Question

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

A. 2
B. 3
C. 4
D. 5

Answer 1

When an oil drop is spread over water it floats over the surface of the water. This is due to the reason that oil has a lower density than water. For light coming to the water, this oil drop will work as a thin spherical lens. Due to this the light has to pass through three mediums and it will refract and form an image due to three times refraction.
As per the given data,
Object distance from the lens is $24cm$
The radius of the lens 6cm
Depth of the water 18cm
Refractive index of air 1
Refractive index of the thin lens (oil drop) $\dfrac{7}{4}$
Refractive index of water $\dfrac{4}{3}$
Formula used:
Lens formula for spherical surfaces,
$\dfrac{{{\mu }_{2}}}{v}-\dfrac{{{\mu }_{1}}}{u}=\dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{R}$

Complete answer:
Here is the detailed diagram for the whole setup for the better understanding of the terms mentioned in the question,

In this case, the light coming from the object has to undergo refraction due to three mediums interacting with each other.
The case I: (when the light undergo refraction due to the air and spherical surface interaction)
Using the spherical lens formula we can find the length at which the 1st image (intermediate image) will form
$\dfrac{{{\mu }_{2}}}{v}-\dfrac{{{\mu }_{1}}}{u}=\dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{R}$
By upping the values as per the sign conventions,

& \dfrac{7}{4v}-\dfrac{1}{-24}=\dfrac{\dfrac{7}{4}-1}{6} \\\ & \\\ & \Rightarrow \dfrac{7\times 2}{4}=v \\\ & \\\ & \Rightarrow v=21cm \\\ \end{aligned}$$ So the intermediate image will be formed 21 cm away from the lens (oil drop). Case II: (when the refraction between the other side of the lens and water) $$\dfrac{{{\mu }_{2}}}{v}-\dfrac{{{\mu }_{1}}}{u}=\dfrac{{{\mu }_{2}}-{{\mu }_{1}}}{R}$$ The radius of the other side of the thin lens (oil drop) is infinity. By putting the values from the given data, $$\begin{aligned} & ~\dfrac{4}{3v}-\dfrac{7}{4(21)}=\dfrac{\dfrac{4}{3}-\dfrac{7}{4}}{\infty } \\\ & \\\ & \Rightarrow v=16cm \\\ \end{aligned}$$ The final image will be produced 16cm below the lens. The image will be formed $$18-16=2cm$$ from the bottom level of the water. **Thus the correct option is Option A.** **Note:** The refraction of light is the bending of light over the edges of the medium when it passes from one medium to another. This bending occurs due to the change in the speed and wavelength of the wave of light while moving in the medium of different refractive index. While calculations the sign convention must be considered with the standard formula of the lens.