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Question #43894

Difficulty: 8/10 | 🟢 Free

The Question

An RL circuit consists of a battery of EMF V0V_0, a switch SS, a resistor RR, and an inductor LL in series. A long cylindrical capacitor with inner radius aa and outer radius bb is connected in parallel with the inductor. The capacitance is negligible and does not affect the RL circuit's transient current. A small biconvex electro-optic lens, with both radii of curvature equal to RcR_c, is placed between the cylindrical plates at a radial distance r0r_0 from the central axis. The refractive index of the lens material varies with the local electric field magnitude EE as n(t)=n0+βE(t)n(t) = n_0 + \beta E(t), where n0=1.5n_0 = 1.5 and β\beta is a positive constant. Given that βV0r0ln⁡(b/a)=0.5\frac{\beta V_0}{r_0 \ln(b/a)} = 0.5. At t=0t=0, the switch SS is closed. A stationary point object is placed on the principal axis at a distance 3Rc3 R_c in front of the lens. Assuming the lens is small enough that the electric field is uniform across it, find the speed of the image at time t=LRln⁡2t = \frac{L}{R} \ln 2.

Options

  • A.

    1849RcRL\frac{18}{49} \frac{R_c R}{L}

  • B.

    8441RcRL\frac{8}{441} \frac{R_c R}{L}

  • C.

    29RcRL\frac{2}{9} \frac{R_c R}{L}

  • D.

    3625RcRL\frac{36}{25} \frac{R_c R}{L}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Optical Instrument Magnification|Electric Field Symmetry|Transient Current in RL Circuit

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of electromagnetic transients, Gauss's Law in cylindrical symmetry, and dynamic optics.

The primary 'Lethal Collision' occurs when students attempt to find the velocity of the image.

A common misconception is blindly applying the standard kinematic magnification formula vimage=m2vobjectv_{image} = m^2 v_{object} or assuming dvdt=m2dfdt\frac{dv}{dt} = m^2 \frac{df}{dt}.

However, differentiating the thin lens formula 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f} with a stationary object (u=constu = \text{const}) and a time-varying focal length yields dvdt=(v/f)2dfdt\frac{dv}{dt} = (v/f)^2 \frac{df}{dt}.

Using m2m^2 instead of (v/f)2(v/f)^2 leads directly to Trap Option B.

Additionally, the problem requires correctly linking the inductor's transient voltage to the electric field via Gauss's Law (E∝1/rE \propto 1/r) to find the dynamic refractive index.

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Question Information

SubjectPhysics
ChapterRay Optics and Optical Instruments - Class 12
ConceptOptical Instrument Magnification
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree