JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Geometrical Optics

A transparent thin film of uniform thickness and refractive index

n_{1} = 1.4

is coated on the convex spherical surface of radius Rat one end of a long solid glass cylinder of refractive index

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

f_{1}

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

f_{2}

from the film. Then

Answer: (A), (C)

Solution

**Air to glass (finding

f_{1}

):**

First refraction (air to film):

\frac{n _{1}}{v _{1}} - \frac{1}{\infty} = \frac{n _{1} - 1}{R}

, giving

v_{1} = \frac{n _{1} R}{n _{1} - 1} = \frac{1.4 R}{0.4} = 3.5 R

.

Second refraction (film to glass):

\frac{n _{2}}{v _{2}} - \frac{n _{1}}{v _{1}} = \frac{n _{2} - n _{1}}{R}

. Substituting:

\frac{1.5}{v _{2}} - \frac{1.4}{3.5 R} = \frac{0.1}{R}

, so

\frac{1.5}{v _{2}} = \frac{0.1}{R} + \frac{0.4}{R} = \frac{0.5}{R}

, giving

v_{2} = 3 R

. So

∣ f_{1} ∣ = 3 R

, option (A).

**Glass to air (finding

f_{2}

):**

First refraction (glass to film):

\frac{n _{1}}{v _{1}} - \frac{n _{2}}{\infty} = \frac{n _{1} - n _{2}}{- R}

, giving

v_{1} = \frac{n _{1} R}{n _{2} - n _{1}} = \frac{1.4 R}{0.1} = 14 R

.

Second refraction (film to air):

\frac{1}{v _{2}} - \frac{n _{1}}{v _{1}} = \frac{1 - n _{1}}{- R}

. Substituting:

\frac{1}{v _{2}} - \frac{1.4}{14 R} = \frac{0.4}{R}

, so

\frac{1}{v _{2}} = \frac{0.4}{R} + \frac{0.1}{R} = \frac{0.5}{R}

, giving

v_{2} = 2 R

. So

∣ f_{2} ∣ = 2 R

, option (C).

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options