.
Numerical:
Bubble rises in liquid. Upward terminal velocity:
\( v_t = \frac{2 r^2 g (\rho_L - \rho_{air})}{9 \eta} \)
\( r = 0.2 \) mm = \( 2 \times 10^{-4} \) m.
\( \rho_L = 900, \rho_{air} = 1.29 \Rightarrow \Delta \rho \approx 898.7 \) kg/m³.
\( \eta = 0.1 \) Ns/m².
\( v_t = \frac{2 \times (2 \times 10^{-4})^2 \times 9.8 \times 898.7}{9 \times 0.1} \)
\( v_t = \frac{2 \times 4 \times 10^{-8} \times 9.8 \times 898.7}{0.9} \)
\( v_t \approx \frac{7.045 \times 10^{-4}}{0.9} \approx 7.83 \times 10^{-4} \) m/s \( = 0.783 \) mm/s.
Q.30. What are Fraunhofer diffraction and Fresnel diffraction? A plane wavefront of light... calculate the distance between the slits.
Fraunhofer Diffraction: Source and screen are effectively at infinite distance from the aperture (using lenses). Plane wavefronts.
Fresnel Diffraction: Source and screen are at finite distance. Spherical/Cylindrical wavefronts.
Numerical:
The question asks for "distance between the slits", implying Young's Double Slit Experiment (Interference), despite mentioning "incident onto a slit".
Let separation of 10 bright fringes = 2 cm = 0.02 m.
Width of 10 fringes \( 10\beta = 0.02 \Rightarrow \beta = 0.002 \) m.
Fringe width \( \beta = \frac{\lambda D}{d} \).
\( d = \frac{\lambda D}{\beta} \)
\( \lambda = 5500 \times 10^{-10} \) m, \( D = 2 \) m.
\( d = \frac{5.5 \times 10^{-7} \times 2}{2 \times 10^{-3}} = 5.5 \times 10^{-4} \) m \( = 0.55 \) mm.
Q.31. What is a transformer? State working principle... distinguish between Step up and Step down transformer.
Definition: A device used to change the voltage of alternating current.
Principle: Mutual Induction.
Distinction:
Step-up: Increases voltage (\(N_s > N_p\)), decreases current.
Step-down: Decreases voltage (\(N_s < N_p\)), increases current.