The CBSE conducted the Class 12 Physics Board Exam on February 21, 2025, from 10:30 AM to 1:30 PM. The Physics theory paper has 70 marks, while 30 marks are allocated for the practical assessment.
The question paper includes multiple-choice questions (1 mark each), short-answer questions (2-3 marks each), and long-answer questions (5 marks each).
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CBSE Class 12 Physics 55-7-3 Question Paper and Detailed Solutions PDF is available for download here.
CBSE Class 12 2025 Physics 55-7-3 Question Paper with Solution PDF
| CBSE Class 12 Physics Question Paper With Answer Key | Download | Check Solutions |
In the circuit shown, the charge on the left plate of the 20 µF capacitor is \( -50\,\mu C \). The charge on the right plate of the 12 µF capacitor is:
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Germanium crystal is doped at room temperature with a minute quantity of boron. The charge carriers in the doped semiconductor will be:
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A steady current \( I \) is passed through a conductor at room temperature for time \( t \). It is observed that its temperature rises by \( 0.5^\circC \). If \( 2I \) current is passed through the conductor (at room temperature) for the same duration, the rise in its temperature will be approximately:
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The dimensions of ‘self-inductance’ are:
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Isotones are the nuclides having:
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A metal rod of length 50 cm is held vertically and moved with a velocity of 10 m/s towards east. The horizontal component of the Earth's magnetic field at the place is 0.4 G. The emf induced across the ends of the rod is:
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Given:
- Length of rod, \( l = 50\, cm = 0.5\, m \)
- Velocity, \( v = 10\, m/s \)
- Magnetic field, \( B = 0.4\, G = 0.4 \times 10^{-4}\, T \)
The emf induced in the rod (using motional emf formula): \[ \varepsilon = B \cdot l \cdot v \cdot \sin\theta \]
Since the rod is vertical, the velocity is eastward and magnetic field is horizontal (north), they are perpendicular: \[ \theta = 90^\circ \Rightarrow \sin\theta = 1 \] \[ \varepsilon = 0.4 \times 10^{-4} \cdot 0.5 \cdot 10 = 2 \times 10^{-4}\, V = 0.2\, mV \]
\[ \boxed{0.2\, mV} \] Quick Tip: Use \( \varepsilon = B l v \sin\theta \) when a conductor moves in a magnetic field. Always convert Gauss to Tesla and cm to meters in SI units.
The frequency of a photon of energy \( 1.326 \, eV \) is:
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An alternating current is given by \( I = I_0 \cos(100\pi t) \). The least time the current takes to decrease from its maximum value to zero will be:
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A rectangular coil of area \( A \) is kept in a uniform magnetic field \( \vec{B} \) such that the plane of the coil makes an angle \( \alpha \) with \( \vec{B} \). The magnetic flux linked with the coil is:
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The minimum energy required to free the electron from the ground state of the hydrogen atom is:
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The energy required to remove (ionize) an electron from the ground state of a hydrogen atom is equal to the energy of the electron in the ground state.
From Bohr’s theory: \[ E_n = -\frac{13.6}{n^2}\, eV \]
For the ground state, \( n = 1 \): \[ E_1 = -13.6\, eV \]
To free the electron, we must supply at least \( 13.6\, eV \), which is the ionization energy of hydrogen.
\[ \boxed{Minimum energy required = 13.6\, eV} \] Quick Tip: The ionization energy of hydrogen is the energy required to move the electron from \( n = 1 \) to \( n = \infty \), which is always \( 13.6\, eV \).
A capacitor and an inductor are connected in series across an AC source of voltage of variable frequency. The frequency is increased continuously. The nature of the circuit before and after the resonance will be:
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A p-n junction diode is forward biased. As a result:
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Assertion (A): EM waves do not require a medium for their propagation.
Reason (R): EM waves are transverse waves.
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Assertion (A): A charged particle is moving with velocity \( \vec{v} \) in the x–y plane, making an angle \( \theta \ (0 < \theta < \frac{\pi}{2}) \) with the x-axis. If a uniform magnetic field \( \vec{B} \) is applied in the region, along y-axis, the particle will move in a helical path with its axis parallel to the x-axis.
Reason (R): The direction of the magnetic force acting on a charged particle moving in a magnetic field is along the velocity of the particle.
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Let us analyze the situation step-by-step:
Step 1: Nature of the velocity and magnetic field
- The particle moves in the x–y plane.
- The velocity vector makes an angle \( \theta \in (0, \frac{\pi}{2}) \) with the x-axis, so it has both x and y components.
- The magnetic field \( \vec{B} \) is applied along the y-axis.
Thus: \[ \vec{v} = v_x \hat{i} + v_y \hat{j}, \quad \vec{B} = B \hat{j} \]
Step 2: Determine magnetic force
The magnetic force is given by: \[ \vec{F} = q \vec{v} \times \vec{B} \]
\[ \vec{v} \times \vec{B} = (v_x \hat{i} + v_y \hat{j}) \times (B \hat{j}) = v_x B (\hat{i} \times \hat{j}) + v_y B (\hat{j} \times \hat{j}) \] \[ = v_x B \hat{k} + 0 = v_x B \hat{k} \]
So only the x-component of velocity contributes to the force, and the force acts in the z-direction (out of plane).
Step 3: Path of the particle
The velocity component along the magnetic field (v_y) experiences no force.
The component perpendicular to the field (v_x) causes the particle to undergo circular motion in the z–x plane, not in the y–z or x–y plane.
Thus, the actual path is a helical motion around the y-axis, because:
- Motion due to \( v_y \) is uniform linear motion along \( y \)
- Motion due to \( v_x \) is circular in the x–z plane.
Therefore, the axis of the helix is along the y-axis, not the x-axis.
Step 4: Analyze the statements
- Assertion (A) is false because the axis of the helical motion is along the y-axis, not x-axis.
- Reason (R) is also false because magnetic force is not along velocity, but perpendicular to it. Quick Tip: A particle moves in a helical path when its velocity has a component parallel to the magnetic field. The axis of the helix is along the direction of the magnetic field. Magnetic force is always perpendicular to velocity.
Assertion (A): The minimum negative potential applied to the anode in a photoelectric experiment at which photoelectric current becomes zero, is called cut-off voltage.
Reason (R): The threshold frequency for a metal is the minimum frequency of incident radiation below which emission of photoelectrons does not take place.
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Assertion (A): A ray of light is incident normally on the face of a prism. The emergent ray will graze along the opposite face of the prism when the critical angle at the glass-air interface is equal to the angle of the prism.
Reason (R): The refractive index of a prism depends on the angle of the prism.
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Calculate the value of the current passing through the battery in the given circuit diagram.
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In a Young’s double-slit experiment, the intensity at the central maximum in the interference pattern on the screen is \( I_0 \). Find the intensity at a point on the screen where the path difference between the interfering waves is \( \frac{\lambda}{6} \).
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Define the ‘distance of closest approach’. An \( \alpha \)-particle of kinetic energy \( K \) is bombarded on a thin gold foil. Derive an expression for the ‘distance of closest approach’.
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Using the mirror equation and the formula of magnification, deduce that “the virtual image produced by a convex mirror is always diminished in size and is located between the pole and the focus.”
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(b) A convex lens of focal length 10 cm, a concave lens of focal length 15 cm and a third lens of unknown focal length are placed coaxially in contact. If the focal length of the combination is +12 cm, find the nature and focal length of the third lens, if all lenses are thin. Will the answer change if the lenses were thick?
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Let: \[ f_1 = +10 \, cm \quad (convex lens) \] \[ f_2 = -15 \, cm \quad (concave lens) \] \[ f_3 = ? \quad (unknown lens) \] \[ f_{eq} = +12 \, cm \]
Using the lens combination formula for lenses in contact: \[ \frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} \]
Substitute known values: \[ \frac{1}{12} = \frac{1}{10} + \left(-\frac{1}{15}\right) + \frac{1}{f_3} \]
\[ \frac{1}{12} = \frac{1}{10} - \frac{1}{15} + \frac{1}{f_3} \]
\[ \frac{1}{12} = \left(\frac{3 - 2}{30}\right) + \frac{1}{f_3} = \frac{1}{30} + \frac{1}{f_3} \]
\[ \frac{1}{f_3} = \frac{1}{12} - \frac{1}{30} = \frac{5 - 2}{60} = \frac{3}{60} = \frac{1}{20} \]
\[ \Rightarrow f_3 = 20 \, cm \]
Since \( f_3 > 0 \), the third lens is convex.
Would the answer change if the lenses were thick?
Yes. If the lenses were thick, their optical centers would not coincide, and separation between lenses would matter, which affects the equivalent focal length due to spacing. So the result would be different. Quick Tip: For thin lenses in contact, the reciprocal of the net focal length is the algebraic sum of the reciprocals of individual focal lengths: \( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} \).
Draw energy band diagrams of n-type and p-type semiconductors at temperature \( T > 0 \, K \). Show the donor/acceptor energy levels with the order of difference of their energies from the bands.
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Explain the process of formation of ‘depletion layer’ and ‘potential barrier’ in a p-n junction region of a diode, with the help of a suitable diagram. Which feature of junction diode makes it suitable for its use as a rectifier?
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An electric field \( \vec{E} \) is given by:
\[ \vec{E} = \begin{cases} +100\, \hat{i} \dfrac{N}{C} & for x > 0
-100\, \hat{i} \dfrac{N}{C} & for x < 0 \end{cases} \]
A right circular cylinder of length \( 10\, cm \) and radius \( 2\, cm \), is placed such that its axis coincides with the x-axis and its two faces are at \( x = -5\, cm \) and \( x = 5\, cm \). Calculate:
(a) the net outward flux through the cylinder, and
(b) the net charge inside the cylinder.
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Answer the following giving reasons:
(a) The maximum kinetic energy of the photoelectrons is independent of the intensity of incident radiation.
(b) Photoelectric current increases with the increase in the intensity of the incident radiation.
(c) The stopping potential \( V_0 \) varies linearly with the frequency \( \nu \) of the incident radiation for a given photosensitive surface.
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(a)
(i) Write Biot–Savart’s law in vector form.
(ii) Two identical circular coils A and B, each of radius \( R \), carrying currents \( I \) and \( \sqrt{3}I \) respectively, are placed concentrically in XY and YZ planes respectively.
Find the magnitude and direction of the net magnetic field at their common centre.
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(i) A rectangular loop of sides \( l \) and \( b \) carries a current \( I \) clockwise. Write the magnetic moment \( \vec{m} \) of the loop and show its direction in a diagram.
(ii) The loop is placed in a uniform magnetic field \( \vec{B} \) and is free to rotate about an axis which is perpendicular to \( \vec{B} \). Prove that the loop experiences no net force, but a torque \( \vec{\tau} = \vec{m} \times \vec{B} \).
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Name the electromagnetic radiations and write their frequency range:
(a) Used to take the photograph of the bones
(b) Produced by hot bodies
(c) Used in television communication systems
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An ac source of voltage \( V = V_0 \sin \omega t \) is connected to a circuit element \( X \). It is observed that the current flowing through \( X \) varies as \( I = I_0 \sin\left(\omega t - \frac{\pi}{2}\right) \).
(a) Identify the element \( X \) and write the expression for its reactance.
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Plot a graph to show the variation of reactance of the element with the frequency of the applied voltage.
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Write the mathematical forms of three postulates of Bohr’s theory of the hydrogen atom. Using them, prove that for an electron revolving in the \( n^{th} \) orbit:
(a) the radius of the orbit is proportional to \( n^2 \), and
(b) the total energy of the atom is proportional to \( \left( \frac{1}{n^2} \right) \).
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Bohr's Postulates:
Electrons revolve in discrete circular orbits around the nucleus without radiating energy.
Angular momentum of the electron is quantized:
\[ mvr = \frac{nh}{2\pi} \]
Radiation is emitted or absorbed when an electron jumps between orbits:
\[ E = h\nu = E_i - E_f \]
(a) Radius of the orbit \( r \propto n^2 \)
Centripetal force is provided by Coulomb force: \[ \frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{r^2} \Rightarrow mv^2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{r} \quad \cdots(1) \]
From Bohr’s quantization: \[ mvr = \frac{nh}{2\pi} \Rightarrow v = \frac{nh}{2\pi mr} \quad \cdots(2) \]
Substitute (2) in (1): \[ m \left( \frac{nh}{2\pi mr} \right)^2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{r} \Rightarrow \frac{n^2 h^2}{4\pi^2 m r^2} = \frac{e^2}{4\pi\varepsilon_0 r} \]
Solve for \( r \): \[ r \propto \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} \Rightarrow r_n \propto n^2 \]
(b) Total energy \( E \propto -\frac{1}{n^2} \)
Kinetic energy: \[ K = \frac{1}{2} mv^2 = \frac{e^2}{8\pi\varepsilon_0 r} \quad (from centripetal force) \]
Potential energy: \[ U = -\frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{r} \]
Total energy: \[ E = K + U = \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r} = -\frac{e^2}{8\pi\varepsilon_0 r} \]
But \( r \propto n^2 \Rightarrow E \propto -\frac{1}{n^2} \) Quick Tip: To prove \( r \propto n^2 \) and \( E \propto -1/n^2 \), combine Coulomb's force law with Bohr's angular momentum quantization: Use \( mvr = \frac{nh}{2\pi} \) and \( \frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \cdot \frac{e^2}{r^2} \)
Consider the contribution of the following two factors I and II in resistivity of a metal:
I. Relaxation time of electrons
II. Number of electrons per unit volume
The resistivity of a metal increases with increase in its temperature because:
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A steady current flows in a copper wire of non-uniform cross-section. Consider the following three physical quantities:
I. Electric field
II. Current density
III. Drift speed
Then at the different points along the wire:
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The temperature coefficient of resistance of nichrome is \(1.70 \times 10^{-4} \, ^\circC^{-1}\). In order to increase the resistance of a nichrome wire by 8.5%, the temperature of the wire should be increased by:
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A wire of length \( 0.5 \, m \) and cross-sectional area \( 1.0 \times 10^{-7} \, m^2 \) is connected to a battery of \( 2 \, V \) that maintains a current of \( 1.5 \, A \) in it. The conductivity of the material of the wire (in \( \Omega^{-1} \, m^{-1} \)) is:
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Consider two cylindrical conductors A and B, made of the same metal connected in series to a battery. The length and the radius of B are twice that of A. If \( \mu_A \) and \( \mu_B \) are the mobility of electrons in A and B respectively, then \( \frac{\mu_A}{\mu_B} \) is:
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An object is placed in front of a convex spherical glass surface (\( n = 1.5 \) and radius of curvature \( R \)) at a distance of \( 4R \) from it. As the object is moved slowly close to the surface, the image formed is:
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A double-convex lens, made of glass of refractive index 1.5, has focal length 10 cm. The radius of curvature of its each face, is:
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A small bulb is placed at the bottom of a tank containing a transparent liquid (refractive index \( n \)) to a depth \( H \). The radius of the circular area of the surface of liquid, through which the light from the bulb can emerge out, is \( R \). Then \( \left( \frac{R}{H} \right) \) is:
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A parallel beam of light is incident on a face of a prism with refracting angle 60°. The angle of minimum deviation is found to be 30°. The refractive index of the material of the prism is close to:
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The angle of minimum deviation for a ray of light incident on a thin prism, made of crown glass (\( n = 1.52 \)) is \( \delta_m \). If the prism was made of dense flint glass (\( n = 1.62 \)) instead of crown glass, the angle of minimum deviation will:
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With the help of a labelled diagram, explain the principle of working of a moving coil galvanometer. Write the purpose of using (i) radial magnetic field, and (ii) soft iron core, in it.
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Define current sensitivity of a galvanometer. “Increasing the current sensitivity may not necessarily increase the voltage sensitivity.” Give reason.
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Write Ampere’s circuital law in mathematical form and explain the terms used.
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As the current carrying solenoid is made longer, the magnetic field produced outside it approaches zero. Why?
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A flexible loop of irregular shape carrying current when located in an external magnetic field, changes to a circular shape. Give reason.
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A galvanometer of resistance \( G \) is converted into a voltmeter to measure up to \( V \) volts, by connecting a resistance \( R_1 \) in series with the coil. If \( R_1 \) is replaced by \( R_2 \), then it can only measure up to \( \frac{V}{2} \) volt. Find the value of the resistance \( R_3 \) (in terms of \( R_1 \) and \( R_2 \)) needed to convert it into a voltmeter that can read up to 2V.
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Let the current for full-scale deflection of the galvanometer be \( I \).
Using Ohm’s law for full-scale deflection:
- For voltage \( V \) with resistance \( R_1 \), we have: \[ V = I(G + R_1) \quad \Rightarrow \quad I = \frac{V}{G + R_1} \]
- For voltage \( \frac{V}{2} \) with resistance \( R_2 \), we have: \[ \frac{V}{2} = I(G + R_2) \quad \Rightarrow \quad I = \frac{V}{2(G + R_2)} \]
Equating the two expressions for current \( I \): \[ \frac{V}{G + R_1} = \frac{V}{2(G + R_2)} \]
Cancel \( V \) and solve: \[ \frac{1}{G + R_1} = \frac{1}{2(G + R_2)} \Rightarrow 2(G + R_2) = G + R_1 \]
\[ 2G + 2R_2 = G + R_1 \Rightarrow R_1 = G + 2R_2 - G = 2R_2 \Rightarrow R_2 = \frac{R_1}{2} \]
Now, to measure up to \( 2V \), the total resistance required is: \[ 2V = I(G + R_3) \Rightarrow I = \frac{2V}{G + R_3} \]
Set equal to previous \( I \) from \( V = I(G + R_1) \Rightarrow I = \frac{V}{G + R_1} \): \[ \frac{2V}{G + R_3} = \frac{V}{G + R_1} \Rightarrow \frac{2}{G + R_3} = \frac{1}{G + R_1} \]
Solving: \[ 2(G + R_1) = G + R_3 \Rightarrow 2G + 2R_1 = G + R_3 \Rightarrow R_3 = G + 2R_1 \]
Thus, substitute \( G = R_1 - 2R_2 \) from earlier: \[ R_3 = (R_1 - 2R_2) + 2R_1 = 3R_1 - 2R_2 \]
% Final Answer
Required Resistance: \( \boxed{R_3 = 3R_1 - 2R_2} \) Quick Tip: When a galvanometer is converted into a voltmeter, use Ohm’s law in different configurations and equate currents for consistent deflection. Eliminate \( G \) wherever possible using given voltage levels.
Explain with the help of a labelled ray diagram the formation of final image by an astronomical telescope at infinity. Write the expression for its magnifying power.
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The total magnification produced by a compound microscope is 20. The magnification produced by the eyepiece is 5. When the microscope is focussed on a certain object, the distance between the objective and eyepiece is observed to be 14 cm. Calculate the focal lengths of the objective and the eyepiece. (Given that the least distance of distinct vision = 25 cm)
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Given:
Total magnification \( M = 20 \)
Magnification of eyepiece \( M_e = 5 \)
Length of microscope tube \( L = 14 \, cm \)
Least distance of distinct vision \( D = 25 \, cm \)
Step 1: Relation between magnifications: \[ M = M_o \cdot M_e \] \[ \Rightarrow M_o = \frac{M}{M_e} = \frac{20}{5} = 4 \]
Step 2: For objective lens: \[ M_o = \frac{v_o}{u_o} \approx \frac{L - f_e}{f_o} \Rightarrow 4 = \frac{14 - f_e}{f_o} \Rightarrow f_o = \frac{14 - f_e}{4} \quad \cdots (1) \]
Step 3: For eyepiece (final image at least distance of distinct vision): \[ M_e = 1 + \frac{D}{f_e} = 5 \Rightarrow \frac{D}{f_e} = 4 \Rightarrow f_e = \frac{D}{4} = \frac{25}{4} = 6.25 \, cm \]
Step 4: Substitute in equation (1): \[ f_o = \frac{14 - 6.25}{4} = \frac{7.75}{4} = 1.9375 \, cm \]
Final Answers:
Focal length of eyepiece: \( f_e = 6.25 \, cm \)
Focal length of objective: \( f_o = 1.94 \, cm (approx) \) Quick Tip: In compound microscopes, total magnification is the product of the magnifications due to the objective and eyepiece. For comfortable viewing, the image is often formed at the least distance of distinct vision.
Two coherent light waves, each of intensity \( I_0 \) superpose each other and produce interference pattern on a screen. Obtain the expression for the resultant intensity at a point where the phase difference between the waves is \( \phi \). Write its maximum and minimum possible values.
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Let the electric field of two coherent waves be: \[ E_1 = E_0 \sin(\omega t), \quad E_2 = E_0 \sin(\omega t + \phi) \]
The resultant electric field due to superposition: \[ E = E_1 + E_2 = E_0 \sin(\omega t) + E_0 \sin(\omega t + \phi) \]
Using the identity: \[ \sin A + \sin B = 2 \sin\left(\frac{A + B}{2} \right) \cos\left(\frac{A - B}{2} \right) \]
We get: \[ E = 2E_0 \cos\left(\frac{\phi}{2}\right) \sin\left(\omega t + \frac{\phi}{2} \right) \]
Since intensity is proportional to the square of the amplitude: \[ I = |E|^2 = \left(2E_0 \cos\left(\frac{\phi}{2} \right) \right)^2 = 4E_0^2 \cos^2\left(\frac{\phi}{2} \right) \]
Given that the individual wave intensity \( I_0 = E_0^2 \), \[ I = 4I_0 \cos^2\left(\frac{\phi}{2} \right) \]
Maximum intensity: When \( \phi = 0 \), \[ I_{max} = 4I_0 \cos^2(0) = 4I_0 \]
Minimum intensity: When \( \phi = \pi \), \[ I_{min} = 4I_0 \cos^2\left(\frac{\pi}{2} \right) = 0 \] Quick Tip: Resultant intensity in interference is based on the square of net amplitude. Use \( I = 4I_0 \cos^2(\phi/2) \) for two waves of equal intensity. Maximum occurs when waves are in phase and minimum when out of phase.
In a single slit diffraction experiment, the aperture of the slit is 3 mm and the separation between the slit and the screen is 1.5 m. A monochromatic light of wavelength 600 nm is normally incident on the slit. Calculate the distance of (I) first order minimum, and (II) second order maximum, from the centre of the screen.
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A parallel plate capacitor with plate area \( A \) and plate separation \( d \) has a capacitance \( C_0 \). A slab of dielectric constant \( K \) having area \( A \) and thickness \( \frac{d}{4} \) is inserted in the capacitor, parallel to the plates. Find the new value of its capacitance.
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You are provided with a large number of \( 1\, \muF \) identical capacitors and a power supply of \( 1200\,V \). The dielectric medium used in each capacitor can withstand up to \( 200\,V \) only. Find the minimum number of capacitors and their arrangement, required to build a capacitor system of equivalent capacitance of \( 2\,\muF \) for use with this supply.
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An electric dipole of dipole moment \( \vec{p} \) consists of point charges \( +q \) and \( -q \), separated by \( 2a \). Derive an expression for the electric potential in terms of its dipole moment at a point at a distance \( x \gg a \) from its centre and lying:
(I) along its axis, and
(II) along its bisector (equatorial) line.
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An electric dipole of dipole moment \( \vec{p} = (0.8\,\hat{i} + 0.6\,\hat{j}) \times 10^{-29} \,Cm \) is placed in an electric field \( \vec{E} = 1.0 \times 10^7\,\hat{k} \,V/m \). Calculate the magnitude of the torque acting on it and the angle it makes with the x-axis at this instant.
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