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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

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?

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.

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?

Contact lenses are placed right on the eyeball, so the distance from the eye to an object (or image) is the same as the distance from the lens to that object (or image). A certain person can see distant objects well, but his near point is \(45.0 \mathrm{~cm}\) from his eyes instead of the usual \(25.0 \mathrm{~cm}\). (a) Is this person nearsighted or farsighted? (b) What type of lens (converging or diverging) is needed to correct his vision? (c) If the correcting lenses will be contact lenses, what focal length lens is needed and what is its power in diopters?

A transparent rod \(30.0 \mathrm{~cm}\) long is cut flat at one end and rounded to a hemispherical surface of radius \(10.0 \mathrm{~cm}\) at the other end. A small object is embedded within the rod along its axis and halfway between its ends, \(15.0 \mathrm{~cm}\) from the flat end and \(15.0 \mathrm{~cm}\) from the vertex of the curved end. When the rod is viewed from its flat end, the apparent depth of the object is \(8.20 \mathrm{~cm}\) from the flat end. What is its apparent depth when the rod is viewed from its curved end?

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