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What should be the index of refraction of a transparent sphere in order for paraxial rays from an infinitely distant object to be brought to a focus at the vertex of the surface opposite the point of incidence?

Short Answer

Expert verified
The index of refraction should be 0.5.

Step by step solution

01

Understanding the Problem

We have a transparent sphere and need to find the index of refraction that will focus paraxial rays from an infinitely distant object on the opposite vertex.
02

Identify Relevant Formula

The paraxial focal point condition for a sphere is given by the Lensmaker's Equation for a spherical refractive surface: \( n_2/v - n_1/u = (n_2-n_1)/R \), where \( n_1 \) is the refractive index of air (\(1\)), \( u \) is infinity (rays coming from infinitely distant object), \( v \) is the focal length, \( R \) is the radius of the sphere, and \( n_2 \) is the sought index of refraction for the sphere.
03

Simplify the Equation

Given \( u = \infty \), the term \( n_1/u \) becomes \(0\). The equation simplifies to \( n_2/v = (n_2 - 1)/R \).
04

Identify Condition

To focus at the vertex, \( v = -R \), because the focus should be at the edge of the sphere's surface opposite the entry surface.
05

Substitute Known Values

Substitute \( v = -R \) in the simplified equation: \( n_2/(-R) = (n_2 - 1)/R \).
06

Solve for Index of Refraction

Simplify and solve the equation: \(-n_2/R = (n_2 - 1)/R \). Cancel \( R \), and simplify to find \( -n_2 = n_2 - 1 \). Thus, \( 2n_2 = 1 \), which leads to \( n_2 = 0.5 \).

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Key Concepts

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

Paraxial Rays
Paraxial rays are rays of light that are close to and nearly parallel to the principal axis of an optical system. In other words, these rays travel at small angles relative to the axis that passes through the center of a lens or mirror. This simplification is often used in optical calculations to make the mathematics more manageable and provide accurate results.

Paraxial rays are a crucial part of understanding how lenses and mirrors work, especially when we're considering the focusing of light. These rays behave predictably according to the Gaussian optics approximation, which assumes that angles are small and calculations are simplified. We often refer to problems involving paraxial rays as 'paraxial optics issues.'

By approximating light paths with paraxial rays, we can use simplified mathematical formulas to predict how light will behave. The equations we derive are much simpler than those used for general, non-paraxial rays, and they serve as the foundation for designing lenses and optical systems.
Lensmaker's Equation
The Lensmaker's Equation is a fundamental formula used in optics to determine the focal length of a lens based on the curvature of its surfaces and the refractive index of the material. For a simple lens in air, the equation is given by:

\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

where \( n \) is the refractive index of the lens material, \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces, and \( f \) is the focal length.

In the given exercise, we modified this equation to unify how a sphere, rather than a traditional lens, focuses light. This special case of the Lensmaker's Equation for spheres simplifies the problem by treating one radius, \( R \), as constant and focusing on the refractive index.

The modified equation becomes especially useful in engineering spherical optics, like those found in camera lenses or eyeglass lenses, where understanding how light refracts through curved, transparent materials is essential.
Sphere Optics
Sphere optics deals with the refraction and focusing properties of spherical shapes. When examining a transparent sphere, we must consider how light bends as it passes through the sphere's surface. This is guided by geometric optics principles, where light changes direction at the interface of different refractive indices.

The simplicity of a sphere offers a special focusing condition. Light coming from infinity will focus at a point on the opposite side of the sphere, directly along the line connecting the two points. This exercise capitalizes on this principle by using the unique curvature and uniform index of a sphere to determine where light will converge.

In sphere optics, understanding the relationships between the refractive index, radius of curvature, and focal point is crucial. By adjusting these variables, we can design spheres to act as lenses, controlling the path of light in various sophisticated optical systems. Optical engineers often use these principles to create highly precise instruments for medical, scientific, and consumer uses.

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Most popular questions from this chapter

A speck of dirt is embedded \(3.50 \mathrm{~cm}\) below the surface of a sheet of ice \((n=1.309)\). What is its apparent depth when viewed at normal incidence?

An insect \(3.75 \mathrm{~mm}\) tall is placed \(22.5 \mathrm{~cm}\) to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude \(13.0 \mathrm{~cm}\), and the index of refraction of the lens material is \(1.70 .\) (a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted? (b) Repeat part (a) if the lens is reversed.

A certain microscope is provided with objectives that have focal lengths of \(16 \mathrm{~mm}, 4 \mathrm{~mm}\), and \(1.9 \mathrm{~mm}\) and with eyepieces that have angular magnifications of \(5 x\) and \(10 \times\). Each objective forms an image \(120 \mathrm{~mm}\) beyond its second focal point. Determine (a) the largest overall angular magnification obtainable and (b) the smallest overall angular magnification obtainable.

A ray of light passes through a prism in a principle plane the deviation being equal to angle of incidence which is equal to \(2 \alpha\). It is given that \(\boldsymbol{\alpha}\) is the angle of prism. Show that \(\cos ^{2} \alpha=\frac{\left(\mu^{2}-1\right)}{8}\). where \(\mu\) is the refractive index of the material of prism.

A Glass Red. Both ends of a glass rod with index of refraction \(1.60\) are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is \(6.00 \mathrm{~cm}\), and the radius of curvature at the right end is \(12.0 \mathrm{~cm}\). The length of the rod between vertices is \(40.0 \mathrm{~cm}\). The object for the surface at the left end is an arrow that lies \(23.0 \mathrm{~cm}\) to the left of the vertex of this surface. The arrow is \(1.50 \mathrm{~mm}\) tall and at right angles to the axis. (a) What constitutes the object for the surface at the right end of the rod? (b) What is the object distance for this surface? (c) Is the object for this surface real or virtual? (d) What is the position of the final image? (e) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (f) What is the height of the final image?

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