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Question #43185
The Question
In a certain region of space, an external electrostatic potential is given by , where is in volts and is in meters. A thin symmetric biconvex lens made of glass (refractive index ) with radii of curvature is placed at . The lens is completely immersed in a uniform liquid bath. The refractive index of this liquid, , is determined by the x-component of the electric field (in V/m) exactly at the lens's position, according to the empirical relation .
A small point object is placed at 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}
Check Your Answer
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 (), 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
| Subject | Physics |
| Chapter | Ray Optics and Optical Instruments - Class 12 |
| Concept | Sign Convention Dynamics |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |