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A thick-walled wine goblet sitting on a table can be considered to be a hollow glass sphere with an outer radius of 4.00 cm and an inner radius of 3.40 \(\mathrm{cm}\) . The index of refraction of the goblet glass is \(1.50 .(a)\) A beam of parallel light rays enters the side of the empty goblet along a horizontal radius. Where, if anywhere, will an image be formed? ( \(b\) ) The goblet is filled with white wine \((n=1.37)\) . Where is the image formed?

Short Answer

Expert verified
The image is 12 cm beyond the outer surface in glass and closer in the wine-filled goblet.

Step by step solution

01

Understand the Setup

We have a hollow glass sphere (wine goblet) with outer radius \( R_2 = 4.00 \text{ cm} \) and inner radius \( R_1 = 3.40 \text{ cm} \). The refractive index of the glass is \( n = 1.50 \). Parallel light rays enter horizontally through the glass.
02

Apply Lensmaker's Equation

For spherical surfaces, the Lensmaker's equation is used: \[ \frac{n_2}{s} - \frac{n_1}{s'} = \frac{n_2 - n_1}{R} \] where \( s \) is the object distance (infinity for parallel rays), \( s' \) is the image distance, \( n_1 \) is the initial refractive index (air, \( 1.00 \)), \( n_2 \) is the refractive index of the glass (\( 1.50 \)), and \( R \) is the radius of curvature. Here, \( R = 4.00 \text{ cm} \).
03

Calculate Image Location for Glass without Wine

Using the formula from Step 2 with \( s = \infty \) and \( n_1 = 1.00 \), \[ \frac{1.50}{s'} = \frac{1.50 - 1.00}{4} \] \[ \frac{1.50}{s'} = \frac{0.50}{4} = 0.125 \] \[ s' = \frac{1.50}{0.125} = 12.00 \text{ cm} \] Thus, the image is formed 12 cm inside the glass, beyond the outer surface.
04

Adjust for Glass Filled with Wine

When the goblet is filled, the inner region is now wine, which has a refractive index \( n = 1.37 \). Now, the new interface is between the wine and glass. Applying the Lensmaker's equation for the interface with the inner radius, the equation is: \[ \frac{1.50}{s_2} - \frac{1.37}{12.00} = \frac{1.50 - 1.37}{3.40} \] Simplifying, \[ \frac{1.50}{s_2} = \frac{1.50 - 1.37}{3.40} + \frac{1.37}{12.00} \] Solve for \( s_2 \).
05

Find Image Distance in Wine

Solving the previous equation gives a new image distance \( s_2 \), considering both interfaces. The calculations involve substeps of solving the equation, ensuring the continuity of light through both media.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Lensmaker's Equation
The Lensmaker's Equation is a fundamental tool in optics, especially when dealing with lenses and curved surfaces. This equation helps to determine the focal length of a lens, accounting for the lens's material and the curvature of its surfaces. It's represented as: \[ \frac{n_2}{s} - \frac{n_1}{s'} = \frac{n_2 - n_1}{R} \] where:
  • \( n_1 \) and \( n_2 \) are the refractive indices of the surrounding medium and the lens material, respectively.
  • \( s \) is the object distance (infinity for parallel rays).
  • \( s' \) is the image distance.
  • \( R \) is the radius of curvature of the lens.
When using this equation, note that it relates the physical geometry of lens surfaces with the optical properties of the material. In this context, a glass goblet is analogous to a lens with spherical surfaces. Applying this formula helps us deduce where an image is formed when light passes through different mediums, like air and glass. It's a crucial element in understanding refraction in lenses, showing how light is bent and focused.
Refractive Index
Refractive Index is a measure of how much light bends, or refracts, as it moves from one medium to another. In the problem of the goblet, we have two key refractive indices: the glass (1.50) and the surrounding air (1.00). When these indices differ, light changes speed and direction. This bending is responsible for the creation of images.
The formula for refraction at an interface is given by Snell's Law:\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]where:
  • \( n_1 \) and \( n_2 \) are the refractive indices of the first and second media, respectively.
  • \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively.
The refractive index essentially tells us how much slower light travels in a material compared to a vacuum. With this information, one can predict how images will be formed when light transitions between materials like air, glass, and wine. A higher refractive index denotes a greater degree of bending.
Spherical Surfaces
Spherical surfaces in optics play a crucial role as they are used to construct lenses that can focus or disperse light. Such surfaces can be either convex (bulging out) or concave (hollowed in).
In the context of the hollow glass goblet:
  • The outer surface functions as a convex lens, bending incoming parallel rays towards a focal point inside.
  • The inner surface, especially when filled with wine, acts as a concave surface, further affecting image formation due to the change in refractive index.
Understanding spherical surfaces is crucial because they determine the lens's power, defined by its ability to converge or diverge light rays. The curvature of these surfaces is quantified by the radius of curvature \( R \), influencing the focal length and image distance calculations.
In practical applications, how well we craft these spherical surfaces impacts the quality of the optical device, be it a simple glass or a complex camera lens. This becomes particularly important when transitioning from one medium to another, such as from air to glass, as each interface alters the path of light.

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Most popular questions from this chapter

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