/*! 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 29 A contact lens is made of plasti... [FREE SOLUTION] | 91Ó°ÊÓ

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A contact lens is made of plastic with an index of refraction of \(1.50 .\) The lens has an outer radius of curvature of \(+2.00 \mathrm{~cm}\) and an inner radius of curvature of \(+2.50 \mathrm{~cm}\). What is the focal length of the lens?

Short Answer

Expert verified
The focal length of the lens is 0.2 meters.

Step by step solution

01

Understand provided values and the lens maker's formula

Given, lens material's index of refraction(n)=1.5, outer radius of curvature(\(R_1\))=+2.00cm=+0.02m, inner radius of curvature(\(R_2\))=+2.50cm=+0.025m. We'll be using the LensMaker's Formula which is \(1/f = (n-1) * (1/R1 - 1/ R2)\), where 'f' is the focal length.
02

Substitute the values into the LensMaker's Formula

Substitute the values of n, R1, and R2 in the LensMaker's formula to calculate the focal length of the lens \(1/f = (1.5 - 1) * (1/0.02 - 1/0.025)\)
03

Calculate the focal length

On solving, we get \(1/f = (0.5) * (50 -40) = 5\). Hence, the focal length is \(f = 1/5 = 0.2\) meters.

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Key Concepts

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

Index of Refraction
The index of refraction, often denoted by 'n', is a measure of how much the speed of light is reduced within a particular material, compared to its speed in a vacuum. It provides us with an understanding of how much light bends, or refracts, when it enters the material. Materials with a higher index of refraction will bend light more, as the light travels slower through them compared to materials with a lower index of refraction.

The index of refraction is crucial when designing lenses because it determines the bending ability of the lens material. In the given problem, the index of refraction of the plastic used in the contact lens is 1.50. This value implies that light travels 1.50 times slower in the plastic than it does in a vacuum. In other words, the higher refractive index, the more powerful the lens for a given curvature, leading to a shorter focal length.
Radius of Curvature
The radius of curvature of a lens is the radius of the 'spherical' surface portions that make up the lens. Simply put, it's the radius of the imaginary sphere where the lens's curved surface segments could come from. This radius is implicated in determining the strength and type of lens. The radius can be positive or negative, indicating the direction in which the lens curves. A positive radius corresponds to a surface that curves away from the light source (convex), while a negative radius means it curves towards the light (concave).

In our example, the contact lens has an outer radius of curvature of +2.00 cm and an inner radius of curvature of +2.50 cm. The sign convention (+) suggests that both surfaces are convex. The varying radii lead to different amounts of bending of light rays and are instrumental in calculating the lens's focal length using the lens maker's formula.
Focal Length of a Lens
The focal length of a lens is the distance from the lens to its focus, where parallel rays of light converge after passing through the lens. In simple terms, it's the measure of how strongly the lens converges or diverges light. A short focal length indicates that the lens has a strong bending power, quickly bringing light rays together, ideal for sharper focusing over short distances. Conversely, a long focal length suggests a weaker bending power suitable for focusing on distant objects.

The focal length depends on the lens's index of refraction and radii of curvature, as illustrated by the lens maker's formula: \(1/f = (n-1) * (1/R_1 - 1/R_2)\). Using the given radii of curvature and the index of refraction for the plastic, we can solve for the contact lens's focal length, finding it to be 0.2 meters. This focal length provides us with the understanding of how the lens will behave when it's worn, particularly how it will focus light on the retina to correct vision.

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Most popular questions from this chapter

A woman holds a tube of lipstick \(10.0 \mathrm{~cm}\) from a spherical mirror and notices that the image of the tube is upright and half its normal size, (a) Determine the position of the image. Is it in front of or behind the mirror? (b) Calculate the focal length of the mirror.

An object \(2.00 \mathrm{~cm}\) high is placed \(40.0 \mathrm{~cm}\) to the left of a converging lens having a focal length of \(30.0 \mathrm{~cm}\). A diverging lens having a focal length of \(-20.0 \mathrm{~cm}\) is placed \(110 \mathrm{~cm}\) to the right of the converging lens. (a) Determine the final position and magnification of the final image. (b) Is the image upright or inverted? (c) Repeat parts (a) and (b) for the case where the second lens is a converging lens having a focal length of \(+20.0 \mathrm{~cm}\).

A \(2.00-\mathrm{cm}\) -high object is placed \(3.00 \mathrm{~cm}\) in front of a concave mirror. If the image is \(5.00 \mathrm{~cm}\) high and virtual, what is the focal length of the mirror?

A convergent lens with a \(50.0-\mathrm{mm}\) focal length is used to focus an image of a very distant scene onto a flat screen \(35.0 \mathrm{~mm}\) wide. What is the angular width \(\alpha\) of the scene included in the image on the screen?

Two converging lenses, each of focal length \(15.0 \mathrm{~cm}\), are placed \(40.0 \mathrm{~cm}\) apart, and an object is placed \(30.0 \mathrm{~cm}\) in front of the first lens. Where is the final image formed. and what is the magnification of the system?

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