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Chapter 34: Problem 123

One end of a long glass rod \((n=1.5)\) is a convex surface of radius \(6.0 \mathrm{~cm}\). An object is located in air along the axis of the rod, at a distance of \(10 \mathrm{~cm}\) from the convex end. (a) How far apart are the object and the image formed by the glass rod? (b) Within what range of distances from the end of the rod must the object be located in order to produce a virtual image?

Short Answer

Expert verified
The object and image are 1.82 cm apart; for a virtual image, the object must be within 12 cm.

Step by step solution

01

Identify Given Quantities

We have a glass rod with refractive index \( n = 1.5 \). Its convex tip has a radius of curvature \( R = 6.0 \) cm. The object is at a distance \( u = 10 \) cm from the convex surface.
02

Apply Lens Maker's Formula

Using the formula for a spherical surface, \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R} \), where \( n_1 = 1 \) (air), \( n_2 = 1.5 \) (glass), \( u = -10 \) cm, substitute the given values:
03

Substitute and Solve for the Image Distance

We substitute into the equation: \( \frac{1.5}{v} - \frac{1}{-10} = \frac{1.5-1}{6} \). Solving gives: \( \frac{1.5}{v} = \frac{1}{10} + \frac{0.5}{6} \). Calculate \( v \).
04

Calculate Resulting Image Distance

Solving the equation: \( \frac{1.5}{v} = \frac{1}{10} + \frac{1}{12} \) \( = \frac{5}{60} + \frac{6}{60} = \frac{11}{60} \). \( v = \frac{1.5 \times 60}{11} \approx 8.18 \) cm.
05

Find Distance Between Object and Image

The object and image are separated by \( |u - v| = |10 - 8.18| = 1.82 \) cm.
06

Determine Condition for Virtual Image

A virtual image occurs when the object is located closer than the principal focus. The principal focal length \( f \) is calculated as \( f = \frac{R}{n-1} = \frac{6.0}{0.5} = 12 \) cm.
07

Range for Virtual Image

The object must be located within 12 cm from the convex end to produce a virtual image, so the range is closer than 12 cm from the rod's surface.

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Key Concepts

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

Refractive Index
In optical physics, the refractive index is a crucial concept that tells us how fast light travels through different materials. It's a number that compares the speed of light in a vacuum to its speed in a given medium.
The formula for refractive index (\( n \)) is:
  • \( n = \frac{c}{v} \)where \( c \)is the speed of light in vacuum and \( v \)is the speed of light in the medium.
When light enters a medium like glass, with a refractive index greater than 1, it slows down and bends towards the normal. In our exercise setup, the refractive index of 1.5 for glass means light travels through the glass rod at 2/3 the speed it would in a vacuum.

Understanding the refractive index helps us calculate how and where images form when light passes through objects like lenses and rods.
Lens Maker's Formula
The Lens Maker's Formula is a powerful tool that lets us connect the shape and material of a lens to its optical properties. This formula is applied here to understand the interaction of light with the spherical surface of the glass rod.
For a single spherical surface, the formula is:
  • \(\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R} \)where:
    • \( n_1 \) is the refractive index of the medium from which light is coming (1 for air).
    • \( n_2 \) is the refractive index of the medium into which light is entering (1.5 for glass).
    • \( u \) is the object distance.
    • \( v \) is the image distance.
    • \( R \) is the radius of curvature.

Applying this, we find the point at which the image is formed by setting up and solving this equation with the convex surface. It's all about how the light bends when it passes from air into glass, guided by the material's curvature and refractive index. Understanding this formula is key to solving many optical physics problems involving lenses and curved surfaces.
Spherical Surface
A spherical surface is a curved shape like the one found at the end of the glass rod in our problem. It's important because it causes light to bend in a specific way, forming images at points that can be calculated using known principles and formulas.
Spherical surfaces have characteristics like:
  • They can have positive or negative curvature depending on their convex or concave nature.
  • The radius of curvature (\( R \)) indicates how curved the surface is, impacting how much the light rays bend when entering or exiting the medium.
In our exercise, the glass rod's convex spherical surface with a radius of 6 cm bends light to form real or virtual images.
By applying principles like the Lens Maker's Formula to spherical surfaces, we can determine the specific locations and nature of these images. This aids in understanding both practical applications like lenses in eyeglasses and more theoretical concepts like how different media interact with light.

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Most popular questions from this chapter

A moth at about eye level is \(10 \mathrm{~cm}\) in front of a plane mirror; you are behind the moth, \(30 \mathrm{~cm}\) from the mirror. What is the distance between your eyes and the apparent position of the moth's image in the mirror?

A glass sphere has radius \(R=5.0 \mathrm{~cm}\) and index of refraction 1.6. A paperweight is constructed by slicing through the sphere along a plane that is \(2.0 \mathrm{~cm}\) from the center of the sphere, leaving height \(h=3.0 \mathrm{~cm}\). The paperweight is placed on a table and viewed from directly above by an observer who is distance \(d=8.0 \mathrm{~cm}\) from the tabletop (Fig. 34-39). When viewed through the paperweight, how far away does the tabletop appear to be to the observer?

A grasshopper hops to a point on the central axis of a spherical mirror. The absolute magnitude of the mirror's focal length is \(40.0 \mathrm{~cm}\), and the lateral magnification of the image produced by the mirror is \(+0.200 .\) (a) Is the mirror convex or concave? (b) How far from the mirror is the grasshopper?

Figure \(34-56\) shows a beam expander made with two coaxial converging lenses of focal lengths \(f_{1}\) and \(f_{2}\) and separation \(d=f_{1}+f_{2}\). The device can expand a laser beam while keeping the light rays in the beam parallel to the central axis through the lenses. Suppose a uniform laser beam of width \(W_{i}=2.5 \mathrm{~mm}\) and intensity \(I_{i}=9.0 \mathrm{~kW} / \mathrm{m}^{2}\) enters a beam expander for which \(f_{1}=12.5\) \(\mathrm{cm}\) and \(f_{2}=30.0 \mathrm{~cm}\). What are (a) \(W_{f}\) and (b) \(I_{f}\) of the beam leaving the expander? (c) What value of \(d\) is needed for the beam expander if lens 1 is replaced with a diverging lens of focal length \(f_{1}=-26.0 \mathrm{~cm} ?\)

Two plane mirrors are placed parallel to each other and \(40 \mathrm{~cm}\) apart. An object is placed \(10 \mathrm{~cm}\) from one mirror. Determine the (a) smallest, (b) second smallest, (c) third smallest (occurs twice), and (d) fourth smallest distance between the object and image of the object.
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