CBSE_Class_12_Physics_Model_Paper_2024

General Instructions :
Read the following instructions carefully and follow them :
(i) This question paper contains 33questions. Allquestions are compulsory.
(ii) This question paper is divided intofive sections – Sections A, B, C, D and E.
(iii) In Section A – Questions no. 1 to 16 are Multiple Choice type questions. Each question
carries 1 mark.
(iv) In Section B – Questions no. 17 to 21 are Very Short Answer type questions. Each
question carries 2 marks.
(v) In Section C – Questions no. 22 to 28 are Short Answer type questions. Each question
carries 3 marks.
(vi) In Section D – Questions no. 29 and 30 are case study-based questions. Each question
carries 4 marks.
(vii) In Section E – Questions no. 31 to 33 are Long Answer type questions. Each question
carries 5 marks.
(viii) There is no overall choice given in the question paper. However, an internal choice has
been provided in few questions in all the Sections except Section A.
(ix) Kindly note that there is a separate question paper for Visually Impaired candidates.
(x) Use of calculators is not allowed.
You may use the following values of physical constants wherever necessary :
c = 3 ´ 108 m/s
h = 6.63 ´ 10–34 Js
e = 1.6 ´ 10–19 C
m0 = 4p ´ 10–7 T m A–1
e0 = 8.854 ´ 10–12 C2 N–1 m–2
0 4
1
pe
= 9 ´ 109 N m2 C–2
Mass of electron (me) = 9.1 ´ 10–31 kg
Mass of neutron = 1.675 ´ 10–27kg
Mass of proton = 1.673 ´ 10–27kg
Avogadro’s number = 6.023 ´ 1023per gram mole
Boltzmann constant = 1.
38 ´ 10–23 JK–1

SECTION A

Here are the solutions for the given questions in Section A:

1. Two particles A and B of the same mass but having charges q and 4q respectively, are accelerated from rest through different potential differences VAV_A and VBV_B such that they attain the same kinetic energies. The value of VAVB\frac{V_A}{V_B} is:

Answer:
We know that the kinetic energy gained by a charged particle when accelerated through a potential difference VV is given by: K.E.=qVK.E. = qV

For two particles A and B to have the same kinetic energy: qAVA=qBVBq_A V_A = q_B V_B

Given that qA=qq_A = q and qB=4qq_B = 4q, we get: qVA=4qVBqV_A = 4qV_B

Thus, VA=4VBV_A = 4V_B

So, VAVB=4\frac{V_A}{V_B} = 4

Correct answer: (D) 4

2. A coil of resistance 20 Ω and self-inductance 10 mH is connected to an AC source of frequency 1000π\frac{1000}{\pi} Hz. The phase difference between the current in the circuit and the source voltage is:

Answer:
The phase difference in an RL circuit is given by: tan⁡(ϕ)=XLR\tan(\phi) = \frac{X_L}{R}

Where XL=2πfLX_L = 2 \pi f L is the inductive reactance, and RR is the resistance of the coil.

Given:

  • R=20 ΩR = 20 \, \Omega
  • L=10 mH=10×10−3 HL = 10 \, \text{mH} = 10 \times 10^{-3} \, \text{H}
  • Frequency f=1000πf = \frac{1000}{\pi} Hz

Now, calculate XLX_L: XL=2πfL=2π×1000π×10×10−3=20 ΩX_L = 2 \pi f L = 2 \pi \times \frac{1000}{\pi} \times 10 \times 10^{-3} = 20 \, \Omega

Now, calculate the phase difference: tan⁡(ϕ)=XLR=2020=1\tan(\phi) = \frac{X_L}{R} = \frac{20}{20} = 1

Therefore, ϕ=45∘\phi = 45^\circ

Correct answer: (D) 45°

3. Isotones are nuclides having:

Answer:
Isotones are atoms that have the same number of neutrons but different numbers of protons.

Correct answer: (A) same number of neutrons but different number of protons

4. A bulb is rated (100 W, 110 V). It is operated by a current of 1.0 A supplied by a step-down transformer. If the input voltage and efficiency of the transformer are 220 V and 0.9 respectively, the input current drawn from the mains is:

Answer:
Power output from the transformer = Power of the bulb = 100 W
Efficiency η=0.9\eta = 0.9, so the power input to the transformer is: Power input=Power outputη=1000.9=111.11 W\text{Power input} = \frac{\text{Power output}}{\eta} = \frac{100}{0.9} = 111.11 \, \text{W}

The input voltage is 220 V, so the input current is: Iinput=Power inputInput voltage=111.11220=0.505 AI_{\text{input}} = \frac{\text{Power input}}{\text{Input voltage}} = \frac{111.11}{220} = 0.505 \, \text{A}

Rounding off, the closest option is: Iinput≈0.5 AI_{\text{input}} \approx 0.5 \, \text{A}

Correct answer: (C) 5/9 A

5. Which of the following substances has relative magnetic permeability μr>>1\mu_r >> 1?

Answer:
Nickel is a ferromagnetic material, which has a high relative magnetic permeability (greater than 1). Other substances like aluminum and copper chloride have much lower permeability.

Correct answer: (C) Nickel

Summary of answers:

  1. (D) 4
  2. (D) 45°
  3. (A) same number of neutrons but different number of protons
  4. (C) 5/9 A
  5. (C) Nickel

Let me know if you’d like more clarification or further questions!

6. Which of the following statements is correct for the alpha particle scattering experiment?

Options:

  • (A) For angle of scattering θ≫0\theta \gg 0θ≫0, the impact parameter is small.
  • (B) For angle of scattering θ≈π\theta \approx \piθ≈π, the impact parameter is large.
  • (C) The number of alpha particles undergoing head-on collision is small.
  • (D) The experiment provides an estimate of the upper limit to the size of the target atom.

Solution: In the alpha particle scattering experiment (also known as Rutherford’s scattering experiment), the key idea is that when the scattering angle θ\thetaθ is large (close to π\piπ, i.e., 180°), the impact parameter (the perpendicular distance from the path of the particle to the center of the nucleus) is large.

  • For small scattering angles, the impact parameter is smaller, as the particles come closer to the nucleus.
  • The experiment indeed provides information about the size of the nucleus, specifically an estimate of the upper limit of the target atom’s size (i.e., the radius of the nucleus).

Correct answer: (D) The experiment provides an estimate of the upper limit to the size of the target atom.

7. A straight wire of length 1.0 m is placed along the x-axis, in a region with magnetic field B⃗=(3i^+2j^) T\vec{B} = (3\hat{i} + 2\hat{j}) \, \text{T}B=(3i^+2j^​)T. A current of 2.0 A flows in the wire along the +x direction. The magnetic force acting on the wire is:

Options:

  • (A) 2.0 N, along z-axis
  • (B) 2.0 N, along -z-axis
  • (C) 4.0 N, along z-axis
  • (D) 4.0 N, along -z-axis

Solution: The magnetic force F⃗\vec{F}F on a current-carrying wire is given by:F⃗=I (L⃗×B⃗)\vec{F} = I \, (\vec{L} \times \vec{B})F=I(L×B)

Where:

  • I=2.0 AI = 2.0 \, \text{A}I=2.0A is the current,
  • L⃗=i^ m\vec{L} = \hat{i} \, \text{m}L=i^m is the length vector of the wire along the x-axis,
  • B⃗=3i^+2j^ T\vec{B} = 3\hat{i} + 2\hat{j} \, \text{T}B=3i^+2j^​T is the magnetic field.

Now, compute the cross product L⃗×B⃗\vec{L} \times \vec{B}L×B:L⃗×B⃗=i^×(3i^+2j^)=i^×3i^+i^×2j^=0+2k^=2k^\vec{L} \times \vec{B} = \hat{i} \times (3\hat{i} + 2\hat{j}) = \hat{i} \times 3\hat{i} + \hat{i} \times 2\hat{j} = 0 + 2\hat{k} = 2\hat{k}L×B=i^×(3i^+2j^​)=i^×3i^+i^×2j^​=0+2k^=2k^

Thus, the magnetic force:F⃗=2.0 A×2k^=4.0k^ N\vec{F} = 2.0 \, \text{A} \times 2\hat{k} = 4.0 \hat{k} \, \text{N}F=2.0A×2k^=4.0k^N

So the force is 4.0 N along the z-axis.

Correct answer: (C) 4.0 N, along z-axis

8. The electric field E⃗\vec{E}E associated with an electromagnetic wave is represented by Ey=E0sin⁡(kx−ωt)E_y = E_0 \sin(kx – \omega t)Ey​=E0​sin(kx−ωt). Which of the following statements is correct?

Options:

  • (A) The wave is propagating along the +x-axis.
  • (B) The wave is propagating along the +z-axis.
  • (C) The magnetic field B⃗\vec{B}B of the wave is acting along the +y-axis.
  • (D) The magnetic field B⃗\vec{B}B of the wave is acting along the -x-axis.

Solution: For an electromagnetic wave, the electric field E⃗\vec{E}E and magnetic field B⃗\vec{B}B are perpendicular to each other and also perpendicular to the direction of propagation. The general form of the electric field is:Ey=E0sin⁡(kx−ωt)E_y = E_0 \sin(kx – \omega t)Ey​=E0​sin(kx−ωt)

This shows that the wave is varying along the x-direction, meaning the wave is propagating along the +x-axis.

The magnetic field B⃗\vec{B}B will be perpendicular to both E⃗\vec{E}E and the propagation direction (x-axis). Therefore, the magnetic field will be along the +z-axis.

Correct answer: (A) The wave is propagating along the +x-axis.

9. A point object is placed in air at a distance of 4R on the principal axis of a convex spherical surface of radius of curvature RRR, separating two mediums, air and glass. As the object is moved towards the surface, the image formed is:

Options:

  • (A) always real
  • (B) always virtual
  • (C) first virtual and then real
  • (D) first real and then virtual

Solution: The formation of images in a convex surface depends on the object’s position relative to the focal point of the lens or mirror. The focal length fff of the surface is related to the radius of curvature by:f=R2f = \frac{R}{2}f=2R​

  • Initially, the object is placed at a distance of 4R4R4R from the surface. Using the mirror equation or refraction equation, this will produce a real image.
  • As the object moves closer to the surface, the image may become virtual after crossing the focal point.

So, the image is first real and then virtual as the object moves towards the surface.

Correct answer: (D) first real and then virtual

Summary of answers:

  1. (D) The experiment provides an estimate of the upper limit to the size of the target atom.
  2. (C) 4.0 N, along z-axis
  3. (A) The wave is propagating along the +x-axis.
  4. (D) first real and then virtual

10. An electron makes a transition from orbit n=2n = 2n=2 to orbit n=1n = 1n=1 in Bohr’s model of the hydrogen atom. Consider the change in magnitudes of its kinetic energy (K) and potential energy (U).

Options:

  • (A) K increases and U decreases
  • (B) K decreases and U increases
  • (C) Both K and U decrease
  • (D) Both K and U increase

Solution: In Bohr’s model, the electron’s kinetic and potential energies are related to the principal quantum number nnn. For an electron in a higher orbit (e.g., n=2n = 2n=2), the potential energy UUU is less negative and the kinetic energy KKK is lower. As the electron transitions to a lower orbit (e.g., n=1n = 1n=1), it moves closer to the nucleus, which increases the magnitude of the negative potential energy (i.e., UUU becomes more negative), and the kinetic energy increases (since it has to balance the attractive Coulomb force).

  • Kinetic energy (K) increases because the electron moves to a lower energy state and has greater speed.
  • Potential energy (U) decreases (more negative) because the electron is closer to the nucleus.

Correct answer: (B) K decreases and U increases

11. Which of the following statements is not true for a p-n junction diode under reverse bias?

Options:

  • (A) The current is almost independent of the applied voltage.
  • (B) Holes flow from p-side to n-side.
  • (C) Electric field in the depletion region increases.
  • (D) n-side of the junction is connected to +ve terminal and p-side to –ve terminal of the battery.

Solution:

  • Reverse bias means that the p-side is connected to the negative terminal of the battery, and the n-side is connected to the positive terminal.
  • Under reverse bias, the current is ideally very small (almost negligible) because the depletion region widens, and there is little charge carrier flow.
  • The electric field in the depletion region increases as the reverse voltage increases, which further restricts the flow of carriers.

Now, let’s examine the statements:

  • (A): True, because in reverse bias, the current remains very small and is nearly constant as the reverse voltage is increased.
  • (B): This is incorrect. In reverse bias, the holes do not flow from the p-side to the n-side. Instead, they are pulled away from the junction.
  • (C): True. The electric field in the depletion region increases in reverse bias.
  • (D): True. In reverse bias, the n-side is connected to the positive terminal, and the p-side to the negative terminal.

Correct answer: (B) Holes flow from p-side to n-side.

12. A parallel plate capacitor is charged by a battery. The battery is then disconnected, and the plates of the charged capacitor are then moved farther apart. In the process:

Options:

  • (A) The charge on the capacitor increases.
  • (B) The potential difference across the plates decreases.
  • (C) The capacitance of the capacitor increases.
  • (D) The electrostatic energy stored in the capacitor increases.

Solution: When the battery is disconnected, the charge QQQ on the capacitor remains constant because no external circuit is providing or removing charge. Now, when the plates are moved farther apart:

  • The capacitance CCC decreases because capacitance is inversely proportional to the distance between the plates: C=ϵ0AdC = \frac{\epsilon_0 A}{d}C=dϵ0​A​, where ddd is the distance between the plates.
  • Since the charge remains constant, and capacitance decreases, the voltage VVV across the plates increases (as Q=C⋅VQ = C \cdot VQ=C⋅V).
  • The electrostatic energy stored in the capacitor is given by U=Q22CU = \frac{Q^2}{2C}U=2CQ2​. Since CCC decreases, the energy stored in the capacitor increases.

Correct answer: (D) The electrostatic energy stored in the capacitor increases.

13 to 16: Assertion-Reason Type Questions

13. Assertion (A): The electric field inside a conductor is always zero when the conductor is in electrostatic equilibrium.

Reason (R): In electrostatic equilibrium, there are no net forces acting on the charges inside the conductor.

Options:

  • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • (C) Assertion (A) is true, but Reason (R) is false.
  • (D) Both Assertion (A) and Reason (R) are false.

Solution:

  • In electrostatic equilibrium, the free charges inside a conductor move in such a way that they cancel out any internal electric field, thus the electric field inside the conductor becomes zero.
  • The reason provided is also correct: no net force acts on the charges in electrostatic equilibrium, hence there is no electric field inside the conductor.

Correct answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

14. Assertion (A): The refractive index of a medium always decreases when the medium is heated.

Reason (R): The speed of light in a medium increases with increasing temperature.

Options:

  • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • (C) Assertion (A) is true, but Reason (R) is false.
  • (D) Both Assertion (A) and Reason (R) are false.

Solution:

  • The refractive index nnn of a medium is given by n=cvn = \frac{c}{v}n=vc​, where ccc is the speed of light in vacuum and vvv is the speed of light in the medium.
  • When the temperature increases, the speed of light vvv in most materials increases (because the material becomes less dense), and thus the refractive index decreases.

Correct answer: (C) Assertion (A) is true, but Reason (R) is false.

15. Assertion (A): The electric field inside a uniformly charged spherical shell is zero.

Reason (R): The net charge inside the shell is zero.

Options:

  • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • (C) Assertion (A) is true, but Reason (R) is false.
  • (D) Both Assertion (A) and Reason (R) are false.

Solution:

  • The electric field inside a uniformly charged spherical shell is indeed zero, which can be explained by Gauss’s Law.
  • The statement about the net charge inside the shell being zero is not the reason; the result arises from the symmetry of the charge distribution.

Correct answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

16. Assertion (A): In a photoelectric effect experiment, if the intensity of light is increased, the kinetic energy of the emitted electrons increases.

Reason (R): The kinetic energy of the emitted electrons depends on the frequency of the incident light, not its intensity.

Options:

  • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • (C) Assertion (A) is true, but Reason (R) is false.
  • (D) Both Assertion (A) and Reason (R) are false.

Solution:

  • The kinetic energy of emitted electrons in the photoelectric effect depends on the frequency of the incident light, as described by the Einstein photoelectric equation K.E.=hν−ϕK.E. = h\nu – \phiK.E.=hν−ϕ, where hνh\nuhν is the energy of the photon and ϕ\phiϕ is the work function.
  • Increasing the intensity of light increases the number of emitted electrons but does not affect their kinetic energy.

Correct answer: (C) Assertion (A) is true, but Reason (R) is false.

Summary of answers:

  • 10: (B) K decreases and U increases
  • 11: (B) Holes flow from p-side to n-side.
  • 12: (D) The electrostatic energy stored in the capacitor increases.
  • 13: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • 14: (C) Assertion (A) is true, but Reason (R) is false.
  • 15: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • 16: (C) Assertion (A) is true, but Reason (R) is false.

13. Assertion (A): The current density J⃗\vec{J}J at a point in a conducting wire is in the direction of the electric field E⃗\vec{E}E at that point.

Reason (R): A conducting wire obeys Ohm’s law.

Answer:

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Explanation:
The current density J⃗\vec{J}J in a conductor is proportional to the electric field E⃗\vec{E}E as described by the relation J⃗=σE⃗\vec{J} = \sigma \vec{E}J=σE, where σ\sigmaσ is the conductivity of the material. According to Ohm’s law, the current density is directly proportional to the electric field, and they are in the same direction. Therefore, the assertion is true, and the reason correctly explains it.

14. Assertion (A): The torque acting on a current-carrying coil is maximum when it is suspended in a radial magnetic field.

Reason (R): The torque tends to rotate the coil on its own axis.

Answer:

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Explanation:
In a radial magnetic field, the torque on the coil is maximum when the coil is positioned such that its plane is perpendicular to the magnetic field lines. The torque causes the coil to rotate around its axis, and this is the reason the torque is maximum in a radial magnetic field. The assertion and reason are both true, and the reason correctly explains why the torque is maximum.

15. Assertion (A): Although the surfaces of a goggle lens are curved, it does not have any power.

Reason (R): In the case of goggles, both the curved surfaces are curved on the same side and have equal radii of curvature.

Answer:

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Explanation:
In goggles, the lens surfaces are often curved on both sides in the same direction with equal radii of curvature. This design results in no significant optical power because the overall effect of the curvature cancels out. The lens behaves almost like a flat lens, and as a result, it does not have focusing power. The assertion is correct, and the reason explains why the lens doesn’t have power.

16. Assertion (A): Nuclear fission reactions are responsible for energy generation in the Sun.

Reason (R): Light nuclei fuse together in the nuclear fission reactions.

Answer:

  • Assertion (A) is true, but Reason (R) is false.

Explanation:
The Sun generates energy through nuclear fusion, not fission. In nuclear fusion, light nuclei such as hydrogen fuse to form heavier nuclei like helium, releasing a vast amount of energy. On the other hand, nuclear fission involves the splitting of heavy nuclei, which is not the process occurring in the Sun. The assertion is true, but the reason is incorrect because it refers to fusion, not fission.

SECTION B

17. Two halves of a silicon crystal (A and B) are doped with arsenic and boron respectively, forming a p-n junction in it. A battery is connected across it as shown in the figure.

(a) Will the junction be forward biased or reverse biased? Give reason.

  • Forward Bias: In forward bias, the positive terminal of the battery is connected to the p-type side (doped with boron) and the negative terminal is connected to the n-type side (doped with arsenic).
    • This helps reduce the width of the depletion region at the p-n junction, allowing current to flow through the junction.
  • Reverse Bias: In reverse bias, the positive terminal of the battery is connected to the n-type side (arsenic-doped) and the negative terminal to the p-type side (boron-doped).
    • This increases the width of the depletion region, preventing current from flowing through the junction.

Answer:

  • If the positive terminal is connected to the p-side (boron-doped), the junction is forward biased.
  • If the positive terminal is connected to the n-side (arsenic-doped), the junction is reverse biased.

(b) Draw V-I graph for this arrangement.

  • Forward Bias (V-I graph):
    • When the voltage is increased in forward bias (positive terminal to p-side), the current remains very small until the threshold voltage is reached (around 0.7 V for a silicon diode). After this point, the current increases sharply.
    • So, the V-I graph for forward bias has a steep rise after the threshold voltage.
  • Reverse Bias (V-I graph):
    • In reverse bias, the current remains zero until the reverse voltage reaches a critical value (called breakdown voltage). After that, the current rises sharply due to breakdown.
    • The V-I graph for reverse bias shows almost no current until breakdown voltage.

V-I Graph Summary:

  1. Forward Bias:
    • Starts with little current.
    • After the threshold voltage (~0.7 V for silicon), current increases rapidly.
  2. Reverse Bias:
    • Current is zero initially.
    • After a certain voltage (breakdown voltage), current increases sharply.

Simplified Steps:

  1. Forward Bias:
    • Positive terminal to p-side (boron).
    • Negative terminal to n-side (arsenic).
    • Current increases after 0.7 V (for silicon).
  2. Reverse Bias:
    • Positive terminal to n-side.
    • Negative terminal to p-side.
    • Current remains near zero until breakdown.

18. A long straight horizontal wire is carrying a current III. At an instant, an alpha particle at a distance rrr from it is travelling with speed vvv parallel to the wire in a direction opposite to the current. Find the magnitude and direction of the force experienced by the particle at this instant.

Answer:

The force on the alpha particle is given by:F=μ0IevπrF = \frac{\mu_0 I e v}{\pi r}F=πrμ0​Iev​

Where:

  • e=1.6×10−19 Ce = 1.6 \times 10^{-19} \, \text{C}e=1.6×10−19C (charge of an electron),
  • μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}μ0​=4π×10−7T\cdotpm/A (permeability of free space),
  • III is the current in the wire,
  • vvv is the velocity of the alpha particle,
  • rrr is the distance of the particle from the wire.

Direction: The force will act towards the wire because the alpha particle is moving opposite to the direction of current (positive charge).

19. A point light source rests on the bottom of a bucket filled with a liquid of refractive index μ=1.25\mu = 1.25μ=1.25 up to a height of 10 cm. Calculate:

(a) the critical angle for the liquid-air interface

Answer:

The critical angle θc\theta_cθc​ is given by:sin⁡θc=1μ\sin \theta_c = \frac{1}{\mu}sinθc​=μ1​

For the liquid with μ=1.25\mu = 1.25μ=1.25, the critical angle is:sin⁡θc=11.25=0.8\sin \theta_c = \frac{1}{1.25} = 0.8sinθc​=1.251​=0.8 θc=sin⁡−1(0.8)=53.13∘\theta_c = \sin^{-1}(0.8) = 53.13^\circθc​=sin−1(0.8)=53.13∘

(b) Radius of the circular light patch formed on the surface by light emerging from the source

Answer:

The radius rrr of the circular light patch is:r=h⋅tan⁡θcr = h \cdot \tan \theta_cr=h⋅tanθc​

Where h=0.1 mh = 0.1 \, \text{m}h=0.1m (height of liquid), and θc=53.13∘\theta_c = 53.13^\circθc​=53.13∘.

Substitute the values:r=0.1⋅tan⁡(53.13∘)=0.1⋅1.333=0.133 m=13.3 cmr = 0.1 \cdot \tan(53.13^\circ) = 0.1 \cdot 1.333 = 0.133 \, \text{m} = 13.3 \, \text{cm}r=0.1⋅tan(53.13∘)=0.1⋅1.333=0.133m=13.3cm

Thus, the radius of the light patch is 13.3 cm.

20. State Huygens’ principle. Using it, draw a diagram showing the details of the passage of a plane wave from a denser into a rarer medium.

Answer:

Huygens’ Principle states that:

  • Every point on a wavefront can be considered as a source of secondary wavelets. These wavelets propagate forward and the new wavefront is the envelope of these secondary wavelets.

Diagram Description:

  1. A plane wavefront approaches the boundary between two media (denser to rarer).
  2. Each point on the wavefront acts as a secondary source of wavelets.
  3. The wavefront bends away from the normal as it passes into the rarer medium.

(For drawing, imagine the wavefront bending outward as it moves from a denser medium like water into a rarer medium like air.)

21. (a) A cell is connected across an external resistance 12 Ω12 \, \Omega12Ω and supplies 0.25 A0.25 \, \text{A}0.25A current. When the external resistance is increased by 4 Ω4 \, \Omega4Ω, the current reduces to 0.2 A0.2 \, \text{A}0.2A. Calculate:

(i) The emf, and

Answer:

From the two conditions:

  1. E=0.25×(12+r)\mathcal{E} = 0.25 \times (12 + r)E=0.25×(12+r)
  2. E=0.2×(16+r)\mathcal{E} = 0.2 \times (16 + r)E=0.2×(16+r)

Equating both equations:0.25×(12+r)=0.2×(16+r)0.25 \times (12 + r) = 0.2 \times (16 + r)0.25×(12+r)=0.2×(16+r)

Solving for r=4 Ωr = 4 \, \Omegar=4Ω (internal resistance), substitute rrr back into the first equation:E=0.25×(12+4)=4 V\mathcal{E} = 0.25 \times (12 + 4) = 4 \, \text{V}E=0.25×(12+4)=4V

Thus, the emf E\mathcal{E}E is 4 V4 \, \text{V}4V and the internal resistance rrr is 4 Ω 4 \, \Omega4Ω.

OR (b) Two point charges of 3 mC3 \, \text{mC}3mC and 4 mC4 \, \text{mC}4mC are kept in air at (0.3 m,0)(0.3 \, \text{m}, 0)(0.3m,0) and (0,0.3 m)(0, 0.3 \, \text{m})(0,0.3m) in the x-y plane. Find the magnitude and direction of the net electric field produced at the origin (0,0)(0, 0)(0,0).

Answer:

  1. Electric field due to charge q1=3 mCq_1 = 3 \, \text{mC}q1​=3mC at (0.3 m,0)(0.3 \, \text{m}, 0)(0.3m,0):E1=k∣q1∣r12=9×109×3×10−3(0.3)2=3×108 N/CE_1 = \frac{k |q_1|}{r_1^2} = \frac{9 \times 10^9 \times 3 \times 10^{-3}}{(0.3)^2} = 3 \times 10^8 \, \text{N/C}E1​=r12​k∣q1​∣​=(0.3)29×109×3×10−3​=3×108N/CThe direction is along the positive x-axis.
  2. Electric field due to charge q2=4 mCq_2 = 4 \, \text{mC}q2​=4mC at (0,0.3 m)(0, 0.3 \, \text{m})(0,0.3m):E2=k∣q2∣r22=9×109×4×10−3(0.3)2=4×108 N/CE_2 = \frac{k |q_2|}{r_2^2} = \frac{9 \times 10^9 \times 4 \times 10^{-3}}{(0.3)^2} = 4 \times 10^8 \, \text{N/C}E2​=r22​k∣q2​∣​=(0.3)29×109×4×10−3​=4×108N/CThe direction is along the positive y-axis.
  3. Net electric field: Since the electric fields are perpendicular, we use the Pythagorean theorem:Enet=E12+E22=(3×108)2+(4×108)2=5×108 N/CE_{\text{net}} = \sqrt{E_1^2 + E_2^2} = \sqrt{(3 \times 10^8)^2 + (4 \times 10^8)^2} = 5 \times 10^8 \, \text{N/C}Enet​=E12​+E22​​=(3×108)2+(4×108)2​=5×108N/CDirection: The direction is 45∘45^\circ45∘ to both the x-axis and y-axis.

Thus, the magnitude of the net electric field is 5×108 N/C5 \times 10^8 \, \text{N/C}5×108N/C, and the direction is along the line making a 45∘45^\circ45∘ angle with both the x and y axes.

Summary of Answers:

  1. Question 18: The magnitude of the force on the alpha particle is μ0Ievπr\frac{\mu_0 I e v}{\pi r}πrμ0​Iev​, and it acts towards the wire.
  2. Question 19:
    • (a) Critical angle is 53.13°.
    • (b) Radius of the circular light patch is 13.3 cm.
  3. Question 20: Huygens’ principle and refraction diagram.
  4. Question 21:
    • (a) Emf = 4 V, internal resistance = 4 Ω.
    • (b) Magnitude of net electric field = 5×108 N/C5 \times 10^8 \, \text{N/C}5×108N/C, direction is at 45° to both axes.

SECTION C

22. A small circular loop of area 6π cm26\pi \, \text{cm}^26πcm2 is placed inside a long solenoid at its center such that its axis makes an angle of 60∘60^\circ60∘ with the axis of the solenoid. The number of turns per cm is 10 in the solenoid. The current in the solenoid changes uniformly from 5 A to zero in 10 ms. Calculate the emf induced in the loop.

Answer:

We can use Faraday’s Law of Induction to calculate the induced emf. The formula for the induced emf is:E=−NdΦdt\mathcal{E} = -N \frac{d\Phi}{dt}E=−NdtdΦ​

Where:

  • Φ\PhiΦ is the magnetic flux through the loop,
  • NNN is the number of turns of the loop,
  • dΦdt\frac{d\Phi}{dt}dtdΦ​ is the rate of change of flux.

Step 1: Calculate the magnetic field inside the solenoid.

The magnetic field inside a solenoid is given by:B=μ0nIB = \mu_0 n IB=μ0​nI

Where:

  • nnn is the number of turns per unit length (turns/cm),
  • III is the current in the solenoid.

Given:

  • n=10 turns/cm=102 turns/mn = 10 \, \text{turns/cm} = 10^2 \, \text{turns/m}n=10turns/cm=102turns/m,
  • Initial current I=5 AI = 5 \, \text{A}I=5A,
  • Permeability of free space μ0=4π×10−7 T\cdotpm/A\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}μ0​=4π×10−7T\cdotpm/A.

Substitute these values into the formula for BBB:B=(4π×10−7)×(102)×5=2×10−4 TB = (4\pi \times 10^{-7}) \times (10^2) \times 5 = 2 \times 10^{-4} \, \text{T}B=(4π×10−7)×(102)×5=2×10−4T

Step 2: Calculate the magnetic flux through the loop.

The flux through the loop is given by:Φ=B⋅A⋅cos⁡(θ)\Phi = B \cdot A \cdot \cos(\theta)Φ=B⋅A⋅cos(θ)

Where:

  • A=6π cm2=6π×10−4 m2A = 6\pi \, \text{cm}^2 = 6\pi \times 10^{-4} \, \text{m}^2A=6πcm2=6π×10−4m2 (area of the loop),
  • θ=60∘\theta = 60^\circθ=60∘ is the angle between the magnetic field and the normal to the loop’s plane.

Substitute values:Φ=(2×10−4)×(6π×10−4)×cos⁡(60∘)\Phi = (2 \times 10^{-4}) \times (6\pi \times 10^{-4}) \times \cos(60^\circ)Φ=(2×10−4)×(6π×10−4)×cos(60∘)Φ=(2×10−4)×(6π×10−4)×12\Phi = (2 \times 10^{-4}) \times (6\pi \times 10^{-4}) \times \frac{1}{2}Φ=(2×10−4)×(6π×10−4)×21​Φ=6π×10−8 Wb\Phi = 6\pi \times 10^{-8} \, \text{Wb}Φ=6π×10−8Wb

Step 3: Calculate the change in flux.

The current in the solenoid changes from 5 A5 \, \text{A}5A to 0 A0 \, \text{A}0A, so the magnetic field decreases to zero. Hence, the change in flux is:ΔΦ=6π×10−8 Wb−0=6π×10−8 Wb\Delta \Phi = 6\pi \times 10^{-8} \, \text{Wb} – 0 = 6\pi \times 10^{-8} \, \text{Wb}ΔΦ=6π×10−8Wb−0=6π×10−8Wb

Step 4: Calculate the induced emf.

The time for this change is Δt=10 ms=10×10−3 s\Delta t = 10 \, \text{ms} = 10 \times 10^{-3} \, \text{s}Δt=10ms=10×10−3s.

Thus, the induced emf is:E=−dΦdt=−6π×10−810×10−3\mathcal{E} = – \frac{d\Phi}{dt} = – \frac{6\pi \times 10^{-8}}{10 \times 10^{-3}}E=−dtdΦ​=−10×10−36π×10−8​E=−1.88×10−3 V\mathcal{E} = – 1.88 \times 10^{-3} \, \text{V}E=−1.88×10−3V

So, the induced emf is approximately 1.88 mV1.88 \, \text{mV}1.88mV.

23. Two point charges of 10 mC10 \, \text{mC}10mC and 20 mC20 \, \text{mC}20mC are located at points (−4 cm,0,0)(-4 \, \text{cm}, 0, 0)(−4cm,0,0) and (5 cm,0,0)(5 \, \text{cm}, 0, 0)(5cm,0,0) respectively, in a region with electric field E=2Ar\mathbf{E} = 2A rE=2Ar, where A=2×106 N\cdotpC−1m2A = 2 \times 10^6 \, \text{N·C}^{-1} \text{m}^2A=2×106N\cdotpC−1m2 and r\mathbf{r}r is the position vector of the point under consideration. Calculate the electrostatic potential energy of the system.

Answer:

The electrostatic potential energy of a system of point charges is given by:U=kq1q2rU = \frac{k q_1 q_2}{r}U=rkq1​q2​​

Where:

  • k=9×109 N\cdotpm2/C2k = 9 \times 10^9 \, \text{N·m}^2/\text{C}^2k=9×109N\cdotpm2/C2,
  • q1=10 mC=10×10−3 Cq_1 = 10 \, \text{mC} = 10 \times 10^{-3} \, \text{C}q1​=10mC=10×10−3C,
  • q2=20 mC=20×10−3 Cq_2 = 20 \, \text{mC} = 20 \times 10^{-3} \, \text{C}q2​=20mC=20×10−3C,
  • rrr is the distance between the two charges.

Step 1: Calculate the distance between the two charges.

The distance between the charges is the difference in their x-coordinates:r=∣5 cm−(−4 cm)∣=9 cm=0.09 mr = |5 \, \text{cm} – (-4 \, \text{cm})| = 9 \, \text{cm} = 0.09 \, \text{m}r=∣5cm−(−4cm)∣=9cm=0.09m

Step 2: Calculate the electrostatic potential energy.

Now, using the formula for electrostatic potential energy:U=9×109×(10×10−3)×(20×10−3)0.09U = \frac{9 \times 10^9 \times (10 \times 10^{-3}) \times (20 \times 10^{-3})}{0.09}U=0.099×109×(10×10−3)×(20×10−3)​U=9×109×200×10−60.09U = \frac{9 \times 10^9 \times 200 \times 10^{-6}}{0.09}U=0.099×109×200×10−6​U=1.8×1060.09U = \frac{1.8 \times 10^6}{0.09}U=0.091.8×106​U=2×107 JU = 2 \times 10^7 \, \text{J}U=2×107J

So, the electrostatic potential energy of the system is 2×107 J2 \times 10^7 \, \text{J}2×107J.

Summary of Answers:

  1. Question 22: The induced emf in the loop is 1.88 mV1.88 \, \text{mV}1.88mV.
  2. Question 23: The electrostatic potential energy of the system is 2×107 J2 \times 10^7 \, \text{J}2×107J.

24. (a) The radius of a conducting wire AB uniformly decreases from its one end A to another end B. It is connected across a battery. How will:

(i) Electric field change?

  • Answer: The electric field will remain constant from end A to end B.
  • Reason: The electric field EEE depends on the potential difference and the length of the wire. Since the battery provides a constant voltage, the electric field remains uniform along the wire.

(ii) Current density change?

  • Answer: The current density will increase from end A to end B.
  • Reason: Current density J=IAJ = \frac{I}{A}J=AI​, where AAA is the cross-sectional area. As the radius decreases, the area AAA decreases, so the current density increases.

(iii) Mobility of electrons change?

  • Answer: The mobility of electrons will remain constant from end A to end B.
  • Reason: Mobility is a material property and does not depend on the geometry of the wire. It remains the same throughout the wire.

Summary of Answers:

  • Electric field: Constant.
  • Current density: Increases (due to decreasing radius).
  • Mobility of electrons: Constant.

Question

(b) Two large plane sheets P1P_1P1​ and P2P_2P2​ having charge densities +σ+\sigma+σ and −3σ-3\sigma−3σ respectively are arranged parallel to each other as shown in the figure. Find the net electric field (E⃗\vec{E}E) at points A, B, and C.

Answer

To find the net electric field at points A, B, and C due to the two parallel sheets, we use the principle of superposition.

Let:

  • σ\sigmaσ be the surface charge density on sheet P1P_1P1​,
  • −3σ-3\sigma−3σ be the surface charge density on sheet P2P_2P2​,
  • ϵ0\epsilon_0ϵ0​ be the permittivity of free space.

For an infinite plane sheet of charge, the electric field at a point due to a sheet with charge density σ\sigmaσ is given by:E=σ2ϵ0E = \frac{\sigma}{2\epsilon_0}E=2ϵ0​σ​

Where:

  • E1=σ2ϵ0E_1 = \frac{\sigma}{2\epsilon_0}E1​=2ϵ0​σ​ is the electric field due to sheet P1P_1P1​,
  • E2=3σ2ϵ0E_2 = \frac{3\sigma}{2\epsilon_0}E2​=2ϵ0​3σ​ is the electric field due to sheet P2P_2P2​.

Electric Field at Point A (Between the Sheets)

  • At point A, between the two sheets, the electric fields from both sheets will point in the same direction (both away from sheet P1P_1P1​ and towards sheet P2P_2P2​).
  • Therefore, the total electric field at point A is:

EA=E1+E2=σ2ϵ0+3σ2ϵ0=4σ2ϵ0=2σϵ0E_A = E_1 + E_2 = \frac{\sigma}{2\epsilon_0} + \frac{3\sigma}{2\epsilon_0} = \frac{4\sigma}{2\epsilon_0} = \frac{2\sigma}{\epsilon_0}EA​=E1​+E2​=2ϵ0​σ​+2ϵ0​3σ​=2ϵ0​4σ​=ϵ0​2σ​

Electric Field at Point B (Outside the Sheets, on the Side of P1P_1P1​)

  • At point B, the electric field due to sheet P1P_1P1​ will point away from it, and the electric field due to sheet P2P_2P2​ will point towards it. These fields point in opposite directions.
  • Therefore, the total electric field at point B is:

EB=E1−E2=σ2ϵ0−3σ2ϵ0=−2σ2ϵ0=−σϵ0E_B = E_1 – E_2 = \frac{\sigma}{2\epsilon_0} – \frac{3\sigma}{2\epsilon_0} = -\frac{2\sigma}{2\epsilon_0} = -\frac{\sigma}{\epsilon_0}EB​=E1​−E2​=2ϵ0​σ​−2ϵ0​3σ​=−2ϵ0​2σ​=−ϵ0​σ​

So, the electric field at point B is directed towards P2P_2P2​, with magnitude σϵ0\frac{\sigma}{\epsilon_0}ϵ0​σ​.

Electric Field at Point C (Outside the Sheets, on the Side of P2P_2P2​)

  • At point C, the electric field due to sheet P1P_1P1​ will point away from it, and the electric field due to sheet P2P_2P2​ will point towards it. These fields again point in opposite directions.
  • Therefore, the total electric field at point C is:

EC=E2−E1=3σ2ϵ0−σ2ϵ0=2σ2ϵ0=σϵ0E_C = E_2 – E_1 = \frac{3\sigma}{2\epsilon_0} – \frac{\sigma}{2\epsilon_0} = \frac{2\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}EC​=E2​−E1​=2ϵ0​3σ​−2ϵ0​σ​=2ϵ0​2σ​=ϵ0​σ​

So, the electric field at point C is directed towards P2P_2P2​, with magnitude σϵ0\frac{\sigma}{\epsilon_0}ϵ0​σ​.

Summary of Results:

  • At point A (between the sheets):
    EA=2σϵ0E_A = \frac{2\sigma}{\epsilon_0}EA​=ϵ0​2σ​ (directed from P1P_1P1​ to P2P_2P2​).
  • At point B (outside the sheets, on the side of P1P_1P1​):
    EB=−σϵ0E_B = -\frac{\sigma}{\epsilon_0}EB​=−ϵ0​σ​ (directed towards P2P_2P2​).
  • At point C (outside the sheets, on the side of P2P_2P2​):
    EC=σϵ0E_C = \frac{\sigma}{\epsilon_0}EC​=ϵ0​σ​ (directed towards P2P_2P2​).

25. In photoelectric effect experiment:

(a) Variation of photocurrent with collector plate potential for a given surface for different intensities of incident radiation. Do the curves meet at any point? If so, why?

  • Answer:
    • Photocurrent (the current due to emitted photoelectrons) depends on the intensity of the incident radiation and the collector plate potential.
    • Variation with Plate Potential:
      • At low collector plate potentials, the photocurrent increases with the increase in potential until it reaches a saturation point where the current becomes constant.
      • The saturation current represents the maximum number of photoelectrons that can reach the collector. This saturation current is directly proportional to the intensity of the incident radiation.
      • When the collector plate potential is sufficiently negative (beyond the stopping potential), no photoelectrons are able to reach the collector, and the photocurrent becomes zero.
    • For Different Intensities:
      • Higher intensity radiation leads to a higher saturation current, as more electrons are emitted.
      • The curves for different intensities of radiation will show the same shape, but the higher intensity curves will have higher saturation currents.
    • Do the curves meet at any point?
      • Yes, all the curves meet at the stopping potential. This is because at the stopping potential, all the photoelectrons are stopped, and the photocurrent becomes zero for all intensities.
    Conclusion: The curves for different intensities meet at the stopping potential because the photocurrent becomes zero once the electrons are unable to reach the collector.

(b) Variation of photocurrent with intensity of radiation incident on a surface, keeping the frequency and plate potential fixed.

  • Answer:
    • Photocurrent is directly proportional to the intensity of the incident radiation.
    • When the frequency of the radiation and the plate potential are fixed, increasing the intensity of the radiation leads to a proportional increase in the number of photoelectrons emitted. Since photocurrent is the flow of these photoelectrons, it increases as the intensity increases.
    • Conclusion: The photocurrent increases linearly with the intensity of the incident radiation, as long as the frequency is above the threshold frequency for the photoelectric effect.

26. Explain the following, giving proper reasons:

(a) During charging of a capacitor, displacement current exists in the capacitor. But there is no displacement current when it gets fully charged.

  • Answer:
    • Displacement Current is a concept introduced by Maxwell to account for changing electric fields in regions where there is no actual movement of charge (like in the capacitor gap).
    • During charging: As the capacitor charges, the electric field between the plates changes with time, creating a changing electric flux. This changing flux generates a displacement current in the gap between the plates.
    • When fully charged: Once the capacitor is fully charged, the electric field between the plates becomes constant, and no more current flows. The displacement current becomes zero because the electric flux is no longer changing.
    Conclusion: Displacement current exists during charging because the electric field changes with time, but it disappears once the capacitor is fully charged because the electric field becomes constant.

(b) The frequency of microwaves in ovens matches with the resonant frequency of water molecules.

  • Answer:
    • Microwaves are electromagnetic waves with frequencies typically in the range of 2.45 GHz used in microwave ovens.
    • Water molecules have a resonant frequency around 2.45 GHz, which means they absorb energy most efficiently at this frequency. This causes the water molecules to vibrate more, increasing their kinetic energy and generating heat.
    • Reason: The frequency of microwaves is chosen to match the resonant frequency of water molecules because it maximizes the energy transferred to the water, thus efficiently heating the food.
    Conclusion: Microwaves match the resonant frequency of water molecules to efficiently transfer energy and heat the food.

(c) Infrared waves are also known as heat waves.

  • Answer:
    • Infrared radiation (IR) is electromagnetic radiation with wavelengths longer than visible light but shorter than microwaves. It is commonly associated with heat.
    • Reason: Infrared radiation is absorbed by matter and increases the kinetic energy of the particles, causing them to vibrate and thus generate heat. This is why infrared waves are often referred to as heat waves.
    • Examples of infrared radiation include the heat felt from the sun or from a heater, which is primarily infrared radiation.
    Conclusion: Infrared waves are called heat waves because they are absorbed by objects and converted into heat.

Summary of Answers:

  1. (a) The photocurrent variation with collector plate potential for different intensities shows curves that meet at the stopping potential.
  2. (b) The photocurrent increases linearly with the intensity of the incident radiation.
  3. (a) Displacement current exists during charging because the electric field changes, but becomes zero when the capacitor is fully charged (constant electric field).
  4. (b) Microwaves match the resonant frequency of water molecules (2.45 GHz) to efficiently heat the food.
  5. (c) Infrared waves are called heat waves because they increase the kinetic energy of particles and generate heat.

27. A galvanometer is converted into a voltmeter:

(a) The resistance of the galvanometer and

(b) The resistance required to convert it into a voltmeter of range (0 – 2V) volts.

Question: A galvanometer is converted into a voltmeter of range 0−V0 – V0−V volts using a resistor of 9900 Ω9900 \, \Omega9900Ω. If a resistor of 4900 Ω4900 \, \Omega4900Ω is used, the range becomes half, i.e., 0−V20 – \frac{V}{2}0−2V​ volts. Calculate: (a) the resistance of the galvanometer. (b) the resistance required to convert it into a voltmeter of range 0−2V0 – 2V0−2V volts.

Answer:

(a) The resistance of the galvanometer:

  • Let the resistance of the galvanometer be GGG.
  • When the resistor is 9900 Ω9900 \, \Omega9900Ω, the total resistance becomes G+9900G + 9900G+9900 and the range is VVV.
  • When the resistor is 4900 Ω4900 \, \Omega4900Ω, the total resistance becomes G+4900G + 4900G+4900 and the range is V2\frac{V}{2}2V​.

By using the proportionality of voltage to resistance:VV2=G+9900G+4900\frac{V}{\frac{V}{2}} = \frac{G + 9900}{G + 4900}2V​V​=G+4900G+9900​

Simplifying:2=G+9900G+49002 = \frac{G + 9900}{G + 4900}2=G+4900G+9900​

Cross-multiply:2(G+4900)=G+99002(G + 4900) = G + 99002(G+4900)=G+9900 2G+9800=G+99002G + 9800 = G + 99002G+9800=G+9900 2G−G=9900−98002G – G = 9900 – 98002G−G=9900−9800 G=100 ΩG = 100 \, \OmegaG=100Ω

Conclusion: The resistance of the galvanometer is 100 Ω100 \, \Omega100Ω.

(b) The resistance required to convert it into a voltmeter of range 0−2V0 – 2V0−2V volts:

  • Using the same formula:

2VV=G+RnewG+9900\frac{2V}{V} = \frac{G + R_{\text{new}}}{G + 9900}V2V​=G+9900G+Rnew​​

Substitute G=100 ΩG = 100 \, \OmegaG=100Ω:2=100+Rnew100002 = \frac{100 + R_{\text{new}}}{10000}2=10000100+Rnew​​

Multiply both sides by 10000:20000=100+Rnew20000 = 100 + R_{\text{new}}20000=100+Rnew​ Rnew=20000−100R_{\text{new}} = 20000 – 100Rnew​=20000−100 Rnew=19900 ΩR_{\text{new}} = 19900 \, \OmegaRnew​=19900Ω

Conclusion: The resistance required to convert it into a voltmeter of range 0−2V0 – 2V0−2V is 19900 Ω19900 \, \Omega19900Ω.

28. (a) Light incident on a surface separating air from a denser medium:

(a) In which one of the two media (A or B) will light travel faster and why?

Question: A ray of light is incident on a surface separating air from a denser medium AAA of refractive index m1m_1m1​. It is then made incident on the parallel surface of another medium BBB of refractive index m2m_2m2​ at the same angle of incidence. If the angle of refraction in the two media are 30° and 35° respectively, then in which one of the two media (A or B) will light travel faster and why?

Answer:

  • The speed of light in a medium is inversely related to the refractive index:v=cmv = \frac{c}{m}v=mc​where ccc is the speed of light in a vacuum and mmm is the refractive index.
  • Snell’s Law relates the refractive index and the angle of refraction:m1sin⁡i=m2sin⁡rm_1 \sin i = m_2 \sin rm1​sini=m2​sinrwhere iii is the angle of incidence and rrr is the angle of refraction.
  • Given:
    • Angle of refraction in medium AAA = 30° (sin⁡30°=0.5\sin 30° = 0.5sin30°=0.5).
    • Angle of refraction in medium BBB = 35° (sin⁡35°≈0.5736\sin 35° \approx 0.5736sin35°≈0.5736).
  • The refractive index is inversely proportional to the speed of light. Since the angle of refraction in medium B is greater than in medium A, it suggests that medium B has a larger refractive index, meaning light travels slower in medium B.
  • Conclusion: The light will travel faster in medium A because its refractive index m1m_1m1​ is smaller than that of medium B.

(b) The intensity of two interfering waves in Young’s double slit experiment:

Question: The intensity of the two interfering waves in Young’s double slit experiment is I0I_0I0​ each. Find the intensity at a point on the screen where the path difference between the interfering waves is: (i) 2λ2\lambda2λ (ii) 3λ3\lambda3λ

Answer:

  • The intensity in an interference pattern is given by:I=I0(1+cos⁡(Δϕ))I = I_0 \left( 1 + \cos(\Delta \phi) \right)I=I0​(1+cos(Δϕ))where Δϕ\Delta \phiΔϕ is the phase difference.
  • Path difference and phase difference are related by:Δϕ=2πλΔx\Delta \phi = \frac{2\pi}{\lambda} \Delta xΔϕ=λ2π​Δxwhere Δx\Delta xΔx is the path difference and λ\lambdaλ is the wavelength.
  • (i) Path difference = 2λ2\lambda2λ:
    • For Δx=2λ\Delta x = 2\lambdaΔx=2λ, the phase difference is Δϕ=2π\Delta \phi = 2\piΔϕ=2π.
    • The intensity is: I=I0(1+cos⁡(2π))=I0(1+1)=2I0I = I_0 \left( 1 + \cos(2\pi) \right) = I_0 (1 + 1) = 2I_0I=I0​(1+cos(2π))=I0​(1+1)=2I0​
    • Intensity = 2I02I_02I0​.
  • (ii) Path difference = 3λ3\lambda3λ:
    • For Δx=3λ\Delta x = 3\lambdaΔx=3λ, the phase difference is Δϕ=3×2π=6π\Delta \phi = 3 \times 2\pi = 6\piΔϕ=3×2π=6π.
    • The intensity is: I=I0(1+cos⁡(6π))=I0(1+1)=2I0I = I_0 \left( 1 + \cos(6\pi) \right) = I_0 (1 + 1) = 2I_0I=I0​(1+cos(6π))=I0​(1+1)=2I0​
    • Intensity = 2I02I_02I0​.

Conclusion:

  • (i) Intensity at path difference 2λ2\lambda2λ is 2I02I_02I0​.
  • (ii) Intensity at path difference 3λ3\lambda3λ is 2I02I_02I0​.

Summary:

  1. 27 (a) The resistance of the galvanometer is 100 Ω100 \, \Omega100Ω.
  2. 27 (b) The resistance required to convert it into a voltmeter of range 0−2V0 – 2V0−2V is 19900 Ω19900 \, \Omega19900Ω.
  3. 28 (a) Light will travel faster in medium A because its refractive index is smaller than that of medium B.
  4. 28 (b) The intensity at path differences 2λ2\lambda2λ and 3λ3\lambda3λ is 2I02I_02I0​ in both cases.

29. Case Study on Electric and Magnetic Dipoles:

(i) Two identical electric dipoles, each consisting of charges –q and +q separated by distance d, are arranged in the x-y plane such that their negative charges lie at the origin O and positive charges lie at points (d, 0) and (0, d) respectively. The net dipole moment of the system is:

Answer:

For the two dipoles:

  • The dipole moment of the first dipole is:
    p1⃗=q⋅di^\vec{p_1} = q \cdot d \hat{i} p1​​=q⋅di^
  • The dipole moment of the second dipole is:
    p2⃗=q⋅dj^\vec{p_2} = q \cdot d \hat{j} p2​​=q⋅dj^​

The net dipole moment of the system is:pnet⃗=p1⃗+p2⃗=qdi^+qdj^=qd(i^+j^)\vec{p_{\text{net}}} = \vec{p_1} + \vec{p_2} = qd \hat{i} + qd \hat{j} = qd (\hat{i} + \hat{j})pnet​​=p1​​+p2​​=qdi^+qdj^​=qd(i^+j^​)

Thus, the net dipole moment of the system is:
Answer: qd(i^+j^)qd (\hat{i} + \hat{j})qd(i^+j^​)

(ii) E₁ and E₂ are magnitudes of electric field due to a dipole, consisting of charges –q and +q separated by distance 2a, at points r (>> a) (1) on its axis, and (2) on the equatorial plane, respectively. Then E1E2\frac{E_1}{E_2}E2​E1​​ is:

Answer:

  1. Electric field on the axis of the dipole at a distance rrr (where r≫ar \gg ar≫a) is:

E1=14πϵ0⋅2pr3E_1 = \frac{1}{4\pi \epsilon_0} \cdot \frac{2p}{r^3}E1​=4πϵ0​1​⋅r32p​

  1. Electric field on the equatorial plane at the same distance is:

E2=14πϵ0⋅pr3E_2 = \frac{1}{4\pi \epsilon_0} \cdot \frac{p}{r^3}E2​=4πϵ0​1​⋅r3p​

Thus, the ratio E1E2\frac{E_1}{E_2}E2​E1​​ is:E1E2=2p/r3p/r3=2\frac{E_1}{E_2} = \frac{2p / r^3}{p / r^3} = 2E2​E1​​=p/r32p/r3​=2

Answer: The ratio of electric field magnitudes is 2.

(iii) An electric dipole of dipole moment 5.0×10−8 Cm5.0 \times 10^{-8} \, \text{Cm}5.0×10−8Cm is placed in a region where an electric field of magnitude 1.0×103 N/C1.0 \times 10^3 \, \text{N/C}1.0×103N/C acts at a given instant. At that instant the electric field E⃗\vec{E}E is inclined at an angle of 30° to dipole moment P⃗\vec{P}P. The magnitude of torque acting on the dipole, at that instant is:

Answer:

The torque τ\tauτ on a dipole in an electric field is given by:τ=pEsin⁡θ\tau = pE \sin \thetaτ=pEsinθ

Where:

  • p=5.0×10−8 Cmp = 5.0 \times 10^{-8} \, \text{Cm}p=5.0×10−8Cm is the dipole moment.
  • E=1.0×103 N/CE = 1.0 \times 10^3 \, \text{N/C}E=1.0×103N/C is the electric field.
  • θ=30∘\theta = 30^\circθ=30∘ is the angle between the dipole moment and the electric field.

Substituting the values:τ=(5.0×10−8 Cm)×(1.0×103 N/C)×sin⁡30∘\tau = (5.0 \times 10^{-8} \, \text{Cm}) \times (1.0 \times 10^3 \, \text{N/C}) \times \sin 30^\circτ=(5.0×10−8Cm)×(1.0×103N/C)×sin30∘ τ=(5.0×10−8)×(1.0×103)×12=2.5×10−5 Nm\tau = (5.0 \times 10^{-8}) \times (1.0 \times 10^3) \times \frac{1}{2} = 2.5 \times 10^{-5} \, \text{Nm}τ=(5.0×10−8)×(1.0×103)×21​=2.5×10−5Nm

Answer: The magnitude of the torque is 2.5×10−5 Nm2.5 \times 10^{-5} \, \text{Nm}2.5×10−5Nm.

(iv) (a) An electron is revolving with speed vvv around the proton in a hydrogen atom, in a circular orbit of radius rrr. The magnitude of magnetic dipole moment of the electron is:

Answer:

The magnetic dipole moment of the electron is given by:μ=e⋅v⋅r2\mu = \frac{e \cdot v \cdot r}{2}μ=2e⋅v⋅r​

Where:

  • eee is the charge of the electron.
  • vvv is the velocity of the electron.
  • rrr is the radius of the orbit.

Thus, the magnitude of the magnetic dipole moment of the electron in the hydrogen atom is:μ=12evr\mu = \frac{1}{2} evrμ=21​evr

Answer: The magnetic dipole moment of the electron is 12evr\frac{1}{2} evr21​evr.

(iv) (b) A square loop of side 5.0 cm carries a current of 2.0 A. The magnitude of magnetic dipole moment associated with the loop is:

Answer:

For a square loop of side aaa carrying current III, the magnetic dipole moment is given by:μ=I⋅A\mu = I \cdot Aμ=I⋅A

Where:

  • A=a2A = a^2A=a2 is the area of the square loop.

Given:

  • Side of the square loop a=5.0 cm=0.05 ma = 5.0 \, \text{cm} = 0.05 \, \text{m}a=5.0cm=0.05m
  • Current I=2.0 AI = 2.0 \, \text{A}I=2.0A

The area is:A=(0.05)2=2.5×10−3 m2A = (0.05)^2 = 2.5 \times 10^{-3} \, \text{m}^2A=(0.05)2=2.5×10−3m2

Now, the magnetic dipole moment is:μ=(2.0)×(2.5×10−3)=5.0×10−3 Am2\mu = (2.0) \times (2.5 \times 10^{-3}) = 5.0 \times 10^{-3} \, \text{Am}^2μ=(2.0)×(2.5×10−3)=5.0×10−3Am2

Answer: The magnetic dipole moment is 5.0×10−3 Am25.0 \times 10^{-3} \, \text{Am}^25.0×10−3Am2.

Q30. The process of converting AC into DC is called rectification and the device used is called a rectifier. When an AC signal is fed to a junction diode during the positive half-cycle, the diode is forward biased and current flows through it. During the negative half-cycle, the diode is reverse biased and it does not conduct. Thus, the AC signal is rectified. The p-n junction diodes can be used as half-wave and full-wave rectifiers.

(i) Which bulb/bulbs will glow in the given circuit?

Answer:

To determine which bulb or bulbs will glow in the given circuit, let’s consider the working of a rectifier circuit:

  1. Half-Wave Rectifier:
    • In a half-wave rectifier, the diode only allows current to flow during the positive half-cycle of the AC input. The current is blocked during the negative half-cycle. As a result, the bulb will glow only during the positive half-cycle and will remain off during the negative half-cycle. Therefore, the bulb connected to the output of a half-wave rectifier will glow intermittently, only during the positive half of the AC cycle.
  2. Full-Wave Rectifier:
    • In a full-wave rectifier, current flows through the load (e.g., a bulb) during both the positive and negative half-cycles, as both halves of the AC signal are used to provide current. In this case, the bulb will glow continuously because the current is always flowing in the same direction, regardless of whether it’s the positive or negative half of the AC signal.

Conclusion:

In the case of a full-wave rectifier, the bulb will glow continuously, as current flows during both half-cycles.

In the case of a half-wave rectifier, the bulb will glow intermittently during the positive half-cycle.

Question:

(ii) (a) A full-wave rectifier circuit is shown in the figure. The contribution in the output waveform from junction diode D1 is:

Answer:

In a full-wave rectifier, the AC input is passed through a bridge of diodes, typically consisting of four diodes arranged in a specific configuration. When analyzing the contribution from a specific diode (in this case, D1), let’s break down how the full-wave rectifier works:

Working of Full-Wave Rectifier:

  • In a full-wave rectifier, the AC signal is rectified during both the positive and negative halves of the input waveform.
  • During the positive half-cycle of the AC input, diode D1 becomes forward-biased, allowing current to flow through it. Diode D2 is also forward-biased, conducting current to the load.
  • During the negative half-cycle of the AC input, D1 becomes reverse-biased and no current flows through it. However, D3 becomes forward-biased during this cycle, allowing current to flow through it, while D4 also conducts.

Contribution of D1 to the Output Waveform:

  • During the positive half-cycle, D1 conducts and contributes to the output waveform by allowing current to pass through the load resistor. The voltage waveform across the load during this period is in the same direction as the AC input voltage.
  • During the negative half-cycle, D1 does not conduct because it is reverse-biased. Instead, D3 and D4 conduct, but D1 does not contribute directly to the output during this half-cycle.

Conclusion:

  • The contribution of D1 to the output waveform occurs only during the positive half-cycle of the input AC signal. In this period, the output waveform has the same polarity as the input, with D1 allowing current to pass through to the load.

Question:

(b) The output in a half-wave rectifier is:
1
(A) unidirectional without ripple
(B) steady and continuous
(C) unidirectional with ripple
(D) steady but discontinuous

Answer:

The output of a half-wave rectifier is unidirectional with ripple.

Here’s why:

  • In a half-wave rectifier, the AC signal is allowed to pass only during the positive half-cycle of the input. During the negative half-cycle, the diode is reverse biased, and no current flows through the circuit.
  • This results in a unidirectional output because the current flows only in one direction (during the positive half-cycle).
  • However, since the output is only present during half of the input cycle (the positive half), there is a rippling effect where the output voltage has a series of pulses corresponding to the positive half-cycles of the AC input. This causes the output to be pulsating (i.e., with ripple).

Thus, the correct answer is:
(C) unidirectional with ripple.

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Question:

(iii) In a p-n junction diode, the majority charge carriers on p-side and on n-side are, respectively:
1
(A) electrons, electrons
(B) electrons, holes
(C) holes, holes
(D) holes, electrons

Answer:

In a p-n junction diode, the majority charge carriers on the p-side are holes, and on the n-side, the majority charge carriers are electrons.

  • The p-side of the diode is doped with acceptor impurities, creating an abundance of holes (positive charge carriers).
  • The n-side is doped with donor impurities, creating an abundance of electrons (negative charge carriers).

Thus, the correct answer is:
(D) holes, electrons.

Question:

(iv) If the frequency of the half-wave rectifier is 50 Hz, the frequency of the full-wave rectifier is:
1
(A) 25 Hz
(B) 50 Hz
(C) 100 Hz
(D) 200 Hz

Answer:

In a half-wave rectifier, the output frequency is the same as the input AC frequency. So, if the input AC frequency is 50 Hz, the output frequency of the half-wave rectifier will also be 50 Hz.

In a full-wave rectifier, the output frequency is twice the input AC frequency because both halves of the AC cycle (positive and negative) are used to produce output.

So, if the input AC frequency is 50 Hz, the output frequency of the full-wave rectifier will be 100 Hz.

Thus, the correct answer is:
(C) 100 Hz.

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Question:

31. (a)

(i) What are matter waves? A particle of mass m and charge q is accelerated from rest through a potential difference V. Obtain an expression for the de Broglie wavelength associated with the particle.

(ii) Monochromatic light of frequency 5.0 × 10¹⁴ Hz is produced by a source of power output 3.315 mW. Calculate:
(1) Energy of the photon in the beam
(2) Number of photons emitted per second by the source

OR

(b)

(i) State Bohr’s postulates and derive an expression for the energy of an electron in the nth orbit in Bohr’s model of the hydrogen atom.
(ii) Calculate binding energy per nucleon (in MeV) of ₁₂₆C.
Given:

  • m₁₂₆C = 12.000000 u
  • m₁₀n = 1.008665 u
  • m₁₁H = 1.007825 u

Answer:

(a) (i) Matter Waves and de Broglie Wavelength:

  • Matter waves are waves associated with particles having mass and charge. According to de Broglie, any moving particle behaves like a wave, and its wavelength is given by:λ=hp\lambda = \frac{h}{p}λ=ph​where:
    • h is Planck’s constant (6.626 × 10⁻³⁴ J·s)
    • p is the momentum of the particle
    For a particle with mass m and velocity v, the momentum p is:p=mvp = mvp=mvWhen a particle is accelerated through a potential difference V, the kinetic energy gained by the particle is equal to the work done on it, which is:12mv2=qV\frac{1}{2} m v^2 = qV21​mv2=qVSolving for velocity v:v=2qVmv = \sqrt{\frac{2qV}{m}}v=m2qV​​Substituting v into the expression for momentum:p=m⋅2qVm=2mqVp = m \cdot \sqrt{\frac{2qV}{m}} = \sqrt{2mqV}p=m⋅m2qV​​=2mqV​Now, the de Broglie wavelength λ is:λ=hp=h2mqV\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}λ=ph​=2mqV​h​Therefore, the de Broglie wavelength associated with a particle accelerated through a potential difference V is:λ=h2mqV\lambda = \frac{h}{\sqrt{2mqV}}λ=2mqV​h​

(a) (ii) Energy of the Photon and Number of Photons Emitted Per Second:

  1. Energy of the Photon:

The energy of a photon is related to its frequency f by the equation:E=h⋅fE = h \cdot fE=h⋅f

Given:

  • Frequency f = 5.0 × 10¹⁴ Hz
  • Planck’s constant h = 6.626 × 10⁻³⁴ J·s

So, the energy of the photon is:E=(6.626×10−34)×(5.0×1014)=3.313×10−19 JE = (6.626 \times 10^{-34}) \times (5.0 \times 10^{14}) = 3.313 \times 10^{-19} \, \text{J}E=(6.626×10−34)×(5.0×1014)=3.313×10−19J

  1. Number of Photons Emitted Per Second:

The power output of the source is given as 3.315 mW, which is equivalent to 3.315 × 10⁻³ W.

The total energy emitted per second (i.e., the power) is the product of the energy of one photon and the number of photons emitted per second:Power=E×Number of photons per second\text{Power} = E \times \text{Number of photons per second}Power=E×Number of photons per second

Let the number of photons emitted per second be N. Then,3.315×10−3=3.313×10−19×N3.315 \times 10^{-3} = 3.313 \times 10^{-19} \times N3.315×10−3=3.313×10−19×N

Solving for N:N=3.315×10−33.313×10−19=1.0×1016 photons/secN = \frac{3.315 \times 10^{-3}}{3.313 \times 10^{-19}} = 1.0 \times 10^{16} \, \text{photons/sec}N=3.313×10−193.315×10−3​=1.0×1016photons/sec

OR

(b) (i) Bohr’s Postulates and Energy of the Electron in the nth Orbit:

Bohr’s postulates:

  1. The electron in the hydrogen atom revolves around the nucleus in circular orbits without radiating energy.
  2. The electron can only occupy certain allowed orbits, called stationary orbits, where the angular momentum of the electron is quantized and given by:L=nℏ=n(h2π)L = n \hbar = n \left( \frac{h}{2\pi} \right)L=nℏ=n(2πh​)where n is a positive integer, and h is Planck’s constant.
  3. The electron emits or absorbs radiation only when it jumps between allowed orbits.

From Bohr’s model, the radius of the nth orbit is given by:rn=n2h2ε0πme2r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2}rn​=πme2n2h2ε0​​

The energy of the electron in the nth orbit is:En=−ke22rnE_n = -\frac{k e^2}{2r_n}En​=−2rn​ke2​

Substituting the expression for r_n, the energy becomes:En=−me42h2ε02n2E_n = -\frac{m e^4}{2 h^2 \varepsilon_0^2 n^2}En​=−2h2ε02​n2me4​

The total energy is the sum of kinetic and potential energies. The negative sign indicates that the electron is bound to the nucleus.

(b) (ii) Binding Energy per Nucleon of Carbon-12:

The binding energy of a nucleus is the energy required to separate all the nucleons (protons and neutrons) in the nucleus. The binding energy per nucleon is given by:Binding energy per nucleon=Total binding energyNumber of nucleons\text{Binding energy per nucleon} = \frac{\text{Total binding energy}}{\text{Number of nucleons}}Binding energy per nucleon=Number of nucleonsTotal binding energy​

The total binding energy can be calculated using the mass defect, which is the difference between the mass of the separate nucleons and the mass of the nucleus.

Given:

  • Mass of C₁₂ = 12.000000 u
  • Mass of 1 neutron = 1.008665 u
  • Mass of 1 proton = 1.007825 u
  • Number of nucleons (protons + neutrons) in C₁₂ = 12

The mass defect Δm is:Δm=[12×Mass of proton]+[12×Mass of neutron]−Mass of C₁₂\Delta m = [12 \times \text{Mass of proton}] + [12 \times \text{Mass of neutron}] – \text{Mass of } \text{C₁₂}Δm=[12×Mass of proton]+[12×Mass of neutron]−Mass of C₁₂

Then, the binding energy is:Ebinding=Δm×c2E_{\text{binding}} = \Delta m \times c^2Ebinding​=Δm×c2

Finally, the binding energy per nucleon is:Binding energy per nucleon=Ebinding12\text{Binding energy per nucleon} = \frac{E_{\text{binding}}}{12}Binding energy per nucleon=12Ebinding​​

Substitute the values into this formula to obtain the binding energy per nucleon in MeV

  1. (a) (i) With the help of a labelled diagram, explain the working of an AC generator. Obtain the expression for the emf induced at an instant ‘t’.

Answer:

Working of an AC Generator:

An AC generator (also called a dynamo) is a device that converts mechanical energy into electrical energy through the principle of electromagnetic induction. It works by rotating a coil in a magnetic field, which induces an alternating current (AC) in the coil.

The basic working principle of an AC generator is Faraday’s Law of Induction, which states that a change in magnetic flux through a coil induces an electromotive force (emf) in the coil.

Key Components of an AC Generator:

  1. Armature: A coil of wire, usually wound on a soft iron core, which rotates in the magnetic field.
  2. Magnetic Field (B): A permanent magnet or electromagnet that provides the magnetic field in which the armature rotates.
  3. Slip Rings: These provide continuous electrical contact between the rotating armature and the external circuit.
  4. Brushes: Carbon brushes make contact with the slip rings to transfer the induced current from the rotating armature to the external circuit.

Working Explanation:

  1. Rotation of the Armature: The armature (the coil of wire) is rotated by mechanical means (like a turbine or engine) inside the magnetic field. The rotation occurs at a constant angular velocity ω\omegaω.
  2. Magnetic Flux: As the armature rotates, the magnetic flux through the coil changes continuously. The angle between the magnetic field and the plane of the coil changes, leading to a time-varying magnetic flux.
  3. Induced emf: According to Faraday’s law of induction, a time-varying magnetic flux induces an emf (electromotive force) in the coil. The amount of induced emf depends on how fast the magnetic flux changes, which is related to the rate of rotation of the coil.
  4. Slip Rings and Brushes: The slip rings ensure continuous contact between the rotating coil and the external circuit. This allows the emf generated in the coil to be transferred as alternating current (AC) to the external circuit.

Expression for the Induced emf:

Let:

  • BBB be the magnetic field strength,
  • AAA be the area of the coil,
  • NNN be the number of turns in the coil,
  • ω\omegaω be the angular velocity of the rotating coil,
  • θ\thetaθ be the angle between the magnetic field and the normal to the plane of the coil at any instant.

The magnetic flux through the coil at any instant is given by:Φ=BAcos⁡(θ)\Phi = B A \cos(\theta)Φ=BAcos(θ)

Where:

  • AAA is the area of the coil,
  • θ=ωt\theta = \omega tθ=ωt is the angle between the normal to the coil and the magnetic field at time ttt.

The induced emf E\mathcal{E}E is the negative rate of change of magnetic flux:E=−dΦdt\mathcal{E} = -\frac{d\Phi}{dt}E=−dtdΦ​

Substituting the expression for magnetic flux:E=−ddt(BAcos⁡(ωt))\mathcal{E} = -\frac{d}{dt} \left( B A \cos(\omega t) \right)E=−dtd​(BAcos(ωt))

Taking the derivative:E=BAωsin⁡(ωt)\mathcal{E} = B A \omega \sin(\omega t)E=BAωsin(ωt)

Thus, the induced emf is:E=E0sin⁡(ωt)\mathcal{E} = \mathcal{E}_0 \sin(\omega t)E=E0​sin(ωt)

Where E0=BAω\mathcal{E}_0 = B A \omegaE0​=BAω is the maximum (peak) emf.

Conclusion:

The induced emf in an AC generator varies sinusoidally with time, given by:E=E0sin⁡(ωt)\mathcal{E} = \mathcal{E}_0 \sin(\omega t)E=E0​sin(ωt)

Where E0\mathcal{E}_0E0​ is the peak emf, and ω\omegaω is the angular frequency of rotation of the coil.

  • The frequency of the generated emf is directly related to the angular velocity ω\omegaω.
  • The peak emf depends on the strength of the magnetic field, the area of the coil, the number of turns in the coil, and the angular velocity of rotation.

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33. (a)

(i) What are the two main considerations for designing the objective and eyepiece lenses of an astronomical telescope?

  • Objective Lens:
    • Large aperture (to gather more light).
    • Long focal length (to form a sharp image of distant objects).
  • Eyepiece Lens:
    • Short focal length (to magnify the image formed by the objective lens).
    • Designed to make the final image at infinity for comfortable viewing.

Expression for Magnifying Power of the Telescope when Final Image is Formed at Infinity:M=fobjectivefeyepieceM = \frac{f_{\text{objective}}}{f_{\text{eyepiece}}}M=feyepiece​fobjective​​

Where:

  • fobjectivef_{\text{objective}}fobjective​ is the focal length of the objective lens.
  • feyepiecef_{\text{eyepiece}}feyepiece​ is the focal length of the eyepiece lens.

(ii) A ray of light is incident at an angle of 45° at one face of an equilateral triangular prism and passes symmetrically through the prism. Calculate:

(1) The Angle of Deviation Produced by the Prism:

The formula for angle of deviation δ\deltaδ for an equilateral prism:δ=A(n−1n)\delta = A \left( \frac{n – 1}{n} \right)δ=A(nn−1​)

Where A=60∘A = 60^\circA=60∘ (the angle of the prism), and nnn is the refractive index.

(2) The Refractive Index of the Material of the Prism:

Using Snell’s Law, the refractive index nnn of the material of the prism can be calculated:n=sin⁡(A+δ2)sin⁡(A2)n = \frac{\sin \left( \frac{A + \delta}{2} \right)}{\sin \left( \frac{A}{2} \right)}n=sin(2A​)sin(2A+δ​)​

Where A=60∘A = 60^\circA=60∘ and δ\deltaδ is the angle of deviation.

OR

(b)

(i) Describe a Simple Activity to Observe Diffraction Pattern Due to a Single Slit:

  • Objective: Observe diffraction patterns produced by a single slit.
  • Procedure:
    • Use a monochromatic light source (like a laser).
    • Pass light through a single slit and project it onto a screen.
    • Observe the central bright fringe (main maxima) and surrounding dark and bright fringes (minima and secondary maxima).
    • The diffraction pattern is due to interference of light passing through the slit.
    The angular position of minima is given by:asin⁡θ=mλa \sin \theta = m \lambdaasinθ=mλWhere aaa is the slit width, θ\thetaθ is the angle of diffraction, mmm is the order of minima, and λ\lambdaλ is the wavelength.

(ii) Refractive Index of the Liquid in the Experiment:

  • Given:
    • Focal length with liquid f1=45.0f_1 = 45.0f1​=45.0 cm.
    • Focal length without liquid f2=30.0f_2 = 30.0f2​=30.0 cm.
    • Refractive index of lens material n1=1.50n_1 = 1.50n1​=1.50.

The refractive index of the liquid n2n_2n2​ can be calculated using the formula:n2=f1f2×(n1−1)+1n_2 = \frac{f_1}{f_2} \times (n_1 – 1) + 1n2​=f2​f1​​×(n1​−1)+1

This formula takes into account the focal length change caused by the liquid.

Answer Summary:

  • (a) (i):
    • Objective Lens: Large aperture, long focal length.
    • Eyepiece Lens: Short focal length, final image at infinity.
    • Magnifying Power: M=fobjectivefeyepieceM = \frac{f_{\text{objective}}}{f_{\text{eyepiece}}}M=feyepiece​fobjective​​.
  • (ii) (1):
    • Angle of Deviation δ=A(n−1n)\delta = A \left( \frac{n – 1}{n} \right)δ=A(nn−1​), where A=60∘A = 60^\circA=60∘.
    • Use the refractive index to find δ\deltaδ.
  • (ii) (2):
    • Refractive Index: n=sin⁡(A+δ2)sin⁡(A2)n = \frac{\sin \left( \frac{A + \delta}{2} \right)}{\sin \left( \frac{A}{2} \right)}n=sin(2A​)sin(2A+δ​)​.

OR

  • (b) (i):
    • Activity: Observe diffraction pattern from a single slit.
    • Formula: asin⁡θ=mλa \sin \theta = m \lambdaasinθ=mλ.
  • (ii):
    • Refractive Index of Liquid: n2=f1f2×(n1−1)+1n_2 = \frac{f_1}{f_2} \times (n_1 – 1) + 1n2​=f2​f1​​×(n1​−1)+1.