Lensing by oil on water

JEE Advanced 2011 Paper 2, Question 36

Water (with refractive index =\frac{4}{3} ) in a tank is 18 \mathrm{~cm} deep. Oil of refractive index \frac{7}{4} lies on water making a convex surface of radius of curvature R=6 {\rm ~cm} as shown. Consider oil to act as a thin lens. An object ‘S’ is placed 24 \mathrm{~cm} above water surface. The location of its image is at ‘x’ cm above the bottom of the tank. What is ‘x’?

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To find the location of the final image we consider the oil and the water in the tank as a combination of two elements: a lens and a …

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Lenses II: Image formation

Continuing our discussion from the previous chapter, we can use the thin lens equation and our knowledge of the signs of f, to determine the position, orientation, and magnification of the image for different object distances p. This is similar to the analysis we did for mirrors here. The results are summarized in the table below.

Converging lens

The focal length of a converging lens is positive. That means light from infinity will be brought to focus behind the lens. We will begin our analysis there.

Starting from p = \infty

Consider an object kept in front of a converging lens. …

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Lenses I: The thin lens equation

A lens is a refracting element with two curved surfaces. Light changes direction as it refracts through each of these surfaces. If the thickness of the lens is small compared to the radius of curvature of each surface, we can think of light as bending just once at the central plane of the lens. This approximation is justified more rigorously in chapter 27 of the Feynman lectures. For our present discussion we start with equation (27.12) from that book, called the thin lens equation,

(1)   \begin{equation*}   \frac{1}{p} + \frac{1}{q} = \frac{1}{f} ,  \end{equation*}

where p is the object distance from the lens, q is the image distance, and …

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