Optics 7 Question 33
34. Water (with refractive index $=\frac{4}{3}$ ) in a tank is $18 cm$ deep. Oil of refractive index $\frac{7}{4}$ lies on water making a convex surface of radius of curvature $R=6 cm$ as shown. Consider oil to act as a thin lens. An object $S$ is placed $24 cm$ above water surface. The location of its image is at $x cm$ above the bottom of the tank. Then $x$ is.
(2011)
s.
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Solution:
Two refractions will take place, first from spherical surface and the other from the plane surface.
So, applying
$$ \frac{\mu _2}{v}-\frac{\mu _1}{u}=\frac{\mu _2-\mu _1}{R} $$
two times with proper sign convention.
Ray of light is travelling downwards. Therefore, downward direction is taken as positive direction.
$$ \begin{aligned} \frac{7 / 4}{v}-\frac{1.0}{-24} & =\frac{7 / 4-1.0}{+6} \\ \frac{4 / 3}{(18-x)}-\frac{7 / 4}{v} & =\frac{4 / 3-7 / 4}{\propto} \end{aligned} $$
Solving these equations, we get, $x=2 cm$
$\therefore$ Answer is 2 .
