Problem 49 - University Physics with Modern Physics

Question

You sight along the rim of a glass with vertical sides so that the top rim is lined up with the opposite edge of the bottom (Fig. P33.49a). The glass is a (Fhin-walled, hollow cylinder 16.0 \(\mathrm{cm}\) high. The diameter of the top and bottom of the glass is 8.0 \(\mathrm{cm} .\) While you keep your eye in the same position, a friend fills the glass with a transparent liquid, and you then see a dime that is lying at the center of the bottom of the glass \((\) Fig. P33.49b). What is the index of refraction of the liquid?

Step-by-Step Solution

Answer

The index of refraction of the liquid is approximately 4.12.

Step 1: Understand the Setup

The problem involves a glass cylinder of height 16 cm and diameter 8 cm. You observe the top rim lining up with the opposite edge of the bottom from a certain angle. When the liquid fills the glass, you view a dime at the center of the bottom.

Step 2: Apply Geometry

When empty, the light travels straight and your sight line runs tangent to the glass's top rim and bottom edge. The geometry suggests that the light path forms a right triangle with the glass's height as one leg (16 cm), and half of the diameter as the other leg (4 cm).

Step 3: Use Snell's Law

When filled with liquid, the light refracts at the liquid-air interface. The apparent position of the dime is shifted upwards, allowing you to see it. This involves Snell's Law at the point where light exits the liquid: \[ n_1 \sin{\theta_1} = n_2 \sin{\theta_2} \] where \( n_1 \) is the index of refraction of liquid and \( n_2 \) = 1 for air.

Step 4: Calculate the Angle of Incidence

In the earlier setup, the angle of incidence (\( \theta_1 \)) can be deduced from the triangle: \[ \tan{\theta_1} = \frac{4}{16} = \frac{1}{4} \]Thus, \( \theta_1 \) is approximately \( an^{-1}{\frac{1}{4}} \).

Step 5: Determine Critical Angle

For a light ray to just start to see the dime at bottom center, the refracted angle \( \theta_2 \) at the critical point would need to be 90 degrees, when the observer sight aligns, leading to:\[ \sin{\theta_2} = 1 \]

Step 6: Solve Index of Refraction

From Snell's Law and geometry:\[ n_1 \sin{\theta_1} = n_2 \cdot 1 \] Hence, \[ n_1 = \frac{1}{\sin{\theta_1}} \] Using calculated \( \theta_1 \), solve for \( n_1 \).

Step 7: Calculate and Conclude

After solving, \[ \theta_1 = an^{-1}{\frac{1}{4}} \approx 14.04^\circ \]use a calculator, \[ \sin{\theta_1} \approx 0.2425 \] Then, \[ n_1 = \frac{1}{0.2425} \approx 4.12 \]. The index of refraction of the liquid is approximately 4.12.

Key Concepts

Snell's Law

Snell's Law is a fundamental principle in optics that describes how light rays bend when transitioning between different media. It is expressed mathematically as:\[ n_1 \sin{\theta_1} = n_2 \sin{\theta_2} \]where:

• \( n_1 \) and \( n_2 \) are the indices of refraction of the two media.
• \( \theta_1 \) is the angle of incidence in the first medium, and \( \theta_2 \) is the angle of refraction in the second medium.

In the context of the original exercise, Snell's Law helps to determine how light behaves as it moves from the liquid inside the glass (with unknown index \( n_1 \)) into air (where \( n_2 = 1 \)). It explains why the light ray bends and why the apparent position of the object viewed through the liquid is different than if the glass was empty. Successfully applying Snell’s Law can help calculate the refractive index of the liquid, giving insights into its optical properties.

Right Triangle Geometry

Right triangle geometry comes into play when analyzing the path of light in the glass cylinder filled with liquid. Imagine the path of the light as forming a right triangle, where:

• The height of the glass cylinder (16 cm) represents one leg of the triangle.
• Half of the diameter of the glass (4 cm) is the other leg, forming a right angle with the height.
• The hypotenuse represents the light's path from your line of sight to the rim of the bottom surface.

This geometric setup allows us to calculate the angle of incidence, \( \theta_1 \), which is critical for using Snell’s Law. By understanding the properties of right triangles, specifically the relationship of the sides through the tangent function \( \tan{\theta_1} = \frac{4}{16} = \frac{1}{4} \), you can determine \( \theta_1 \) as roughly \( 14.04^\circ \). Mastery of right triangle geometry provides a foundational step in solving the optical problem presented in the exercise.

Critical Angle

The concept of the critical angle is important in understanding total internal reflection, but in this exercise, it pertains to the point at which the light from the dime at the bottom can just start to refract into the air from the denser liquid. This occurs when the refraction angle, \( \theta_2 \), is maximized at 90 degrees, just before the onset of total internal reflection.To establish this critical situation, the equation:\[ \sin{\theta_2} = 1 \]is used. By setting the ball rolling for solving for the unknown index of refraction, measuring \( \theta_1 \) and applying Snell’s Law again, we can infer when and why a certain scenario becomes visible or invisible, governed by these critical angles. It becomes essential in contexts like fiber optics and elaborate visual phenomena in nature.

Sight Line Tangent

The sight line tangent is a straightforward concept that defines the ability to directly observe an object along a line of sight tangent to a curve. In this exercise, your line of sight is initially tangent to the open top of the glass, just reaching the bottom edge of the opposite side. As the liquid is added, the sight line diverges due to the refractive properties of the liquid, adjusting the tangent to accommodate the shift in the light’s pathway. With the glass being a hollow cylinder, the tangent relates to how the light angles shift and produce a visible path that alters the apparent position of objects viewed through the medium. Embracing the roles tangents play in optics describes why certain visual changes occur as media change—a sine qua non of practical optics.

Apparent Position in Refraction

Refraction is a wondrous natural phenomenon that often causes objects under water to appear closer to the surface or shifted visually from their actual location. This effect is called apparent position change. When light travels through different mediums, such as from air into water, it changes velocity, causing a bend in the light’s path. It means the object, like the dime at the bottom of the glass, is seen at a different position when looked at through the refractive medium. In this particular instance, the apparent position of the dime changes due to the light refracting at the surface of the liquid as it transitions into air. As the task involves deducing the liquid's refractive index, the apparent position becomes the key observable feature utilized to back-calculate the optical properties of the liquid.