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Chapter 34: Problem 4

A concave mirror has a radius of curvature of \(34.0 \mathrm{~cm}\). (a) What is its focal length? (b) If the mirror is immersed in water (refractive index 1.33 ), what is its focal length?

Short Answer

Expert verified
The focal length of the concave mirror is \(17.0 \mathrm{~cm}\) in air and \(12.8 \mathrm{~cm}\) when it is immersed in water.

Step by step solution

01

Find the initial focal length

For a mirror, whether it is concave or convex, the radius of curvature \(R\) is twice the focal length \(f\). Hence, we can find the focal length by dividing the radius of curvature by 2. In this case, the radius of curvature \(R\) is given as \(34.0 \mathrm{~cm}\). Therefore, we can calculate the focal length \(f\) as \(f = R/2 = 34.0 \mathrm{~cm} / 2 = 17.0 \mathrm{~cm}\).
02

Calculate the new focal length in water

When a mirror is immersed in a medium other than air, its focal length changes. We can calculate the new focal length \(f^{'}\) using the formula \(f^{'} = f/n\), where \(n\) is the refractive index of the medium. In this case, the refractive index of water is given as \(n = 1.33\). Therefore, the new focal length in water \(f^{'}\) is \(f^{'} = 17.0 \mathrm{~cm} / 1.33 = 12.8 \mathrm{~cm}\).

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Key Concepts

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

Radius of Curvature
In the realm of optics, the radius of curvature is a fundamental concept that signifies the radius of an imaginary circle that has the same curvature as a given optical surface at a specific point. For mirrors, this is particularly important as it defines how parallel rays will converge after reflection.

For concave mirrors, such as in the exercise under consideration, the radius of curvature is essentially twice the distance from the mirror's surface to its focal point鈥攖hat is, the point where light rays converge after reflection. The relationship is mathematically expressed as \( R = 2f \) , where \( R \) is the radius of curvature, and \( f \) is the focal length. Understanding this relationship is crucial for answering part (a) of the exercise, where we directly use it to deduce the focal length from the given radius of curvature.
Optics in Different Media
The interaction of light with different media is a key aspect of optics. When light passes through different media or reflects off surfaces in different media, its behavior changes. Optical phenomena such as reflection, refraction, and dispersion are influenced by the medium in which they occur.

In the given exercise, we see an example of this with a concave mirror being immersed in water. The presence of water, a medium with a different density than air, modifies how the light is reflected by the mirror. This immersion doesn't change the physical curvature of the mirror but changes the effective focal length of the system鈥攖his is because the speed of light and its interaction with the mirror鈥檚 surface are altered due to the optical density of the new medium.
Refractive Index
The refractive index, denoted as \( n \) , is a dimensionless number that describes how light propagates through a medium. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium: \( n = c / v \) , where \( c \) is the speed of light in a vacuum, and \( v \) is the speed of light in the medium. The refractive index determines how much the path of light is bent, or refracted, when entering a material.

In part (b) of our exercise, this concept is used to calculate the new focal length of the concave mirror when immersed in water. Water's refractive index (1.33) indicates that light travels slower in water than in air. By applying the refractive index to the mirror's focal length in air, we can find the focal length in the new medium, highlighting the refractive index's influence on optical systems.

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Most popular questions from this chapter

The science museum where you work is constructing a new display. You are given a glass rod that is surrounded by air and was ground on its left-hand end to form a hemispherical surface there. You must determine the radius of curvature of that surface and the index of refraction of the glass. Remembering the optics portion of your physics course, you place a small object to the left of the rod, on the rod's optic axis, at a distance \(s\) from the vertex of the hemispherical surface. You measure the distance \(s^{\prime}\) of the image from the vertex of the surface, with the image being to the right of the vertex. Your measurements are as follows: $$\begin{array}{l|rrrrrr} \boldsymbol{s}(\mathbf{c m}) & 22.5 & 25.0 & 30.0 & 35.0 & 40.0 & 45.0 \\ \hline s^{\prime}(\mathbf{c m}) & 271.6 & 148.3 & 89.4 & 71.1 & 60.8 & 53.2 \end{array}$$ Recalling that the object-image relationships for thin lenses and spherical mirrors include reciprocals of distances, you plot your data as \(1 / s^{\prime}\) versus \(1 / s .\) (a) Explain why your data points plotted this way lie close to a straight line. (b) Use the slope and \(y\) -intercept of the best-fit straight line to your data to calculate the index of refraction of the glass and the radius of curvature of the hemispherical surface of the rod. (c) Where is the image if the object distance is \(15.0 \mathrm{~cm} ?\)

An object with height \(4.00 \mathrm{~mm}\) is placed \(28.0 \mathrm{~cm}\) to the left of a converging lens that has focal length \(8.40 \mathrm{~cm} .\) A second lens is placed \(8.00 \mathrm{~cm}\) to the right of the converging lens. (a) What is the focal length of the second lens if the final image is inverted relative to the \(4.00-\mathrm{mm}\) -tall object and has height \(5.60 \mathrm{~mm} ?\) (b) What is the distance between the original object and the final image?

A person with a near point of \(85 \mathrm{~cm},\) but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest \(2.0 \mathrm{~cm}\) in front of his eye? (b) What would his near point be if his old glasses were contact lenses with the same power instead?

Three thin lenses, each with a focal length of \(40.0 \mathrm{~cm},\) are aligned on a common axis; adjacent lenses are separated by \(52.0 \mathrm{~cm} .\) Find the position of the image of a small object on the axis, \(80.0 \mathrm{~cm}\) to the left of the first lens.

A thin lens with a focal length of \(6.00 \mathrm{~cm}\) is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, \(25.0 \mathrm{~cm}\) from the eye, and that the lens is very close to the eye.
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