To solve this problem, we need to analyze the geometry of the situation and apply Snell's law of refraction. The scenario involves a circular disc placed inside a hemispherical bowl, and we want to determine the relationship between the diameter of the disc and the radius of the bowl when the bowl is filled with a liquid of a certain refractive index.
Understanding the Geometry
We start by visualizing the setup. The hemispherical bowl has a radius of 'a', and the circular disc has a diameter 'd'. When the disc is just visible to an observer looking over the edge of the bowl, it means that the line of sight from the observer to the edge of the disc is tangent to the surface of the bowl.
Initial Visibility Condition
When the bowl is empty, the observer's line of sight just grazes the edge of the disc. The geometry can be described as follows:
- The center of the disc is at a vertical distance 'a' from the top of the bowl.
- The radius of the disc is 'd/2'.
Using the right triangle formed by the radius of the bowl, the radius of the disc, and the height from the observer's eye to the center of the disc, we can derive a relationship. The tangent line from the observer's eye to the edge of the disc will create a right triangle where:
- The height is 'a' (the radius of the bowl).
- The base is 'd/2' (the radius of the disc).
Applying Snell's Law
When the bowl is filled with a liquid of refractive index 'u', the situation changes. The light rays from the disc will refract as they pass from the liquid into the air. According to Snell's law, the relationship between the angles of incidence and refraction can be expressed as:
n_1 sin(θ_1) = n_2 sin(θ_2)
Here, 'n_1' is the refractive index of the liquid (which is 'u'), and 'n_2' is the refractive index of air (which is approximately 1). The angles θ_1 and θ_2 are the angles of incidence and refraction, respectively.
Finding the Diameter of the Disc
For the disc to be fully visible, the light rays from the edge of the disc must reach the observer's eye after refraction. The critical angle for the light to just graze the edge of the disc can be determined. The geometry gives us:
- The height from the center of the disc to the observer's eye is still 'a'.
- The radius of the disc remains 'd/2'.
Using the right triangle formed by the observer's eye, the center of the disc, and the edge of the disc, we can derive the relationship:
From the geometry, we can express the tangent of the angle of incidence as:
tan(θ_1) = (d/2) / a
Applying Snell's law, we can relate the angles:
u sin(θ_1) = sin(θ_2)
Using the small angle approximation (for small angles, sin(θ) ≈ tan(θ)), we can substitute:
u (d/2) / a = (d/2) / (a + d/2)
After simplifying and rearranging, we arrive at:
d = 2a (u^2 - 1) / (u^2 + 1)
Final Thoughts
This relationship shows how the diameter of the disc is influenced by the refractive index of the liquid and the radius of the bowl. The derivation highlights the interplay between geometry and optics, illustrating how light behaves differently in various media. Understanding these principles is crucial in fields such as physics and engineering, where optics plays a significant role.
