Optics 2 Question 18
18. A transparent thin film of uniform thickness and refractive index $n _1=1.4$ is coated on the convex spherical surface of radius $R$ at one end of a long solid glass cylinder of refractive index $n _2=1.5$, 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
(2014 Adv.)
(a) $\left|f _1\right|=3 R$
(b) $\left|f _1\right|=2.8 R$
(c) $\left|f _2\right|=2 R$
(d) $\left|f _2\right|=1.4 R$
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Answer:
Correct Answer: 18. $(a, c)$
Solution:
- $\frac{1}{f _{\text {film }}}=\left(n _1-1\right) \frac{1}{R}-\frac{1}{R} \Rightarrow f _{\text {film }}=\infty$
(infinite)
$\therefore$ There is no effect of presence of film.
From Air to Glass
Using the equation $\quad \frac{n _2}{v}-\frac{1}{u}=\frac{n _2-1}{R}$
$$ \begin{array}{rlrl} & & \frac{1.5}{v}-\frac{1}{\infty} & =\frac{1.5-1}{R} \Rightarrow v=3 R \\ \therefore \quad f _1 & =3 R \end{array} $$
From Glass to Air Again using the same equation
$$ \begin{gathered} \frac{1}{v}-\frac{n _2}{u}=\frac{1-n _2}{-R} \Rightarrow \frac{1}{v}-\frac{1.5}{\infty}=\frac{1-1.5}{-R} \Rightarrow v=2 R \\ \therefore \quad f _2=2 R \end{gathered} $$
