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← JEE Advanced 2014 · Paper 1
PhysicsMultiple correct+3Question 9 of 60

JEE Advanced 2014 · Paper 1, Physics question 9

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 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 f1f_1 from the film, while rays of light traversing from glass to air get focused at distance f2f_2 from the film. Then

A long glass cylinder (index n₂) whose left end is a convex spherical surface bulging into the air, coated with a thin shaded film of index n₁.
Choose one or more options
Answer and worked solution

Official answer: A, C

The film does nothing. A thin film of uniform thickness has parallel (concentric) faces, so it only displaces rays negligibly. The system is a single air-glass spherical surface of radius RR, centre inside the glass.

Air to glass. n2v−1∞=n2−1R\dfrac{n_2}{v}-\dfrac1\infty=\dfrac{n_2-1}{R}:
v=1.5R0.5=3R.v=\frac{1.5R}{0.5}=3R.
So ∣f1∣=3R|f_1|=3R.

Glass to air. Light now travels towards the air side and the centre is behind it:
1v−1.5∞=1−1.5−R  ⟹  v=2R.\frac1v-\frac{1.5}\infty=\frac{1-1.5}{-R}\implies v=2R.
So ∣f2∣=2R|f_2|=2R.

(A), (C)

Topics:Refraction at spherical surfaces
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