Interpret the setup

• Water fills the tank up to a depth of $18\text{ cm}$.
• Above water, there is oil of refractive index $\mu_o=\dfrac{7}{4}$.
• Water has refractive index $\mu_w=\dfrac{4}{3}$.
• The oil-air upper surface is convex with radius of curvature $R=6\text{ cm}$.
• The object $S$ is $24\text{ cm}$ above the plane water surface.

Since the problem says consider oil to act as a thin lens, the oil layer behaves like a plano-convex lens:

• one surface plane (oil-water interface),
• one surface convex (oil-air interface),
• lens material is oil, surrounding medium is air.

Find focal length of the oil lens

For a thin lens in air, $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$

Here:

• plane surface $\Rightarrow R_1=\infty$,
• convex second surface. Taking sign convention carefully for a plano-convex lens, this gives effective power $\frac{1}{f}=(\mu_o-1)\cdot \frac{1}{R}.$

So,

Image formed by the oil lens

The object is $24\text{ cm}$ above the water surface. Since the oil lens is thin and lies near that level, we take object distance $u=-24\text{ cm}$ (Cartesian sign convention).

Lens formula: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}.$

Substitute $f=8$ and $u=-24$: $\frac{1}{8}=\frac{1}{v}-\left(-\frac{1}{24}\right)=\frac{1}{v}+\frac{1}{24}.$

Therefore, $\frac{1}{v}=\frac{1}{8}-\frac{1}{24}=\frac{3-1}{24}=\frac{2}{24}=\frac{1}{12}.$

So, $v=12\text{ cm}.$

Thus the lens forms an image $12\text{ cm}$ below the water surface.

Refraction at plane water surface

This image acts as a virtual object for water at depth $12\text{ cm}$ below the surface. We now locate the final image seen inside water through the plane interface.

For refraction at a plane surface,

Height above the bottom

Total water depth = $18\text{ cm}$.

Therefore the image is above the bottom by $x=18-16=2\text{ cm}.$

Final answer

$\boxed{2}$

Comparison with stored answer

Stored correct answer = $2$.

Our derived answer matches it.