Animated Solution for Physics - Optics: Water (with refractive index = 4/3) in a tank is 18 cm deep. Oil of refractive index 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.

The problem asks us to find the position of an image formed by a combination of a convex oil surface and a flat water surface. It's a classic example of sequential refraction, where the image formed by the first surface acts as the object for the second surface.

Imagine you are looking down into a tank. At the very top, there's air. Below that, a thin layer of oil forms a convex bump, and below the oil is a deep pool of water. We are given the refractive indices of all three mediums: air ($\mu =1$), oil ($\mu =7/4$), and water ($\mu =4/3$).

An object $S$ is placed $24\text{ cm}$ above the water surface. Since the oil layer is described as a "thin lens", we can assume its thickness is negligible. This means the distance from the object to the convex oil surface is also $24\text{ cm}$. The light rays from the object will first hit the convex air-oil interface, bend, and then hit the flat oil-water interface, bending again to form the final image.

To solve this, we need our trusty formula for refraction at a single spherical surface:

We will apply this formula twice. Since the light is traveling downwards, let's take the downward direction as positive. This sign convention is crucial to avoid silly mistakes!

Air to Oil

Let's look at the first interface. The light travels from air (${\mu }_1=1$) to oil (${\mu }_2=7/4$). The object is above the surface, so the object distance $u=-24\text{ cm}$. The surface is convex towards the rarer medium, meaning its center of curvature lies below it, so $R=+6\text{ cm}$.

Substituting these values into our master equation:

Let's simplify the right side. $7/4-1=3/4$, and dividing that by $6$ gives us $1/8$.

Moving the $1/24$ to the other side:

Cross-multiplying gives us:

So, the first refraction creates an image $I_1$ at a distance of $21\text{ cm}$ below the surface. This image will now act as a virtual object for the second interface.

Oil to Water

Now, the light travels from the oil (${\mu }_1=7/4$) into the water (${\mu }_2=4/3$). Our virtual object $I_1$ is at $+21\text{ cm}$, so $u=+21\text{ cm}$. The interface between the oil and water is flat, which means its radius of curvature is infinity ($R=\infty$).

Plugging these into the formula:

Anything divided by infinity is zero, so the right side vanishes!

Solving for $v_2$:

The final image $I_2$ is formed $16\text{ cm}$ below the surface.

We are almost there! The question asks for the distance of the image from the bottom of the tank. We know the total depth of the water is $18\text{ cm}$.

Since the image is $16\text{ cm}$ from the top, its distance from the bottom $x$ is simply: