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You hold a spherical salad bowl \(60 \mathrm{~cm}\) in front of your face with the bottom of the bowl facing you. The bowl is made of polished metal with a \(35 \mathrm{~cm}\) radius of curvature. (a) Where is the image of your \(5.0-\mathrm{cm}\) -tall nose located? (b) What are the image's size, orientation, and nature (real or virtual)?

Short Answer

Expert verified
The image of your nose is located 20 cm behind the mirror. Its size is 1.67 cm, it's upright and virtual.

Step by step solution

01

Find the focal length.

The radius of curvature is given as \(35 \mathrm{~cm}\). We can find the focal length using the relation \(f = R/2\), which gives \(f = 35 \mathrm{~cm}/2 = 17.5 \mathrm{~cm}\). Since we know it’s a convex mirror, the focal length is considered negative. So, \(f = -17.5 \mathrm{~cm}\).
02

Calculate the object distance.

We are told that the mirror (bowl) is held \(60 \mathrm{~cm}\) from the face. So, the object distance \(d_o\) is \(60 \mathrm{~cm}\). This is a real object so its distance is positive.
03

Calculate the image distance using the mirror equation.

Using the mirror equation, \(1/f = 1/d_o + 1/d_i\), we substitute values and solve for \(d_i\). After substitution, we get an equation \(-1/17.5 = 1/60 + 1/d_i\). Solving for \(d_i\) we get \(d_i = -20 \mathrm{~cm}\). The negative sign indicates that the image is virtual and located on the same side as the light source.
04

Calculate the image's size using the magnification equation.

The magnification equation is \(m = -d_i / d_o = h_i / h_o\). Here, \(d_i = -20 \mathrm{cm}\), \(d_o = 60 \mathrm{~cm}\), and \(h_o = 5.0 \mathrm{cm}\). So, the magnification is \(m = 20 / 60 = 1/3\). Since the magnification is less than 1, our image will be smaller than the object. Using \(h_i = m∗h_o\), we find \(h_i = (1/3) * 5 \mathrm{~cm} = 1.67 \mathrm{~cm}\).
05

Determine the image's orientation and nature.

Because the magnification is positive, the image is upright. And since the image distance \(d_i\) is negative, that means it’s a virtual image. This is expected for a convex mirror.

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Key Concepts

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

Mirror Equation
Understanding the mirror equation is crucial when dealing with optics, especially when analyzing convex mirrors. A crucial formula, it relates the focal length ( f ), the object distance ( d_o ), and the image distance ( d_i ). The equation is given by:
\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]
This formula helps us find where the image will form with respect to the mirror's surface. Solving this equation allows us to determine whether an image is real or virtual, and thus, whether it's something you can actually "catch" on a screen or if it's just a reflection.
The sign conventions are important: in convex mirrors, the focal length is negative, affecting the calculations.
Focal Length
The focal length of a mirror is a key aspect that describes its optical characteristics. For spherical mirrors, the focal length f is half of the radius of curvature (R) :
\[ f = \frac{R}{2} \]
In the case of a convex mirror, often used in automobilists’ mirrors to provide a wider field of view, the focal length is negative. This is because parallel rays appear to "diverge" from a point behind the mirror.
Thus, the focal length provides a measure of how much the mirror will converge or diverge light rays, directly impacting the positioning of the image.
Image Magnification
Image magnification helps us understand how the size of the image relates to the size of the object. It tells us if an image is bigger, smaller, or the same size as the object it reflects.
Magnification (m) can be calculated using:
\[ m = \frac{-d_i}{d_o} = \frac{h_i}{h_o} \]
Where d_i is the image distance, d_o is the object distance, h_i is the image height, and h_o is the object height.
For convex mirrors, the magnification is usually less than 1, indicating the image formed is smaller than the actual object, as it is a feature of diverging surfaces. Importantly, if (m) is positive, the image remains upright relative to the object.
Virtual Image
A virtual image is a fundamental concept in the study of optics, particularly when involving convex mirrors. Unlike real images, virtual images cannot be projected onto a screen.
They appear to be located behind the mirror where light does not actually reach. In convex mirrors, all images are virtual due to the nature of light rays diverging after reflection.
This results in the image distance (d_i) being negative in the mirror equation, confirming the image's virtual nature. Virtual images appear smaller and upright compared to the object in convex mirrors. They are an integral part of certain applications, like vehicle rearview mirrors, where they provide a wider field of view.
Optics
Optics is the branch of physics that deals with light and its interactions with different materials. Understanding how light behaves is vital for several applications, especially mirrors.
Convex mirrors are an essential component in optics; they have unique properties that make them invaluable in various practical scenarios. The study of optics involves exploring concepts like refraction, reflection, and the behavior of light as either a particle or wave. Through optics, you'll learn about how different wavelengths of light interact with materials and their practical applications such as lenses in glasses or cameras. In the context of mirrors, it guides us to comprehensively analyze image orientation, size, and nature. Optics bridges theoretical understanding and practical technology, allowing innovations like telescopes and optical fibers.

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Most popular questions from this chapter

It is your first day at work as a summer intern at an optics company. Your supervisor hands you a diverging lens and asks you to measure its focal length. You know that with a converging lens, you can measure the focal length by placing an object a distance \(s\) to the left of the lens, far enough from the lens for the image to be real, and viewing the image on a screen that is to the right of the lens. By adjusting the position of the screen until the image is in sharp focus, you can determine the image distance \(s^{\prime}\) and then use Eq. (34.16) to calculate the focal length \(f\) of the lens. But this procedure won't work with a diverging lens - by itself, a diverging lens produces only virtual images, which can't be projected onto a screen. Therefore, to determine the focal length of a diverging lens, you do the following: First you take a converging lens and measure that, for an object \(20.0 \mathrm{~cm}\) to the left of the lens, the image is \(29.7 \mathrm{~cm}\) to the right of the lens. You then place a diverging lens \(20.0 \mathrm{~cm}\) to the right of the converging lens and measure the final image to be \(42.8 \mathrm{~cm}\) to the right of the converging lens. Suspecting some inaccuracy in measurement, you repeat the lenscombination measurement with the same object distance for the converging lens but with the diverging lens \(25.0 \mathrm{~cm}\) to the right of the converging lens. You measure the final image to be \(31.6 \mathrm{~cm}\) to the right of the converging lens. (a) Use both lens-combination measurements to calculate the focal length of the diverging lens. Take as your best experimental value for the focal length the average of the two values. (b) Which position of the diverging lens, \(20.0 \mathrm{~cm}\) to the right or \(25.0 \mathrm{~cm}\) to the right of the converging lens, gives the tallest image?

The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns \((\mu \mathrm{m})\) is typical near the center of the eye. We shall model the eye as a sphere \(2.50 \mathrm{~cm}\) in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells \(5.0 \mu \mathrm{m}\) in diameter. (a) What is the smallest object you can resolve at a near point of \(25 \mathrm{~cm} ?\) (b) What angle is subtended by this object at the eye? Express your answer in units of minutes \(\left(1^{\circ}=60 \mathrm{~min}\right),\) and compare it with the typical experimental value of about \(1.0 \mathrm{~min} .\) (Note: There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.)

A dentist uses a curved mirror to view teeth on the upper side of the mouth. Suppose she wants an erect 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 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?

A concave mirror is to form an image of the filament of a headlight lamp on a screen \(8.00 \mathrm{~m}\) from the mirror. The filament is \(6.00 \mathrm{~mm}\) tall, and the image is to be \(24.0 \mathrm{~cm}\) tall. (a) How far in front of the vertex of the mirror should the filament be placed? (b) What should be the radius of curvature of the mirror?

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