Problem 20 - University Physics with Modern Physics

Question

The left end of a long glass rod 6.00 \(\mathrm{cm}\) in diameter has a convex hemispherical surface 3.00 \(\mathrm{cm}\) in radius. The refractive index of the glass is 1.60. Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end: (a) infinitely far, (b) \(12.0 \mathrm{cm} ;(c) 2.00 \mathrm{cm} .\)

Step-by-Step Solution

Answer

(a) 8.00 cm inside, (b) 13.71 cm inside, (c) 5.33 cm outside.

Step 1: Understanding the Lens Maker's Formula

The lens maker's formula is used to determine the position of the image formed by spherical surfaces. It is given by:\[\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\]where \(n_2\) is the refractive index of the glass (1.60), \(n_1\) is the refractive index of air (approximately 1.00), \(u\) is the object distance, \(v\) is the image distance, and \(R\) is the radius of curvature of the hemispherical surface.

Step 2: Solving for an Object at Infinity

When an object is at infinity, the incoming rays are parallel and the image is formed at the focal point of the lens. In this case:\[u = \infty\]Thus, \(\frac{n_1}{u} = 0\). Substituting in the lens maker's formula, we get:\[\frac{n_2}{v} = \frac{n_2 - n_1}{R}\]Substitute \(n_2 = 1.60\), \(n_1 = 1.00\), and \(R = 3.00\, \mathrm{cm}\) to find \(v\):\[v = \frac{n_2 \cdot R}{n_2 - n_1} = \frac{1.60 \times 3.00}{1.60 - 1.00} = 8.00\, \mathrm{cm}\]The image is formed 8.00 cm inside the glass rod.

Step 3: Solving for an Object at 12.0 cm

For an object at a distance of 12.0 cm:\[u = -12.0\, \mathrm{cm}\] (negative since object is on the left)Using the lens maker's formula:\[\frac{1.60}{v} - \frac{1.00}{(-12.0)} = \frac{0.60}{3.00}\]Simplify and solve for \(v\):\[\frac{1.60}{v} + \frac{1}{12} = \frac{0.60}{3}\]\[\frac{1.60}{v} = 0.2 - 0.0833 = 0.1167\]\[v = \frac{1.60}{0.1167} \approx 13.71\, \mathrm{cm}\]The image is formed 13.71 cm inside the glass rod.

Step 4: Solving for an Object at 2.00 cm

For an object at a distance of 2.00 cm:\[u = -2.00\, \mathrm{cm}\]Using the lens maker's formula:\[\frac{1.60}{v} - \frac{1.00}{(-2.00)} = \frac{0.60}{3.00}\]Simplify and solve for \(v\):\[\frac{1.60}{v} + 0.5 = 0.2\]\[\frac{1.60}{v} = 0.2 - 0.5 = -0.3\]\[v = \frac{1.60}{-0.3} = -5.33\, \mathrm{cm}\]The image is formed 5.33 cm to the left of the vertex, outside the glass rod.

Key Concepts

Refractive Index

The refractive index is a fundamental concept in optics that describes how light propagates through a medium. It is denoted by the symbol "n" and is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. For the given problem, we have:

• The refractive index of glass (\(n_2\)) is 1.60.
• The refractive index of air (\(n_1\)) is approximately 1.00.

This indicates that light travels slower in glass than in air. The refractive index affects how much light bends when entering a new medium, which is crucial in lens calculations. A higher refractive index means the light bends more. In the lens maker's formula, refractive index values are used to determine how a lens focuses light. This is essential for calculating image distances in optical devices.

Hemispherical Surface

A hemispherical surface, in this context, refers to the round, convex end of the glass rod described in the problem. This part of the lens has a specific role in the way it refracts light beams passing through it. Here are some key characteristics:

• The radius of curvature (\(R\)) of the hemispherical surface is 3.00 cm. It represents the radius of the sphere from which this hemispherical lens surface is a part.
• Its convex shape means it curves outward, which affects how light converges or diverges upon entering this surface from another medium.

When dealing with such surfaces, understanding the shape and radius of the surface is crucial, as they influence the focus and formation of images through refraction. The lens maker's formula incorporates this radius to find where light will gather to form an image.

Image Distance Calculation

Image distance calculation involves determining where an image will form relative to a lens or curved surface. The main formula used is the lens maker's formula:\[\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\]Here:

• \(v\) is the image distance, the distance from the lens to where the image is formed.
• \(u\) is the object distance, the distance from the object to the lens.
• Using positive and negative signs is crucial. In this problem, a negative object distance indicates the object is placed to the left of the lens.
• As demonstrated in the solutions, setting different object distances shows their impact on image formation within or outside the lens material.

The approach involves substituting known values into the formula to solve for the unknown variable, \(v\). This allows us to see whether the image is real and on the same side as the outgoing light or virtual and needing further understanding of the setup to be visualized.