Start Learning Free

Before You Begin

Question #43126

Difficulty: 8/10 | 🟢 Free

The Question

A point charge QQ is placed on the principal axis of a thin plano-convex lens at a distance uu from its flat surface. The lens has a radius of curvature RR for its convex surface and a maximum central thickness t0t_0. The lens is made of an electro-optic material whose refractive index varies with the local electric field magnitude EE as n(E)=n0+αE2n(E) = n_0 + \alpha E^2, where n0n_0 and α\alpha are positive constants. Assume u≫Rapertureu \gg R_{\text{aperture}} (paraxial approximation) and that the electric field inside the lens is approximately the vacuum field produced by the point charge QQ. The charge QQ also acts as a point source of monochromatic light. Let E0=Q4πϵ0u2E_0 = \frac{Q}{4\pi\epsilon_0 u^2} be the electric field magnitude at the pole of the flat surface. Which of the following expressions correctly represents the effective optical power PP of the lens for paraxial rays emitted by QQ?

Options

  • A.

    n0+αE02−1R+4αE02t0u2\frac{n_0 + \alpha E_0^2 - 1}{R} + \frac{4\alpha E_0^2 t_0}{u^2}

  • B.

    n0+αE02−1R\frac{n_0 + \alpha E_0^2 - 1}{R}

  • C.

    n0+αE02−1R−4αE02t0u2\frac{n_0 + \alpha E_0^2 - 1}{R} - \frac{4\alpha E_0^2 t_0}{u^2}

  • D.

    n0+αE02−1R+2αE02t0u2\frac{n_0 + \alpha E_0^2 - 1}{R} + \frac{2\alpha E_0^2 t_0}{u^2}

Check Your Answer

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

Spherical Surface Refraction|Newton’s Form of Lens Equation|Trajectory of Charged Particle|Vector Cross-Product

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize when standard paraxial optics formulas break down due to spatial variations in the medium.

The 'lethal trap' occurs when a student calculates the modified refractive index at the pole (n1=n0+αE02n_1 = n_0 + \alpha E_0^2) and blindly substitutes it into the standard Lens Maker's Formula, yielding Option B.

They fail to account for the non-uniform electric field creating a radial gradient in the refractive index (a GRIN lens effect).

Because the field is weaker at the edges, the refractive index is lower, which introduces an additional converging phase shift.

The correct analytical approach requires integrating the optical path length (Fermat's Principle) over the spatial profile to extract the true effective focal length.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterRay Optics and Optical Instruments - Class 12
ConceptSpherical Surface Refraction
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree