Introduction:
In this scenario, we have a vessel filled with water up to a certain depth. A point source of light is placed above the water surface. We need to determine the distance of the final image formed from the water surface.

Understanding the situation:
To solve this problem, we need to consider the properties of reflection and refraction of light at the interface between air and water. When light passes from air to water, it undergoes both reflection and refraction.

Reflection of light:
When light strikes a perfectly reflecting plane surface, it follows the laws of reflection. The incident angle is equal to the reflected angle, and the incident ray, reflected ray, and the normal to the surface lie in the same plane.

Refraction of light:
When light passes from air to water, it undergoes refraction due to the change in the medium. The incident ray bends towards the normal as it enters the water. The angle of refraction can be determined using Snell's law: n₁ sinθ₁ = n₂ sinθ₂, where n₁ and n₂ are the refractive indices of the media and θ₁ and θ₂ are the angles of incidence and refraction, respectively.

Image formation:
In this scenario, the light from the point source travels in all directions. Some of the light will reflect off the water surface and form a virtual image above the surface. This image will be formed by the reflected rays and will be at the same distance below the water surface as the source is above it. This is due to the law of reflection.

Distance of the final image:
To determine the distance of the final image from the water surface, we need to consider the path of the light rays. The light rays from the point source will undergo reflection at the water surface and then refraction as they pass through the water.

Calculation:
Let's assume the distance of the final image from the water surface is x. The light ray from the point source will travel a total distance of (h + x) before reaching the final image.

Using Snell's law, we can determine the angle of refraction when the light ray enters the water. The refractive index of water is 4/3, and the angle of incidence is equal to the angle of reflection.

Using the relationship n₁ sinθ₁ = n₂ sinθ₂, we can write:
1 sin(θ₁) = (4/3) sin(θ₂)

Since θ₁ = θ₂:
1 sin(θ₁) = (4/3) sin(θ₁)

Simplifying the equation, we get:
1 = (4/3) sin(θ₁)

Solving for sin(θ₁), we find:
sin(θ₁) = 3/4

Using the inverse sine function, we can determine the value of θ₁:
θ₁ = sin^(-1)(3/4)

Knowing the value of θ₁, we can calculate the distance x using trigonometry. The distance traveled by the light ray in the water can be given by:
d = x tan(θ₁)

Solving for x, we get:
x = d / tan(θ₁)

Substituting the value of θ₁, we find:
x = d / tan(sin^

Options : a) h + 5/2 d. (b) h- 3/2d (c) h+4/3d (d) h +3/2d