Image Formation by Oil Lens on Water Surface

To find the position of the image formed by the system, we consider the refraction of light through the two interfaces forming the thin oil lens:

1. Refraction at the first (air-oil) spherical surface:

   • Refractive index of air: $\mu_1 = 1.0$
   • Refractive index of oil: $\mu_2 = \frac{7}{4}$
   • Object distance: $u_1 = -24\text{ cm}$
   • Radius of curvature of the convex surface: $R_1 = +6\text{ cm}$ (center of curvature lies in the medium of transmission, i.e., downwards)

   Using the spherical surface refraction formula:
   $\frac{\mu_2}{v_1} - \frac{\mu_1}{u_1} = \frac{\mu_2 - \mu_1}{R_1}$

   Substituting the given values:
   $\frac{7/4}{v_1} - \frac{1}{-24} = \frac{\frac{7}{4} - 1}{6}$
   $\frac{7}{4v_1} + \frac{1}{24} = \frac{3/4}{6} = \frac{1}{8}$
   $\frac{7}{4v_1} = \frac{1}{8} - \frac{1}{24} = \frac{3 - 1}{24} = \frac{2}{24} = \frac{1}{12}$
   $v_1 = \frac{7}{4} \times 12 = 21\text{ cm}$

2. Refraction at the second (oil-water) plane surface:

   • The oil lens is considered thin, so the refraction at the flat interface occurs at effectively the same location as the water surface.
   • The image formed by the first surface acts as a virtual object for the plane oil-water interface.
   • Object distance for the second surface: $u_2 = +v_1 = +21\text{ cm}$
   • Refractive index of oil: $\mu_2 = \frac{7}{4}$
   • Refractive index of water: $\mu_3 = \frac{4}{3}$
   • Radius of curvature of the plane surface: $R_2 = \infty$

   Using the refraction formula for a plane surface:
   $\frac{\mu_3}{v_2} - \frac{\mu_2}{u_2} = 0 \implies \frac{\mu_3}{v_2} = \frac{\mu_2}{u_2}$

   Solving for $v_2$:
   $v_2 = u_2 \left(\frac{\mu_3}{\mu_2}\right) = 21 \times \frac{4/3}{7/4} = 21 \times \frac{16}{21} = 16\text{ cm}$

   Therefore, the final image is formed at a distance of $16\text{ cm}$ below the water surface.

3. Position of the image from the bottom of the tank:

   • The depth of the water in the tank is $H = 18\text{ cm}$.
   • The image is formed at a depth of $16\text{ cm}$ below the surface.
   • The distance $x$ of the image from the bottom of the tank is:
     $x = 18\text{ cm} - 16\text{ cm} = 2\text{ cm}$

Thus, the value of $x$ is $2$.