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Question #43134

Difficulty: 8/10 | 🟢 Free

The Question

A point object OO is fixed to the floor of a train car. A convex lens of focal length f=20 cmf = 20 \text{ cm} and mass mm is suspended from the roof of the train car by two parallel massless rigid rods of length l=50 cml = 50 \text{ cm}, forming a parallelogram linkage. This ensures the principal axis of the lens always remains horizontal. When the train is at rest, the lens hangs vertically, and the object OO lies exactly on the principal axis at a distance of u0=60 cmu_0 = 60 \text{ cm} to the left of the lens. The train then accelerates horizontally to the right with a constant acceleration a=34ga = \frac{3}{4}g (where g=10 m/s2g = 10 \text{ m/s}^2). Assuming the lens reaches its new equilibrium position without oscillating, what is the magnitude of the displacement of the image of object OO relative to the train car, compared to its initial position when the train was at rest?

Options

  • A.

    30 cm30 \text{ cm}

  • B.

    20 cm20 \text{ cm}

  • C.

    10 cm10 \text{ cm}

  • D.

    0 cm0 \text{ cm}

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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

Newton’s Form of Lens Equation|Frame of Reference|Pseudo Force Directionality

Why This Question Is Tricky

Common Mistakes

This question tests the application of optical principles within a non-inertial reference frame, specifically targeting 'pseudo-force directionality' and coordinate transformations.

Students often fall into multiple traps: ignoring the pseudo-force entirely, applying it in the wrong direction, or failing to account for the vertical displacement of the focal plane caused by the pendulum's circular arc.

A common critical error is calculating the transverse image height relative to the shifted principal axis (20 cm20\text{ cm}) but forgetting to add the lens's own vertical shift (10 cm10\text{ cm}) to find the true displacement in the laboratory frame.

The problem elegantly demonstrates the decoupling of the frame's coordinate transformation from Newton's Lens Equation (x1x2=f2x_1 x_2 = f^2), revealing that the longitudinal image position remains invariant due to optical reversibility, leaving only a pure vertical displacement.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterRay Optics and Optical Instruments - Class 12
ConceptNewton’s Form of Lens Equation
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree