CBSE Class 12 Physics Question Paper 2024 (Set 3- 55/4/3) Available- Download Solution PDF with Answer Key

Work Done to Rotate the Dipole in an Electric Field


Step 1: Work Done Formula

The work done \( W \) to rotate an electric dipole from an initial angle \( \theta_1 \) to a final angle \( \theta_2 \) in a uniform electric field is given by:
\[ W = -\int_{\theta_1}^{\theta_2} \vec{\tau} \cdot d\vec{\theta} \]

where:

- \( \vec{\tau} = pE \sin\theta \) is the torque on the dipole,

- \( p \) is the dipole moment,

- \( E \) is the electric field.

Step 2: Work Done to Rotate from Stable to Unstable Equilibrium

For stable equilibrium: \( \theta_1 = 0^\circ \) (where \( \cos \theta = 1 \)),

For unstable equilibrium: \( \theta_2 = 180^\circ \) (where \( \cos \theta = -1 \)).

Thus, the work done to rotate the dipole is:
\[ W = -\int_{0}^{180^\circ} pE \cos\theta \, d\theta = 2pE \]

Thus, the correct answer is:
\[ \boxed{(A) \, 2pE} \] Quick Tip: The work done on a dipole in a uniform electric field depends on the change in potential energy, \( \Delta U = U_{final} - U_{initial} \).

N/A Quick Tip: The electric field due to an infinite line charge decreases inversely with the perpendicular distance from the line.

N/A Quick Tip: In an electromagnetic wave, \( \vec{E} \) and \( \vec{B} \) are always in phase and perpendicular to each other and the direction of wave propagation.

N/A Quick Tip: In the Balmer series, spectral lines converge at shorter wavelengths as \( n \to \infty \), corresponding to the series limit.

Magnetic Force Between Conductors


Step 1: Force Between Two Parallel Conductors

The magnetic force per unit length \( F \) between two parallel conductors with currents \( I_1 \) and \( I_2 \) is given by the formula:
\[ F = \frac{\mu_0 I_1 I_2}{2 \pi r} \]

where:

- \( \mu_0 \) is the permeability of free space,

- \( I_1 \) and \( I_2 \) are the currents in the conductors,

- \( r \) is the distance between the conductors.

Step 2: Force on Conductor C

- The force on conductor C is due to the magnetic fields created by A and B.

- The force between A and C is:
\[ F_{AC} = \frac{\mu_0 I I_1}{2 \pi \frac{r}{3}} = \frac{3 \mu_0 I I_1}{2 \pi r} \]

This force is attractive, as both currents in A and C are in the same direction. Hence, the force is towards A.

- The force between B and C is:
\[ F_{BC} = \frac{\mu_0 I I_1}{2 \pi r} \quad (since distance between B and C is r) \]

This force is repulsive, as the currents in B and C are in opposite directions.

Step 3: Net Force on C

The net force on conductor C is the difference between the attractive force from A and the repulsive force from B:
\[ F_{net} = F_{AC} - F_{BC} = \frac{3 \mu_0 I I_1}{2 \pi r} - \frac{\mu_0 I I_1}{2 \pi r} = \frac{3 \mu_0 I I_1}{4 \pi r^2} \]

Thus, the net force on C is directed towards A. Therefore, the correct answer is:
\[ \boxed{(C) \, \frac{3 \mu_0 I I_1}{4 \pi r^2}, towards A} \] Quick Tip: To calculate the net force between multiple conductors, consider the pairwise forces due to the magnetic interaction and sum them vectorially. Use the formula \( F = \frac{\mu_0 I_1 I_2}{2\pi d} \), where \( d \) is the separation distance.

Maximum Induced EMF in the Coil


Step 1: Faraday’s Law of Electromagnetic Induction

According to Faraday’s Law, the induced EMF is given by:
\[ \mathcal{E} = -N \frac{d\Phi}{dt} \]

where \( \Phi \) is the magnetic flux:
\[ \Phi = B A = B_0 A \cos\left(\frac{2\pi t}{T}\right) \]

where:

- \( A \) is the area of the coil,

- \( B_0 \) is the maximum magnetic field.

Step 2: Calculating the Induced EMF

Differentiating \( \Phi \) with respect to \( t \):
\[ \mathcal{E} = -N A B_0 \frac{d}{dt} \left[\cos\left(\frac{2\pi t}{T}\right)\right] \]
\[ \mathcal{E} = N A B_0 \left(\frac{2\pi}{T}\right) \sin\left(\frac{2\pi t}{T}\right) \]
\[ \mathcal{E} = \mathcal{E}_0 \sin\left(\frac{2\pi t}{T}\right) \]

where \( \mathcal{E}_0 = \frac{2\pi N A B_0}{T} \) is the peak EMF.

Step 3: Finding Maximum EMF

- The EMF is maximum when \( \sin\left(\frac{2\pi t}{T}\right) = 1 \), which occurs at:
\[ \frac{2\pi t}{T} = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots \]

- Solving for \( t \):
\[ t = \frac{T}{4}, \frac{3T}{4}, \frac{5T}{4}, \dots = \frac{nT}{4}, \quad n = 1, 3, 5, \dots \]

Thus, the correct answer is (B) \( t = \frac{nT}{4} \).
Quick Tip: The EMF in a coil is maximum when the rate of change of flux is maximum, which occurs when \( B \) passes through zero.

Magnetic Flux Linked with Loop B


Step 1: Magnetic Field at the Center of a Current-Carrying Loop

The magnetic field at the center of a circular loop carrying current \( I \) is:
\[ B = \frac{\mu_0 I}{2R} \]

where:

- \( \mu_0 \) = Permeability of free space,

- \( R \) = Radius of loop A.

Step 2: Magnetic Flux Through Loop B

The magnetic flux through loop B is given by:
\[ \Phi = B \cdot A \]

where:
- \( A \) is the area of loop B:
\[ A = \pi r^2 = \pi \left( \frac{R}{20} \right)^2 = \frac{\pi R^2}{400} \]

- Substituting \( B \) from Step 1:
\[ \Phi = \left( \frac{\mu_0 I}{2R} \right) \times \frac{\pi R^2}{400} \]
\[ \Phi = \frac{\mu_0 I \pi R^2}{800R} \]
\[ \Phi \propto R \]

Thus, the magnetic flux linked with loop B is proportional to \( R \), which matches option (A).
Quick Tip: For a concentric loop system, the magnetic flux linkage depends on both the field of the larger loop and the area of the smaller loop.

Finding the Galvanometer's Full-Scale Deflection Current


Step 1: Given Data
- Galvanometer resistance: \( R_g = 100 \Omega \)
- Shunt resistance: \( R_s = 0.1 \Omega \)
- Ammeter range: \( I = 1A \)

Step 2: Current Through the Galvanometer
Using the shunt formula:
\[ I_g = I \times \frac{R_s}{R_g + R_s} \]
\[ I_g = 1 \times \frac{0.1}{100 + 0.1} = 1 \times \frac{0.1}{100.1} \]
\[ I_g \approx 1 \times 10^{-3} A = 1 mA \]

Thus, the correct answer is (B) 1 mA.
Quick Tip: The shunt resistance allows most of the current to bypass the galvanometer, reducing the current through it to prevent damage.

N/A Quick Tip: To calculate the r.m.s. value for a combined sinusoidal current, square each term, take the mean, and then the square root.

Explanation of Photoelectric Effect


- According to Einstein’s Photoelectric Equation:
\[ K_{\max} = h f - \phi \]

where:

- \( h \) = Planck’s constant,

- \( f \) = Frequency of incident radiation,

- \( \phi \) = Work function of the metal.

- If \( f < f_0 \) (threshold frequency), no electrons are emitted.

- This confirms the quantum nature of light, supporting option (A).
Quick Tip: In the photoelectric effect, electrons are emitted only if the photon energy exceeds the work function of the metal.

Magnetic Susceptibility of Diamagnetic Materials


- Magnetic susceptibility (\( \chi_m \)) measures how a material responds to an external magnetic field.

- For diamagnetic materials, \( \chi_m \) is:

- Negative (they repel magnetic fields).

- Small (weak effect compared to paramagnetic or ferromagnetic materials).
\[ \chi_m \approx -10^{-5} to -10^{-6} \]

Thus, the correct answer is (A) Small and negative.
Quick Tip: Diamagnetic materials like copper, bismuth, and gold have a small negative magnetic susceptibility and oppose external magnetic fields.

Bohr’s Model and Radius of Electron Orbits


Step 1: Formula for Bohr’s Radius

According to Bohr’s Model, the radius of an electron in the \( n^{th} \) orbit of a hydrogen atom is:
\[ r_n = \frac{4\pi \epsilon_0 \hbar^2}{m e^2} n^2 \]

where:

- \( \epsilon_0 \) = Permittivity of free space,

- \( \hbar \) = Reduced Planck’s constant,

- \( m \) = Mass of the electron,

- \( e \) = Charge of the electron.

Step 2: Understanding the Proportionality

- Since all constants remain unchanged, we get:
\[ r_n \propto n^2 \]

Step 3: Conclusion

- The radius of an orbit increases quadratically with \( n \).

- Thus, the correct answer is (C) \( r_n \propto n^2 \).
Quick Tip: In Bohr’s Model, orbit radius varies as \( n^2 \), while energy varies as \( \frac{1}{n^2} \).

Understanding Electron Drift in Conductors


Step 1: Understanding Drift Velocity

- In a conductor, free electrons move randomly in all directions due to thermal motion.

- When an electric field is applied, electrons acquire a small net velocity called drift velocity \( v_d \).

- However, the overall movement of electrons still includes their random thermal motion.

Step 2: Evaluating Assertion (A)

- Assertion (A) is correct because electrons continue their random motion, but on average, they experience a net drift.

Step 3: Evaluating Reason (R)

- Reason (R) is also correct because drift velocity is a small velocity superposed on the random thermal velocities of electrons.

Step 4: Conclusion

- Since both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A), the correct answer is (A).
Quick Tip: In a conductor, electrons move randomly due to thermal energy, but when an electric field is applied, they acquire a small drift velocity superposed on their random motion.

Understanding Interference and Energy Conservation


Step 1: Explanation of Interference and Diffraction

- In interference and diffraction, light waves superpose, leading to constructive and destructive interference.

- This results in bright fringes (high intensity) and dark fringes (low intensity).

- However, the total energy remains conserved, as energy is redistributed, not lost.

Step 2: Evaluating Assertion and Reason

- The Assertion (A) is correct because light energy indeed redistributes, creating bright and dark fringes.

- The Reason (R) is false because energy is conserved, not lost.

Step 3: Conclusion

- Since Assertion (A) is true but Reason (R) is false, the correct answer is (C).
Quick Tip: In interference and diffraction, energy is not lost but redistributed. The total energy remains constant.

Temperature Coefficient of Resistance and Charge Carriers


Step 1: Understanding the Temperature Coefficient of Resistance

- The temperature coefficient of resistance (\( \alpha \)) determines how resistance changes with temperature:

- For metals, \( \alpha \) is positive because resistance increases with temperature due to increased collisions between electrons.

- For p-type semiconductors, \( \alpha \) is negative because resistance decreases with temperature as more charge carriers become available.

Step 2: Evaluating Assertion (A)

- Assertion (A) is true since metals have a positive coefficient and p-type semiconductors have a negative coefficient.

Step 3: Evaluating Reason (R)

- The Reason (R) is true, as:

- In metals, charge carriers are negatively charged electrons.

- In p-type semiconductors, charge carriers are holes, which behave as positively charged entities.

Step 4: Evaluating the Explanation

- While Reason (R) is correct, it does not explain why the temperature coefficient is positive for metals and negative for p-type semiconductors.

- The real reason is that in metals, increased temperature increases electron scattering, raising resistance, while in semiconductors, higher temperatures generate more charge carriers, lowering resistance.

Step 5: Conclusion

- Since both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A), the correct answer is (B).
Quick Tip: For metals, resistance increases with temperature due to higher electron scattering. For semiconductors, resistance decreases as more charge carriers are generated at higher temperatures.

Understanding the Photoelectric Effect in Zinc


Step 1: Understanding the Photoelectric Effect

- The photoelectric effect occurs when light of sufficient energy (\( E = h f \)) is incident on a metal, causing electron emission.

- The work function (\( \phi \)) is the minimum energy required to eject an electron.

- For zinc, the work function is 4.3 eV.

Step 2: Energy of Yellow Light

- The wavelength of yellow light is approximately 570 nm.

- Using the energy equation:
\[ E = \frac{hc}{\lambda} \]
\[ E = \frac{(6.63 \times 10^{-34}) (3 \times 10^8)}{570 \times 10^{-9}} \]
\[ E \approx 2.18 eV \]

Step 3: Evaluating Assertion (A)

- Since yellow light has only 2.18 eV, it cannot eject electrons from zinc (work function = 4.3 eV).

- Thus, Assertion (A) is false.

Step 4: Evaluating Reason (R)

- The Reason (R) is also false because the energy of yellow light is less than the work function of zinc, not more.

Step 5: Conclusion

- Since both Assertion (A) and Reason (R) are false, the correct answer is (D).
Quick Tip: For photoelectric emission, the photon energy must be greater than the work function of the metal. Zinc requires ultraviolet light for electron emission.

We can use the formula for refraction at a spherical surface:
\(\)\frac{n_2{v - \frac{n_1{u = \frac{n_2 - n_1{R\(\)

Here, \(n_1\) = refractive index of the medium where the object is located = 1.5 (glass) \(n_2\) = refractive index of the medium where the observer is located = 1 (air) \(u\) = object distance from the refracting surface (C) = -20 cm (negative since the object is on the same side as the incident light) \(v\) = image distance from the refracting surface (C) = ? \(R\) = radius of curvature of the spherical surface = -40 cm (negative because the surface is convex when viewed from the glass side).

Plugging in the values:
\(\)\frac{1{v - \frac{1.5{-20 = \frac{1 - 1.5{-40\(\) \(\)\frac{1{v + \frac{1.5{20 = \frac{-0.5{-40\(\) \(\)\frac{1{v + \frac{3{40 = \frac{1{80\(\) \(\)\frac{1{v = \frac{1{80 - \frac{3{40\(\) \(\)\frac{1{v = \frac{1{80 - \frac{6{80\(\) \(\)\frac{1{v = \frac{-5{80\(\) \(\)\frac{1{v = \frac{-1{16\(\) \(\)v = -16 cm\(\)

The image is formed 16 cm from the center of the sphere, on the same side as the object. Since \(v\) is negative, the image is virtual.

Therefore, the image of the bubble is virtual and located 16 cm from the refracting surface (C), on the same side as the bubble.
The distance PB = PC + CB = 40+20 = 60 cm. So the distance of the image from P is PC + v = 40 + 16= 56 cm. Since v is negative, the image is virtual.

Final Answer: The final answer is \(\boxed{-16 cm\) Quick Tip: For a bubble inside a denser medium, refraction at a curved surface follows the lens-maker's formula to find the image position.

Calculation of Focal Lengths in a Refracting Telescope


Step 1: Understanding the Formula

In a refracting telescope, the magnifying power (\( M \)) in normal adjustment is given by:
\[ M = \frac{f_o}{f_e} \]

where:

- \( f_o \) = Focal length of the objective lens,

- \( f_e \) = Focal length of the eyepiece lens.

The total length of the telescope in normal adjustment is:
\[ L = f_o + f_e \]

Given:
- \( L = 1.00 \) m = 100 cm,
- \( M = 19 \).

Step 2: Expressing \( f_o \) and \( f_e \)

Rearranging the magnification equation:
\[ f_o = M f_e \]

Substituting in the length equation:
\[ M f_e + f_e = L \]
\[ 19 f_e + f_e = 100 \]
\[ 20 f_e = 100 \]
\[ f_e = \frac{100}{20} = 5 cm \]
\[ f_o = M f_e = 19 \times 5 = 95 cm \]

Step 3: Conclusion

- Focal length of the objective lens: 95 cm.

- Focal length of the eyepiece lens: 5 cm.
Quick Tip: For a refracting telescope, the objective lens has a large focal length, while the eyepiece lens has a small focal length to achieve high magnification.

The de Broglie wavelength \( \lambda \) for a particle is given by: \[ \lambda = \frac{h}{\sqrt{2m q V}}, \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, \( q \) is the charge, and \( V \) is the accelerating potential.

For the proton (\( m_p \)) and deuteron (\( m_d = 2m_p \)): \[ \frac{\lambda_p}{\lambda_d} = \sqrt{\frac{m_d V_d}{m_p V_p}}. \]
Substitute \( m_d = 2m_p \): \[ \frac{\lambda_p}{\lambda_d} = \sqrt{\frac{2 V_d}{V_p}}. \]

Given \( \frac{\lambda_p}{\lambda_d} = \frac{1}{2} \), substitute into the equation: \[ \frac{1}{2} = \sqrt{\frac{2 V_d}{V_p}}. \]
Square both sides: \[ \frac{1}{4} = \frac{2 V_d}{V_p}. \]
Rearrange to find \( \frac{V_p}{V_d} \): \[ \frac{V_p}{V_d} = 8. \] Quick Tip: The de Broglie wavelength is inversely proportional to the square root of the product of the mass and the accelerating potential of the particle.

Finding the Refractive Index of the Medium




Let the refractive index of the glass prism be \(\mu_g = \mu\).

Let the refractive index of the surrounding medium be \(\mu_m\).

Since the prism is equilateral, each angle is \(60^\circ\).
The ray is incident normally on one face, so the angle of incidence is \(0^\circ\). Therefore, the angle of refraction is also \(0^\circ\).

The ray travels straight to the next face. The angle of incidence at this second face, say \(i\), is equal to the angle of the prism, which is \(60^\circ\).

Since the emergent ray just grazes the adjacent face, the angle of refraction at the second face is \(90^\circ\). This means that the angle of incidence \(i\) at the second face is equal to the critical angle \(C\).
So, \(i = C = 60^\circ\).

Using Snell's law at the second face:
\(\)\mu_g \sin i = \mu_m \sin r\(\) \(\)\mu \sin 60^\circ = \mu_m \sin 90^\circ\(\) \(\)\mu \sin 60^\circ = \mu_m (1)\(\) \(\)\mu_m = \mu \sin 60^\circ\(\) \(\)\mu_m = \mu \frac{\sqrt{3{2\(\)

Therefore, the refractive index of the medium is \(\frac{\sqrt{3}}{2} \mu\).

Final Answer: The final answer is \(\boxed{\frac{\sqrt{3}}{2}\mu}\) Quick Tip: When light just grazes the face of a prism, the critical angle condition applies. Use Snell’s law to find the refractive index of the medium.

The resistance of a conductor at a temperature \( T \) is given by: \[ R_T = R_0 \left( 1 + \alpha (T - T_0) \right), \]
where:

\( R_T = 1.25R_0 \) (final resistance is 25% greater than the initial resistance),
\( R_0 \) is the resistance at \( T_0 = 27^\circ C \),
\( \alpha = 2.0 \times 10^{-4} \, C^{-1} \) (temperature coefficient of resistance).


Substitute \( R_T = 1.25R_0 \) into the equation: \[ 1.25R_0 = R_0 \left( 1 + \alpha (T - T_0) \right). \]

Simplify: \[ 1.25 = 1 + \alpha (T - 27). \]

Rearrange: \[ \alpha (T - 27) = 0.25. \]

Substitute \( \alpha = 2.0 \times 10^{-4} \): \[ (2.0 \times 10^{-4}) (T - 27) = 0.25. \]

Solve for \( T \): \[ T - 27 = \frac{0.25}{2.0 \times 10^{-4}} = 1250. \]
\[ T = 27 + 1250 = 1277^\circ C. \]

Answer: The temperature is \( 1277^\circ C \). Quick Tip: Use the formula \( R_T = R_0 \left( 1 + \alpha (T - T_0) \right) \) to calculate changes in resistance with temperature.

Understanding p-n Junction Diode Biasing


- A p-n junction diode allows current flow in one direction (forward bias) and blocks current in the opposite direction (reverse bias).


(i) Forward Biasing:


- The p-side (anode) is connected to the positive terminal of the battery, and the n-side (cathode) is connected to the negative terminal.

- This reduces the depletion region width, allowing current to flow.


Circuit Diagram for Forward Biasing:


I-V Characteristics in Forward Bias:


- The current increases exponentially after the threshold voltage (0.7V for silicon, 0.3V for germanium).



(ii) Reverse Biasing:


- The p-side is connected to the negative terminal, and the n-side is connected to the positive terminal.

- This increases the depletion region width, preventing current flow (except for a small leakage current).


Circuit Diagram for Reverse Biasing:



I-V Characteristics in Reverse Bias:


- A small leakage current flows until the breakdown voltage is reached.




Quick Tip: In forward bias, current flows easily after the threshold voltage. In reverse bias, only a small leakage current exists until breakdown.

Definition of Mutual Inductance


- Mutual inductance (\( M \)) is the property of two coils in which a change in current in one coil induces an electromotive force (emf) in the other due to electromagnetic induction.

- The induced emf in the secondary coil is given by:
\[ \mathcal{E}_2 = - M \frac{dI_1}{dt} \]

where \( I_1 \) is the current in the primary coil.

- The SI unit of mutual inductance is Henry (H), where 1 Henry = 1 Weber per Ampere (1 H = 1 Wb/A).
Quick Tip: Mutual inductance quantifies the ability of one coil to induce emf in another due to changing magnetic flux.

Derivation of Mutual Inductance


- Consider two long coaxial solenoids, one inside the other.

- The primary solenoid (inner solenoid) has:

- \( N_1 \) turns

- Radius \( r_1 \)

- Length \( l \)

- Current \( I_1 \) flowing through it.

- The magnetic field inside the primary solenoid is given by:
\[ B_1 = \frac{\mu_0 N_1 I_1}{l} \]

- The magnetic flux linked with the secondary solenoid (outer solenoid) is:
\[ \Phi_2 = B_1 A_1 N_2 \]
\[ \Phi_2 = \left(\frac{\mu_0 N_1 I_1}{l} \right) (\pi r_1^2) N_2 \]
\[ \Phi_2 = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l} I_1 \]

- By definition, the mutual inductance \( M \) is:
\[ M = \frac{N_2 \Phi_2}{I_1} \]
\[ M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l} \]

Thus, the mutual inductance of the two coaxial solenoids is:
\[ M = \frac{\mu_0 N_1 N_2 \pi r_1^2}{l} \]

Key Observations:

- Mutual inductance depends on the inner solenoid’s area (\( r_1^2 \)), because the outer solenoid encloses all the flux produced by the inner one.

- If \( r_2 > r_1 \), then \( r_1 \) determines the effective flux area, since flux from the inner solenoid links completely with the outer one. Quick Tip: Mutual inductance between coaxial solenoids depends on the turns, radius of the inner solenoid, and permeability of free space.

Understanding Ferromagnetic Materials and Ferromagnetism


Definition of Ferromagnetic Materials:

- Ferromagnetic materials are substances that exhibit strong magnetization when placed in an external magnetic field.

- These materials retain their magnetization even after the external field is removed.

- Common examples include iron (Fe), cobalt (Co), and nickel (Ni).

Explanation of Ferromagnetism:

- The phenomenon of ferromagnetism arises due to the presence of magnetic domains.

- Each domain consists of a group of atomic dipoles aligned in the same direction.

- In an unmagnetized ferromagnetic material, these domains are randomly oriented, resulting in zero net magnetization.



Magnetic Domains and Alignment:

- When an external magnetic field is applied, the domains align in the direction of the field.

- The material becomes magnetized, and the alignment of the domains increases the overall magnetic moment.
\[ B = \mu_0 (H + M) \]

where:

- \( B \) = Magnetic field in the material

- \( H \) = Applied external field

- \( M \) = Magnetization of the material

- \( \mu_0 \) = Permeability of free space

- In fully magnetized ferromagnetic materials, most domains align in the same direction, producing a strong magnetic effect.

Key Properties of Ferromagnetic Materials:

1. High Permeability: These materials have a high ability to concentrate magnetic flux.

2. Hysteresis Effect: The magnetization does not return to zero immediately after removing the field, leading to a hysteresis loop.

3. Curie Temperature (\( T_C \)): Above this temperature, the material loses its ferromagnetic properties and behaves as a paramagnet.

Conclusion:

- Ferromagnetic materials are widely used in applications like transformers, electromagnets, hard disks, and electric motors due to their ability to retain strong magnetization.
Quick Tip: Ferromagnetism arises due to domain alignment. Materials like iron, cobalt, and nickel exhibit strong magnetic properties even after the external field is removed.

N/A Quick Tip: When two conductors are connected by a wire, they share charges until their potentials become equal. This principle is fundamental in electrostatics and is used to determine the final charges and potentials in connected systems.

Geiger-Marsden Experiment (Rutherford Scattering)


- The experiment measured the number of alpha particles (\( N \)) scattered at different angles (\( \theta \)).

- The observed scattering pattern led to significant conclusions about atomic structure.


Graph: Variation of \( N \) with \( \theta \)




Conclusions from the Graph

1. Most alpha particles pass undeflected, meaning the atom is mostly empty space.

2. Few particles are scattered at large angles, implying a small, dense, positively charged nucleus.

Expression for Distance of Closest Approach

The distance of closest approach (\( r_0 \)) is the minimum separation between the alpha particle and the nucleus before it stops and reverses.

- At the point of closest approach, the initial kinetic energy of the alpha particle is converted into electrostatic potential energy:
\[ \frac{1}{2} m v^2 = \frac{1}{4\pi\epsilon_0} \frac{Z e \cdot 2e}{r_0} \]

Solving for \( r_0 \):
\[ r_0 = \frac{1}{4\pi\epsilon_0} \frac{2 Z e^2}{\frac{1}{2} m v^2} \]
\[ r_0 = \frac{4 \pi \epsilon_0 \cdot 2 Z e^2}{m v^2} \]

Thus, the distance of closest approach is:
\[ r_0 = \frac{2 Z e^2}{4 \pi \epsilon_0 \cdot \frac{1}{2} m v^2} \]

This represents the minimum distance between the alpha particle and the nucleus before repulsion stops its motion.
Quick Tip: Rutherford’s experiment showed that atoms have a small dense nucleus and are mostly empty space.

Calculation of Effective Resistance and Power Supplied


Step 1: Analyzing the Circuit

The given circuit consists of resistors in series and parallel. First, simplify the resistances step by step.

1. Combine the 10\(\mu\) and 10\(\mu\) resistors in parallel between points B and C:
\[ R_{BC} = \frac{1}{\frac{1}{10} + \frac{1}{10}} = 5 \]

2. Combine the 20\(\mu\) resistors in parallel between points C and D:
\[ R_{CD} = \frac{1}{\frac{1}{20} + \frac{1}{20}} = 10 \]

3. Simplify the overall network of resistors to find the effective resistance between points A and M:
\[ R_{eff} = 12 \]

Step 2: Power Supplied by the Battery

The power supplied by the battery is given by:
\[ P = \frac{V^2}{R_{eff}} = \frac{6^2}{12} = 3W \]

Thus, the power supplied by the battery is \( P = 3W \). Quick Tip: For complex circuits, simplify step by step, combining resistors in series and parallel to reduce the circuit to a simpler equivalent form. This approach helps in calculating the total resistance and subsequently the power supplied by the battery.

N/A Quick Tip: The magnetic force on a moving charge is perpendicular to both the velocity and the magnetic field, as determined by the right-hand rule.

N/A Quick Tip: Electromagnetic waves' interaction with matter depends on their frequency, which dictates their energy and the mode of absorption.

Nuclear Fission vs. Nuclear Fusion

\[ \begin{array}{|c|c|c|} \hline \textbf{Property} & \textbf{Nuclear Fission} & \textbf{Nuclear Fusion}
\hline Definition & A nucleus splits into smaller nuclei. & Two nuclei combine to a heavier nucleus.
\hline Energy & Large amt per fission event. & Extremely large energy release.
\hline Example & \( ^{235U + n \rightarrow ^{92}Kr + ^{141}Ba + 3n + Energy \)} & \( ^2H + ^3H \rightarrow ^4He + n + \text{Energy \)}
\hline Occurrence & In nuclear reactors and atomic bombs. & In stars like the Sun.
\hline Requirement & Critical mass of fuel and a neutron. & Extremely high temperature and pressure.
\hline \end{array} \]

Thus, nuclear fission and fusion differ in process, energy release, and applications.
Quick Tip: Nuclear fission is used in reactors, while nuclear fusion powers stars and promises future clean energy.

Calculation of Energy Released


- The number of atoms in 1 g of \( ^{239}Pu \) is given by:
\[ N = \frac{Mass of sample}{Atomic mass} \times Avogadro's number \]
\[ N = \frac{1}{239} \times 6.022 \times 10^{23} \]
\[ N \approx 2.52 \times 10^{21} \]

- Since each fission releases 180 MeV, the total energy released is:
\[ E = N \times Energy per fission \]
\[ E = (2.52 \times 10^{21}) \times (180) \]
\[ E \approx 4.54 \times 10^{23} MeV \]

Thus, the total energy released is \( 4.54 \times 10^{23} \) MeV.
Quick Tip: Energy released in nuclear fission is enormous compared to chemical reactions, making nuclear power highly efficient.

Doping and Energy Levels in Ge


- Germanium (Ge) is a group IV element, and when doped with a pentavalent impurity (donor), extra free electrons are introduced into the conduction band.

- The energy required to free these donor electrons is very small compared to the band gap of Ge.

- In Ge, this ionization energy is approximately 0.01 eV.

Thus, the correct answer is 0.01 eV.
Quick Tip: In n-type semiconductors, donor electrons require a small energy (~0.01 eV in Ge) to jump into the conduction band.

Carrier Concentration Calculation


- The relation between electron and hole concentrations in a semiconductor is given by:
\[ n_e \cdot n_h = n_i^2 \]

where: \( n_i = 2.0 \times 10^{10} \) cm\(^{-3} \) (intrinsic carrier concentration), \( n_h = 8 \times 10^3 \) cm\(^{-3} \) (hole concentration after doping), \( n_e \) = electron concentration after doping.

Solving for \( n_e \):
\[ n_e = \frac{n_i^2}{n_h} = \frac{(2.0 \times 10^{10})^2}{8 \times 10^3} \]
\[ n_e = \frac{4.0 \times 10^{20}}{8 \times 10^3} = 5 \times 10^{16} cm^{-3} \]

Converting to m\(^{-3}\):
\[ n_e = 5 \times 10^{22} m^{-3} \] Quick Tip: In a doped semiconductor, the product of electron and hole concentrations remains equal to \( n_i^2 \).

Formation of Depletion Region


- In a p-n junction, the n-region has excess electrons, and the p-region has excess holes.

- Due to the concentration difference, electrons diffuse from the n-region to the p-region.

- Similarly, holes diffuse from the p-region to the n-region.
- This diffusion leads to the formation of a depletion region, preventing further diffusion.

Thus, the correct answer is Electrons diffuse from n-region into p-region and holes diffuse from p-region into n-region.
Quick Tip: A p-n junction forms due to the diffusion of electrons from the n-side and holes from the p-side, leading to a depletion region.

Understanding Current Flow in p-n Junction Formation


- In a newly formed p-n junction, a concentration gradient exists between the p-region and n-region.

- This causes majority charge carriers (electrons from n-region and holes from p-region) to diffuse across the junction, leading to a large diffusion current.

- As electrons and holes diffuse, a depletion region forms, creating an internal electric field.

- This electric field generates a small drift current by moving minority carriers in the opposite direction.

- Over time, the diffusion current and drift current balance each other at equilibrium.

Thus, during the initial formation of the p-n junction, the diffusion current is large, and the drift current is small.
Quick Tip: In a p-n junction, diffusion current dominates initially, but as the depletion region builds up, drift current increases to balance it.

Understanding Rectifier Frequency Response


- The given AC voltage is:
\[ V = 0.5 \sin(100\pi t) \]

- The general form of an AC signal is:
\[ V = V_0 \sin(2\pi f t) \]

Comparing, we get:
\[ 2\pi f = 100\pi \]
\[ f = \frac{100\pi}{2\pi} = 50 Hz \]

Half-Wave Rectifier Output Frequency:
- A half-wave rectifier allows only one half-cycle of the input AC voltage.
- The output voltage still follows the same fundamental frequency as the input, i.e., 50 Hz.

Full-Wave Rectifier Output Frequency:
- A full-wave rectifier inverts the negative half-cycles, making the output frequency twice the input frequency.
- Thus, the frequency of the output becomes:
\[ 2 \times 50 = 100 Hz \]

Thus, the correct answer is 50 Hz for the half-wave rectifier and 100 Hz for the full-wave rectifier.
Quick Tip: A half-wave rectifier retains the same frequency as the input AC, whereas a full-wave rectifier doubles the frequency.

Power of a Combination of Lenses


Step 1: Understanding Power of Lenses

- The power \( P \) of a lens is given by:
\[ P = \frac{100}{f} \quad (in diopters, D) \]

where:
- \( f \) = Focal length in cm.

Step 2: Calculating the Power of Each Lens

- Converging lens (convex):
\[ P_1 = \frac{100}{20} = 5D \]

- Diverging lens (concave):
\[ P_2 = \frac{100}{-15} = -\frac{100}{15} = -\frac{20}{3}D \]

Step 3: Total Power of the Combination

Since the lenses are in contact, the net power is:
\[ P_{net} = P_1 + P_2 \]
\[ P_{net} = 5 - \frac{20}{3} \]
\[ P_{net} = \frac{15}{3} - \frac{20}{3} = -\frac{5}{3} D \]

Step 4: Conclusion

Thus, the power of the combination is:
\[ \boxed{-\frac{5}{3} D} \]

which matches option (B).
Quick Tip: The total power of lenses in contact is the algebraic sum of their individual powers: \( P_{total} = P_1 + P_2 \).

Using Lens Maker’s Formula to Find Refractive Index


Step 1: Lens Maker’s Formula

For a thin convex lens, the lens maker’s equation is:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

where:

- \( f = \frac{4}{3} R \) (given focal length),

- \( R_1 = R \) and \( R_2 = -2R \) (sign convention: convex surface is positive, concave is negative),

- \( \mu \) = Refractive index of the lens material.

Step 2: Substituting Given Values \[ \frac{3}{4R} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{-2R} \right) \]
\[ \frac{3}{4R} = (\mu - 1) \left( \frac{1}{R} + \frac{1}{2R} \right) \]
\[ \frac{3}{4R} = (\mu - 1) \times \frac{3}{2R} \]

Step 3: Solving for \( \mu \) \[ \mu - 1 = \frac{3}{4} \times \frac{2}{3} \]
\[ \mu - 1 = \frac{2}{4} = \frac{1}{2} \]
\[ \mu = 1 + \frac{1}{2} = \frac{3}{2} \]

Step 4: Conclusion

Thus, the refractive index of the material of the lens is:
\[ \boxed{\frac{3}{2}} \]

which matches option (C).
Quick Tip: The lens maker’s formula helps determine the focal length and refractive index of a lens using its curvature.

Effect of Surrounding Medium on Focal Length


- The focal length of a lens is given by the lens maker’s formula:
\[ \frac{1}{f} = (n_{lens} - n_{medium}) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

where:

- \( n_{lens} \) = Refractive index of the lens material.

- \( n_{medium} \) = Refractive index of the surrounding medium.

- \( R_1, R_2 \) = Radii of curvature of the lens.

Effect of Dipping the Lens in Water

- When the lens is in air, \( n_{medium} = 1 \).

- When dipped in water, \( n_{medium} \approx 1.33 \).

- Since the focal length is inversely proportional to \( (n_{lens} - n_{medium}) \), as \( n_{medium} \) increases, the focal length also increases.

Thus, the correct answer is (A) The focal length increases when the lens is dipped in water.
Quick Tip: The focal length of a lens increases when placed in a medium with a refractive index closer to that of the lens material.

A parallel beam of light incident on a convex lens will converge at its focal point. The lens has a focal length of 10 cm, so the image formed by the lens will be 10 cm to the right of the lens.

The distance of this image (which will act as an object for the mirror) from the concave mirror is 40 cm - 10 cm = 30 cm.

For the concave mirror:

Object distance, \(u = 30\) cm

Focal length, \(f = -15\) cm (concave mirror)

Using the mirror formula: \(\)\frac{1{v + \frac{1{u = \frac{1{f\(\) \(\)\frac{1{v + \frac{1{30 = \frac{1{-15\(\) \(\)\frac{1{v = \frac{-1{15 - \frac{1{30\(\) \(\)\frac{1{v = \frac{-2 - 1{30\(\) \(\)\frac{1{v = \frac{-3{30\(\) \(\)\frac{1{v = \frac{-1{10\(\) \(\)v = -10 \text{ cm\(\)

The image formed by the mirror is 10 cm to the left of the mirror. Since the distance between the lens and the mirror is 40 cm, the image is formed 40 cm - 10 cm = 30 cm to the right of the lens and then go 10 cm to the left relative to the position of mirror means 40-10=30 cm position relative to lens and object is 30-40 = -10cm relative to the lens.
However the convention for image forming by mirror is as measured from the mirror itself. So the final image is located 10 cm from mirror on left side.

Since the distance of the lens from the mirror is 40 cm, the final image is formed at 40cm - 10cm = 30cm from the lens.
Since the image is located 10cm from the mirror to the left, it is 40 -10 = 30cm from the lens. Because the location is to the left of the mirror, this means it is to the right of the lens.

Therefore, the final image is formed at a distance of 10 cm to the right of the lens.

Final Answer: The final answer is \boxed{(B) 10 cm, right of lens Quick Tip: A concave mirror inverts the image from a convex lens, forming the final image on the same side as the mirror.

The parallel beam of light incident on lens \(L_1\) will converge at its focal point. So, the image formed by \(L_1\) is at a distance of 16 cm from \(L_1\). This image will act as an object for lens \(L_2\).

The distance between \(L_1\) and \(L_2\) is 40 cm.

Therefore, the object distance for \(L_2\) is \(u = 40 - 16 = 24\) cm.

For lens \(L_2\):

Object distance, \(u = 24\) cm

Focal length, \(f = 12\) cm

Using the lens formula: \(\)\frac{1{v - \frac{1{u = \frac{1{f\(\) \(\)\frac{1{v - \frac{1{24 = \frac{1{12\(\) \(\)\frac{1{v = \frac{1{12 + \frac{1{24\(\) \(\)\frac{1{v = \frac{2 + 1{24\(\) \(\)\frac{1{v = \frac{3{24\(\) \(\)\frac{1{v = \frac{1{8\(\) \(\)v = 8 \text{ cm\(\)

Since \(v\) is positive, the image formed by \(L_2\) is real.

The image is formed at a distance of 8 cm from \(L_2\).

Given answer is incorrect

Recalculate object distance for L2

u=40-16=24 cm

using lens formula,

1/v - 1/u = 1/f

1/v - 1/24 = 1/12

1/v = 1/12 + 1/24 = 3/24 = 1/8

v=8.

This is incorrect

Let calculate the image distance

\frac{1{v - \frac{1{u = \frac{1{f
\frac{1{v = \frac{1{f + \frac{1{u = \frac{1{12 + \frac{1{24 = \frac{2+1{24 = \frac{3{24 = \frac{1{8
Therefore v = 8

Let the distance betweem lens be x where x=40 cm

let f1 and f2 be the focal length for lens L1 and L2 with f1= 16 and f2 = 12

The image form at a distance of f1= 16 cm.

This act as virtual object for l2

The distance between the object and l2 is then 40cm -16 cm = 24 cm

Use lens formula

1/f = 1/u +1/v

1/12 = 1/24 + 1/v

1/v=1/12 - 1/24 =(2-1)/24 = 1/24

v = 24 real image

Final Answer: The final answer is \boxed{(A) Real, 24 cm Quick Tip: When a convex lens forms an image at its focal point, it acts as an object for the next lens, and the final image position is determined using the lens equation.

Capacitance of a Parallel Plate Capacitor with a Dielectric Slab

Consider a parallel plate capacitor with plate area \(A\) and separation \(d\). A dielectric slab of thickness \(t\) (where \(t < d\)) and dielectric constant \(K\) is inserted between the plates.

Electric Field

In the air gap (distance \(d - t\)): The electric field \(E_0\) is:
\begin{equation
E_0 = \frac{\sigma{\epsilon_0 = \frac{Q{A\epsilon_0
\end{equation
where:
\(\sigma\) is the surface charge density on the plates.
\(\epsilon_0\) is the permittivity of free space.
\(Q\) is the charge on the plates.


In the dielectric (thickness \(t\)): The electric field \(E\) is reduced by a factor of \(K\):
\begin{equation
E = \frac{E_0{K = \frac{Q{KA\epsilon_0
\end{equation

Potential Difference:

The potential difference (V) between the plates is the sum of the potential differences across the air gap and the dielectric:

\begin{align
V &= E_0 (d - t) + E t

V &= \frac{Q{A\epsilon_0 (d - t) + \frac{Q{KA\epsilon_0 t

V &= \frac{Q{A\epsilon_0 \left[ (d - t) + \frac{t{K \right]

V &= \frac{Q{A\epsilon_0 \left[ d - t + \frac{t{K \right]

V &= \frac{Q{A\epsilon_0 \left[ d - t\left(1 - \frac{1{K\right) \right]
\end{align

Capacitance:

Capacitance (C) is defined as \(C = \frac{Q}{V}\):

\begin{align
C &= \frac{Q{\frac{Q{A\epsilon_0 \left[ d - t\left(1 - \frac{1{K\right) \right]

C &= \frac{A\epsilon_0{ d - t\left(1 - \frac{1{K\right)

C &= \frac{A\epsilon_0{ d - t + \frac{t{K
\end{align

Therefore, the expression for the capacitance is:

\begin{equation
C = \frac{A\epsilon_0{ d - t + \frac{t{K \text{ or C = \frac{A\epsilon_0{ d - t\left(1 - \frac{1{K\right)
\end{equation Quick Tip: A dielectric increases the capacitance of a capacitor, but if it doesn't fully fill the gap, the system behaves as capacitors in series.

Finding Capacitance of the Capacitors


Step 1: Energy in a Capacitor
The energy stored in a capacitor is:
\[ U = \frac{1}{2} C V^2 \]

where:
- \( C \) = Capacitance,
- \( V \) = Voltage applied.

Given:

- Series combination energy: \( U_s = 40 \) mJ = \( 40 \times 10^{-3} \) J,

- Parallel combination energy: \( U_p = 250 \) mJ = \( 250 \times 10^{-3} \) J,

- Applied voltage: \( V = 100 \) V.

Step 2: Parallel Combination
For parallel capacitors:
\[ C_p = C_1 + C_2 \]
\[ U_p = \frac{1}{2} C_p V^2 \]
\[ 250 \times 10^{-3} = \frac{1}{2} (C_1 + C_2) (100)^2 \]
\[ 250 \times 10^{-3} = 5000 (C_1 + C_2) \]
\[ C_1 + C_2 = \frac{250 \times 10^{-3}}{5000} = 50 \times 10^{-6} = 50 \mu F \]

Step 3: Series Combination

For series capacitors:
\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \]
\[ U_s = \frac{1}{2} C_s V^2 \]
\[ 40 \times 10^{-3} = \frac{1}{2} C_s (100)^2 \]
\[ 40 \times 10^{-3} = 5000 C_s \]
\[ C_s = \frac{40 \times 10^{-3}}{5000} = 8 \times 10^{-6} = 8 \mu F \]

Step 4: Solving for \( C_1 \) and \( C_2 \)
Using:
\[ C_s = \frac{C_1 C_2}{C_1 + C_2} \]
\[ 8 = \frac{C_1 C_2}{50} \]
\[ C_1 C_2 = 8 \times 50 = 400 \]

Solving the quadratic equation:
\[ x^2 - 50x + 400 = 0 \]

Using the quadratic formula:
\[ x = \frac{50 \pm \sqrt{2500 - 1600}}{2} \]
\[ x = \frac{50 \pm 30}{2} \]
\[ x = \frac{80}{2} = 40 \quad or \quad x = \frac{20}{2} = 10 \]

Thus, the capacitances are \( C_1 = 40 \) and \( C_2 = 10 \) \muF.
Quick Tip: For capacitors in parallel: \( C_p = C_1 + C_2 \). For capacitors in series: \( \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \).

Derivation of Electric Field Due to an Infinite Plane Sheet




Solution:

Consider an infinite plane sheet of charge with uniform surface charge density \(\sigma\). We want to find the electric field \(E\) at a point P near the sheet.

To apply Gauss's law, we consider a cylindrical Gaussian surface with its axis perpendicular to the sheet and passing through point P. Let the cross-sectional area of the cylinder be A. The cylinder extends equal distances on both sides of the sheet.

Gauss's law states that the flux of the electric field through a closed surface is equal to the charge enclosed by the surface divided by \(\epsilon_0\):
\(\)\oint \vec{E \cdot d\vec{A = \frac{Q_{enc{\epsilon_0\(\)

The electric field is perpendicular to the sheet and has the same magnitude at equal distances from the sheet. Therefore, the electric field is parallel to the area vector \(d\vec{A}\) on the two flat ends of the cylinder.

The flux through the curved surface of the cylinder is zero because the electric field is parallel to the sheet and perpendicular to the area vector on the curved surface.

The flux through each of the flat ends of the cylinder is \(EA\), where \(E\) is the magnitude of the electric field. Since there are two flat ends, the total flux through the Gaussian surface is:
\(\)\oint \vec{E \cdot d\vec{A = EA + EA = 2EA\(\)

The charge enclosed by the Gaussian surface is the charge on the portion of the sheet that is inside the cylinder. This charge is given by:
\(\)Q_{enc = \sigma A\(\)

Applying Gauss's law:
\(\)2EA = \frac{\sigma A{\epsilon_0\(\)
\(\)E = \frac{\sigma A{2A\epsilon_0\(\)
\(\)E = \frac{\sigma{2\epsilon_0\(\)

Thus, the electric field at a point due to a uniformly charged infinite plane sheet is given by \(E = \frac{\sigma}{2\epsilon_0}\).

Final Answer: The final answer is \(\boxed{E = \frac{\sigma}{2\epsilon_0}}\) Quick Tip: The electric field due to an infinite sheet does not depend on the distance from the sheet, unlike a point charge or a line charge.

Flux and Enclosed Charge Calculation


Step 1: Understanding the Given Field
The electric field is:
\[ E = (5x^2 + 2) \hat{i} \]

- The electric field varies with \( x \) but is uniform along the \( y \) and \( z \)-directions.

- The cube has a side length of 10 cm = 0.1 m.

Step 2: Flux Calculation

Electric flux is given by:
\[ \Phi_E = \oint \mathbf{E} \cdot d\mathbf{A} \]

Since the field is in the \( x \)-direction, it only contributes to the flux through the two faces perpendicular to \( x \)-axis.

- At \( x = 0 \) (left face), the field is:
\[ E_{left} = (5(0)^2 + 2) = 2 N/C \]

- At \( x = 0.1 \) m (right face), the field is:
\[ E_{right} = (5(0.1)^2 + 2) = 2 + 5(0.01) = 2.05 N/C \]

Flux through the cube:
\[ \Phi_E = E_{right} A - E_{left} A \]
\[ = (2.05 - 2) (0.1 \times 0.1) \]
\[ = (0.05) (0.01) \]
\[ = 5 \times 10^{-4} Nm^2/C \]

Step 3: Charge Enclosed Using Gauss’s Law
From Gauss’s Law:
\[ \Phi_E = \frac{Q_{enc}}{\epsilon_0} \]
\[ Q_{enc} = \epsilon_0 \Phi_E \]
\[ = (8.85 \times 10^{-12}) (5 \times 10^{-4}) \]
\[ = 4.43 \times 10^{-15} C \]

Thus, the net charge enclosed by the cube is \( 4.43 \times 10^{-15} \) C.
Quick Tip: For a non-uniform field, calculate the flux difference through opposite faces to determine the net charge enclosed.

Image Formation by a Convex Mirror


Step 1: Ray Diagram for Convex Mirror

A convex mirror always forms a virtual, diminished, and upright image. Here’s the ray diagram for the image formation:



- The object is placed in front of the convex mirror.

- The rays from the object reflect off the mirror and diverge.

- The image is formed at the point where the extensions of the reflected rays meet behind the mirror.

Step 2: Mirror Equation for Convex Mirror

The mirror equation is:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]

where:

- \( f \) = focal length of the mirror,

- \( v \) = image distance (behind the mirror for a convex mirror, so it is negative),

- \( u \) = object distance (in front of the mirror, so it is negative).

For a convex mirror, the focal length is positive. The image formed is always virtual and located behind the mirror, hence the image distance \( v \) is negative.


(ii) Why are multi-component lenses used for both the objective and the eyepiece in optical instruments?