/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 What should be the index of refr... [FREE SOLUTION] | 91Ó°ÊÓ

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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 1.25.

Step by step solution

01

Understanding the Problem

We need to determine the index of refraction for a transparent sphere such that light rays from an infinitely distant object are focused at the vertex on the opposite side. This scenario resembles a lens focusing light.
02

Apply the Lensmaker's Formula

The Lensmaker's Formula relates the focal length ( f ) of a lens to the radii of curvature ( R_1 , R_2 ) and the index of refraction ( n ). For a sphere, the formula simplifies since R_1 = R and R_2 = -R . Thus, the formula is: \[ \frac{1}{f} = (n-1)\left(\frac{1}{R} - \frac{1}{-R}\right) = (n-1)\left(\frac{2}{R}\right). \]
03

Set the Focal Length to Radius of the Sphere

Since we want the rays to focus at the vertex opposite the point of incidence, the focal length will be equal to the diameter of the sphere, f = 2R. We substitute this into the equation from the Lensmaker's Formula to find n: \[ \frac{1}{2R} = (n-1)\left(\frac{2}{R}\right). \] Simplifying this gives: \[ \frac{1}{2} = 2(n-1). \]
04

Solve for the Index of Refraction

Rearrange the equation \frac{1}{2} = 2(n-1) to solve for n: \[ \frac{1}{2} = 2n - 2. \] Add 2 to both sides: \[ 2.5 = 2n. \] Divide by 2: \[ n = 1.25. \]
05

Final Step: Verification of Solution

Verify that with n = 1.25 , the focal point is indeed at the opposite vertex of the sphere. Given the symmetry of the sphere and the calculations conforming to the formula, the solution is consistent. The index of refraction required for the focus is 1.25.

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Key Concepts

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

Lensmaker's Formula
When designing lenses, it's crucial to determine how they will focus light. This is where the Lensmaker's Formula plays a vital role. The Lensmaker's Formula links the focal length of a lens with the radii of curvature of its two surfaces and the index of refraction. It tells us how the lens will bend light.

For lenses that are spheres, this formula simplifies due to their symmetry. The spherical design means both surfaces have the same radius, but one side curves inwards while the other outwards. Therefore, if we consider a sphere, these radii can be represented as \( R_1 = R \) and \( R_2 = -R \).
  • The Lensmaker's Formula for a sphere becomes: \[ \frac{1}{f} = (n-1)\left(\frac{1}{R} - \frac{1}{-R}\right) = (n-1)\left(\frac{2}{R}\right). \]
This equation indicates that the focal length depends on both the radius of the sphere and the refractive index. Understanding this relationship allows us to manipulate factors to get desired focusing properties.
Focal Length
The focal length is a fundamental concept when discussing lenses. It defines the distance over which initially parallel rays converge or diverge. In the context of a transparent sphere used as a lens, the focal length is essential in predicting where light rays will concentrate.

For our transparent sphere, the focal length is chosen to be equal to its diameter. This means we want light from an infinitely distant object to converge at the point opposite where it enters the sphere. Mathematically, this means:
  • Focal length \( f = 2R \) (where \( R \) is the radius of the sphere).
Plugging this into the simplified Lensmaker's formula provides the conditions needed for focusing, helping us determine the necessary index of refraction. This direct relationship illustrates the physics behind the focusing properties of the lens.
Spherical Lenses
Spherical lenses are a type of optical lens where the surfaces are parts of a sphere, either convex or concave. This shape is simple yet powerful, often used in various optical applications because of its ability to accurately focus light.

Understanding spherical lenses involves appreciating their symmetry and its impact on light behavior. Both principal radii of such lenses are equal, making spherical lenses easier to analyze than their aspherical counterparts.
  • Spherical lenses focus light due to their curved surfaces, which direct incoming light rays to meet at a focal point.
  • The focal point's location can vary based on the lens's curvature and material properties, like the index of refraction.
In our problem, we explored how changing the index of refraction allows a spherical lens to focus at specific distances. This understanding is crucial for applications involving image formation and light manipulation.

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Most popular questions from this chapter

A light bulb is 4.00 \(\mathrm{m}\) from a wall. You are to use a concave mirror to project an image of the bulb on the wall, with the image 2.25 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?

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?

Dental Mirror. A dentist uses a curved mirror to view teeth on the upper side of the mouth. Suppose she wants an enect image with a magnification of 2.00 when the mirror is 1.25 \(\mathrm{cm}\) from a tooth. (Treat this problem as though the object and image lie along a straight line.) (a) What kind of mirror (concave or convex) is needed? Use a ray diagram to decide, without performing any calculations. (b) What must be the focal length and radius of curvature of this mirror?(c) Draw a principal-ray diagram to check your answer in part (b).

A glass plate 3.50 \(\mathrm{cm}\) thick, with an index of refraction of 1.55 and plane parallel faces, is held with its faces horizontal and its lower face 6.00 \(\mathrm{cm}\) above a printed page. Find the position of the image of the page formed by rays making a small angle with the normal to the plate.

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?

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