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Question #43550

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The Question

A compound optical system consists of a fixed objective lens of focal length $f_o$ and an eyepiece of focal length $f_e$ . The eyepiece, having mass $m$ , is mounted at the lower end of a light rigid rod of length $R$ pivoted at its upper end, forming a simple pendulum that swings in a vertical plane containing the optical axis. The optical axis is horizontal and passes through the eyepiece's equilibrium position. The equilibrium distance between the lenses is $L_0 = \frac{3}{2}f_o + \frac{3}{2}f_e$ . A point object is placed on the optical axis at a distance $3f_o$ in front of the objective lens. A small horizontal impulse $J$ is applied to the eyepiece along the optical axis. Assuming small angle oscillations and letting $g$ be the acceleration due to gravity, what is the maximum displacement of the final image from its equilibrium position relative to the ground?

Options

A.
$3 \frac{J}{m} \sqrt{\frac{R}{g}}$
B.
$4 \frac{J}{m} \sqrt{\frac{R}{g}}$
C.
$5 \frac{J}{m} \sqrt{\frac{R}{g}}$
D.
$\frac{J}{m} \sqrt{\frac{R}{g}}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Microscope Tube Length|Equilibrium Stability|Small Angle Approximation

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of mechanical oscillations and geometric optics, specifically targeting the trap of reference frames in longitudinal magnification.

Students frequently memorize the longitudinal magnification formula $\Delta v = m^2 \Delta u$ but misapply it by failing to account for the fact that this gives the image shift relative to the moving lens.

When the lens itself is oscillating, the object (intermediate image) is stationary in the ground frame, meaning the object velocity relative to the lens is $-v_{lens}$ .

Consequently, the image velocity relative to the ground is $v_I = v_{I/lens} + v_{lens} = -m^2 v_{lens} + v_{lens} = (1-m^2)v_{lens}$ .

A common lethal mistake is either ignoring the lens's own motion (yielding a factor of 4) or adding the velocities with the wrong sign (yielding a factor of 5).

Furthermore, the problem requires applying the small-angle approximation to linearize the pendulum's restoring torque to find the mechanical amplitude.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Ray Optics and Optical Instruments - Class 12
Concept	Microscope Tube Length
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free