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JEE Advanced 2014Paper 1
Exam
JEE Advanced
Year
2014
Paper
Paper 1
Subject
Physics
Chapter
Ray Optics

JEE Advanced 2014 · Paper 1 · Physics · Ray Optics

A transparent thin film of uniform thickness and refractive index n...

A transparent thin film of uniform thickness and refractive index n1=1.4{n}_{1} = 1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index n2=1.5{n}_{2} = 1.5 , as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance f1 from the film, while rays of light traversing from glass to air get focused at distance f2 from the film. Then

A transparent thin film of uniform thickness and refractive index n _ 1 = 1.4 is coated on the convex spherical surface

Options

  1. A∣f1∣=3R\left|{f}_{1}\right| = 3R
  2. B∣f1∣=2.8R\left|{f}_{1}\right| = 2.8R
  3. C∣f2∣=2R\left|{f}_{2}\right| = 2R
  4. D∣f2∣=1.4R\left|{f}_{2}\right| = 1.4R

Correct answer

A. ∣f1∣=3R\left|{f}_{1}\right| = 3R

Step-by-step solution

For air to glass
1.5f1=1.4−1R+1.5−1.4R\dfrac{1.5}{{f}_{1}} = \dfrac{1.4 - 1}{R} + \dfrac{1.5 - 1.4}{R}
∴f1=3R∴ {f}_{1} = 3R
For glass to air.
1f2=1.4−1.5−R+1−1.4−R\dfrac{1}{{f}_{2}} = \dfrac{1.4 - 1.5}{- R} + \dfrac{1 - 1.4}{- R}
∴f2=2R∴ {f}_{2} = 2R
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