JEE Challenger
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Refraction at Spherical Boundary of Coated Glass Cylinder

A transparent thin film of uniform thickness and refractive index n1=1.4n_1 = 1.4 is coated on the convex spherical surface of radius RR at one end of a long solid glass cylinder of refractive index n2=1.5n_2 = 1.5, as illustrated in the figure. Rays of light incident parallel to the cylinder's axis traveling through the film from air to glass come to a focus at a distance f1f_1 from the film, whereas rays traversing from glass to air get focused at a distance f2f_2 from the film. Then:

Question Diagram 1

Options

A

∣f1∣=3R|f_1| = 3R

Correct
B

∣f1∣=2.8R|f_1| = 2.8R

C

∣f2∣=2R|f_2| = 2R

Correct
D

∣f2∣=1.4R|f_2| = 1.4R

Step-by-Step Solution

To determine the focal lengths f1f_1 and f2f_2, we consider the refraction of light through a spherical boundary between two media coated with a thin film.

Refraction at a Thin Coated Spherical Boundary

Since the coating is a thin film of uniform and negligible thickness (t→0t \to 0), the total refraction across the two interfaces (air-film and film-glass) can be analyzed by combining the refraction equations at each boundary:

  1. Refraction at the first surface (medium aa to medium bb of radius RR): nbv1−nau=nb−naR\frac{n_b}{v_1} - \frac{n_a}{u} = \frac{n_b - n_a}{R}

  2. Refraction at the second surface (medium bb to medium cc of radius RR): ncv−nbv1=nc−nbR\frac{n_c}{v} - \frac{n_b}{v_1} = \frac{n_c - n_b}{R}

Adding the two equations yields the net refraction formula, showing that the intermediate refractive index nbn_b cancels out: ncv−nau=nc−naR\frac{n_c}{v} - \frac{n_a}{u} = \frac{n_c - n_a}{R}


1. Rays traveling from Air to Glass (f1f_1)

  • Incident medium: Air (na=1n_a = 1)
  • Final medium: Glass (nc=1.5n_c = 1.5)
  • Object distance: Parallel rays imply u=−∞u = -\infty
  • Radius of curvature: The surface is convex towards air, so its center of curvature lies inside the glass cylinder in the direction of light propagation (R=+RR = +R).

Using the refraction formula: 1.5f1−0=1.5−1.0+R\frac{1.5}{f_1} - 0 = \frac{1.5 - 1.0}{+R} 1.5f1=0.5R\frac{1.5}{f_1} = \frac{0.5}{R} f1=3R  ⟹  ∣f1∣=3Rf_1 = 3R \implies |f_1| = 3R

This matches Option (A).


2. Rays traveling from Glass to Air (f2f_2)

  • Incident medium: Glass (na=1.5n_a = 1.5)
  • Final medium: Air (nc=1n_c = 1)
  • Object distance: Parallel rays inside the glass imply u=−∞u = -\infty
  • Radius of curvature: As light travels from glass to air, it encounters a surface that is concave relative to the incident direction. Therefore, the center of curvature lies opposite to the direction of light propagation (R=−RR = -R).

Using the refraction formula: 1.0f2−0=1.0−1.5−R\frac{1.0}{f_2} - 0 = \frac{1.0 - 1.5}{-R} 1.0f2=−0.5−R=0.5R\frac{1.0}{f_2} = \frac{-0.5}{-R} = \frac{0.5}{R} f2=2R  ⟹  ∣f2∣=2Rf_2 = 2R \implies |f_2| = 2R

This matches Option (C).


Conclusion

The correct options are (A) and (C).