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Question #43185

Difficulty: 8/10 | 🟢 Free

The Question

In a certain region of space, an external electrostatic potential is given by V(x)=10x2V(x) = 10 x^2, where VV is in volts and xx is in meters. A thin symmetric biconvex lens made of glass (refractive index ng=1.8n_g = 1.8) with radii of curvature R=20 cmR = 20\text{ cm} is placed at x=5.0 mx = 5.0\text{ m}. The lens is completely immersed in a uniform liquid bath. The refractive index of this liquid, nln_l, is determined by the x-component of the electric field ExE_x (in V/m) exactly at the lens's position, according to the empirical relation nl=1.25−0.0025Exn_l = 1.25 - 0.0025 E_x.

A small point object is placed at x=5.25 mx = 5.25\text{ m} on the principal axis of the lens. Find the final x-coordinate of the image formed by the lens.

Options

  • A.

    5.50\text{ m}

  • B.

    4.75\text{ m}

  • C.

    4.50\text{ m}

  • D.

    5.25\text{ m}

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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

Sign Convention Dynamics|Potential vs Field Relationship|Calculus/Gradient Integration

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to rigorously apply and reconcile two distinct sign conventions.

The first trap lies in electrodynamics: students often forget the negative sign in the potential gradient (E=−dV/dxE = -dV/dx), leading to an inverted electric field and an incorrect refractive index of 1.0 instead of 1.5.

The second, more lethal trap lies in ray optics: the Cartesian sign convention dictates that the positive optical axis aligns with the direction of incident light.

Because the object is placed at a greater x-coordinate than the lens, light travels in the -x direction.

Students who blindly associate the +x direction with the optical positive axis will misinterpret the final negative image distance, placing the image on the wrong side of the lens.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterRay Optics and Optical Instruments - Class 12
ConceptSign Convention Dynamics
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree