/*! 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 62 The focal length of a mirror can... [FREE SOLUTION] | 91Ó°ÊÓ

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The focal length of a mirror can be determined entirely from the shape of the mirror. In contrast, to determine the focal length of a lens, we must know both the shape of the lens and its index of refraction - as well as the index of refraction of the surrounding medium. For instance, when a thin lens is immersed in a liquid, we must modify the thin-lens equation to take into account the refractive properties of the surrounding liquid: $$\frac{1}{f}=\left(\frac{n}{n_{\mathrm{liq}}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right),$$ where \(n_{\text {liq }}\) is the index of refraction of the liquid and \(n\) is the index of refraction of the glass. If you place a glass lens \((n=1.5),\) which has a focal length of \(0.5 \mathrm{~m}\) in air, into a tank of water \((n=1.33),\) what will happen to its focal length? A. Nothing will happen. B. The focal length of the lens will be reduced. C. The focal length of the lens will be increased. D. There is not enough information to answer the question.

Short Answer

Expert verified
C. The focal length of the lens will be increased.

Step by step solution

01

Understanding Parameters

We are given a glass lens with a refractive index \(n = 1.5\) and a focal length \(f = 0.5\,\text{m}\) in air. The lens is placed in water, which has a refractive index \(n_{\text{liq}} = 1.33\).
02

Analyze the Thin-Lens Equation

The thin-lens equation needs to account for the surrounding liquid. The modified equation is: \[ \frac{1}{f'} = \left(\frac{n}{n_{\text{liq}}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] where \(f'\) is the new focal length in the liquid.
03

Compare the Equation Components

In air, \(n_{\text{air}} = 1.0\), and the original focal length equation was: \[ \frac{1}{f} = \left(n - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]. In water, the modified component is \(\frac{n}{n_{\text{liq}}} - 1\), which is less than \(n - 1\).
04

Effect on Focal Length

Since \(\frac{n}{n_{\text{liq}}} - 1\) is smaller than \(n - 1\), the term \(\frac{1}{f'}\) becomes smaller than \(\frac{1}{f}\). This implies \(f' > f\), meaning the focal length increases.
05

Conclusion

Placing the lens in water increases its focal length. Thus, the focal length of the lens will be increased.

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Key Concepts

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

Focal Length
The focal length of a lens is a fundamental concept in optics. It is the distance from the lens where parallel rays of light converge to a single focal point. This length is a key determining factor in the optical power of the lens. A short focal length means that the lens has high optical power and bends light more sharply. Conversely, a longer focal length indicates lower optical power and a gentler curvature of light.
Understanding a lens's focal length helps in manipulating how images are formed, and it is essential in designing optical instruments like cameras, glasses, and microscopes.
Refractive Index
The refractive index, often denoted by the symbol $n$, measures how much a beam of light bends when it enters a material. When light moves from one medium to another, its speed changes, resulting in refraction. This index is calculated by the ratio of the speed of light in a vacuum to the speed of light in the medium.
Knowing the refractive index of a material is crucial because it impacts how much light will be refracted. For example, the refractive index of air is approximately 1.0, while glass typically has an index around 1.5. Water's refractive index is about 1.33. Each material's refractive index helps predict how it will alter the light's path.
Thin-Lens Equation
The thin-lens equation is a fundamental formula used to relate the focal length of a lens to the physical properties of the lens and its environment. The standard form of the thin-lens equation is given by: \[ \frac{1}{f} = \left(n - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \] where \(R_1\) and \(R_2\) are the radii of curvatures of the lens's surfaces, and \(n\) is the refractive index of the lens.
This equation is essential for designing lenses to achieve a desired focal length by selecting appropriate materials and curvatures. In scenarios like lens immersion, modifications are made to the thin-lens equation to consider different surrounding mediums.
Lens Immersion
Lens immersion involves submerging a lens into a different material, altering its optical properties. When a lens is placed in a liquid, the surrounding medium's refractive index must be considered. This is because the velocity of light changes as it transitions between different materials, affecting the focal length of the lens.
Immersing a lens in a medium like water changes the thin-lens equation to account for the new environment. As seen in the problem, by substituting the liquid's refractive index into the equation, you can determine the new focal length in this medium.
Liquid Refraction
Liquid refraction occurs when light travels through a liquid, bending as a result of changes in the medium. This bending is defined by the refractive index of the liquid. In the case of lenses, this means that the lens’s effective focal length can be altered by the liquid it is submerged in.
Understanding liquid refraction is crucial for applications where lenses need to be used underwater or with liquid-based optical systems. This concept explains why a lens might focus light differently under water compared to in the air, due to a relative change in refractive indices between the glass and the liquid, affecting image formation.

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Most popular questions from this chapter

The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about \(8.0 \mathrm{~mm},\) which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the radii of curvature of this lens. (b) If an object \(16 \mathrm{~cm}\) tall were placed \(30.0 \mathrm{~cm}\) from the eye lens, where would the lens focus it and how tall would the image be? Is this image real or virtual? Is it upright or inverted? (Note: The results obtained here are not strictly accurate because the lens is embedded in fluids having refractive indices different from that of air.)

The diameter of Mars is \(6794 \mathrm{~km}\), and its minimum distance from the earth is \(5.58 \times 10^{7} \mathrm{~km} .\) (a) When Mars is at this distance, find the diameter of the image of Mars formed by a spherical, concave telescope mirror with a focal length of \(1.75 \mathrm{~m}\). (b) Where is the image located?

A spherical, concave shaving mirror has a radius of curvature of \(32.0 \mathrm{~cm} .\) (a) What is the magnification of a person's face when it is \(12.0 \mathrm{~cm}\) to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.

The focal length of a mirror can be determined entirely from the shape of the mirror. In contrast, to determine the focal length of a lens, we must know both the shape of the lens and its index of refraction - as well as the index of refraction of the surrounding medium. For instance, when a thin lens is immersed in a liquid, we must modify the thin-lens equation to take into account the refractive properties of the surrounding liquid: $$\frac{1}{f}=\left(\frac{n}{n_{\mathrm{liq}}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right),$$ where \(n_{\text {liq }}\) is the index of refraction of the liquid and \(n\) is the index of refraction of the glass. If you place a concave glass lens into a tank of a liquid that has an index of refraction that is greater than that of the lens, what will happen? A. The lens will no longer be able to create any images. B. The focal length of the lens will become longer. C. The focal length of the lens will become shorter. D. The lens will become a converging lens.

An object is placed \(18.0 \mathrm{~cm}\) from a screen. (a) At what two points between object and screen may a converging lens with a \(3.00 \mathrm{~cm}\) focal length be placed to obtain an image on the screen? (b) What is the magnification of the image for each position of the lens?

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