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A convex spherical mirror with a focal length of magnitude 24.0 cm is placed 20.0 cm to the left of a plane mirror. An object 0.250 cm tall is placed midway between the surface of the plane mirror and the vertex of the spherical mirror. The spherical mirror forms multiple images of the object. Where are the two images of the object formed by the spherical mirror that are closest to the spherical mirror, and how tall is each image?

Short Answer

Expert verified
The two images are located at -7.06 cm and -15.48 cm from the mirror, with heights 0.1765 cm and 0.08225 cm respectively.

Step by step solution

01

Identify the Setup

We have two mirrors: a convex spherical mirror with a focal length \(|f| = 24.0\) cm, and a plane mirror. The distance between the convex mirror and the plane mirror is 20.0 cm. The object is placed midway, i.e., 10.0 cm from each mirror.
02

Use the Mirror Equation for Convex Mirror

The mirror equation is \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). For a convex mirror, the focal length \(f\) is negative, so \(f = -24.0\) cm.The object distance \(d_o\) from the convex mirror is 10.0 cm.
03

Calculate First Image by Convex Mirror

Substitute \(f = -24.0\) cm and \(d_o = 10.0\) cm into the mirror equation:\[ \frac{1}{-24} = \frac{1}{10} + \frac{1}{d_{i1}} \]Solving gives \(d_{i1} = -7.0588\) cm (negative indicating a virtual image behind the mirror).
04

Calculate Image Height for First Image

The magnification \(m\) is given by \(-\frac{d_i}{d_o}\). For the first image:\[m_1 = -\frac{-7.0588}{10} = 0.7059\]The image height \(h_i = m \times h_o\), so:\[h_{i1} = 0.7059 \times 0.250 = 0.1765\] cm.
05

Calculate Second Image by Plane Mirror

The first virtual image by the convex mirror acts as an object for the plane mirror. This image's distance from the plane mirror is \(20.0 + 7.0588 = 27.0588\) cm on the left side, which forms a second image 27.0588 cm behind the plane mirror.
06

Reflected Image by Convex Mirror

The reflected image by the plane mirror acts as another object for the convex mirror at a distance of \(20 + 27.0588 = 47.0588\) cm from it. Use the mirror equation again:\[\frac{1}{-24} = \frac{1}{47.0588} + \frac{1}{d_{i2}}\]Solving gives \(d_{i2} = -15.4839\) cm.
07

Calculate Image Height for Second Image

The magnification \(m\) for the second image is:\[m_2 = -\frac{-15.4839}{47.0588} = 0.329 \] Thus, \(h_{i2} = 0.329 \times 0.250 = 0.08225\) cm.

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Key Concepts

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

Convex Mirror
A convex mirror, sometimes called a diverging mirror, has a surface that curves outward. Unlike concave mirrors, which can focus light, convex mirrors do the opposite by spreading light out. This makes them valuable for providing a wide field of view. You often see them used in places like parking garages or stores for security.
  • Convex mirrors always form virtual images because reflected light from a convex mirror diverges.
  • The images are usually upright and smaller than the actual object.

This is key in the exercise, where the convex mirror's properties help form the images of the object in unique ways.
Mirror Equation
The mirror equation is crucial when dealing with curved mirrors. This algebraic relationship helps us determine where images will form when objects are placed in front of mirrors. The equation is:
  • \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \)
where
- \( f \) is the focal length of the mirror (negative for convex mirrors),
- \( d_o \) is the distance of the object from the mirror, and
- \( d_i \) is the image distance from the mirror.

The sign of each term tells us about the image's nature: real or virtual, upright or inverted. In the example problem, we used this equation twice. First, to find out where the first image forms relative to the convex mirror, and again for the reflected image from the plane mirror.
Image Magnification
Image magnification tells us how much bigger or smaller an image is compared to the actual object. For mirrors, it's represented by:
  • \( m = -\frac{d_i}{d_o} \)
Here,
- \( m \) is magnification,
- \( d_i \) is the image distance, and
- \( d_o \) is the object distance.

Negative magnification implies that the image is inverted, though in a convex mirror scenario, the image will still be upright due to the virtual nature. This was used in the example to calculate how tall the images are that the convex mirror formed, revealing that both images are smaller than the object placed between the mirrors.
Virtual Image
Virtual images are formed when light rays appear to diverge from a point behind the mirror. Unlike real images, which can be projected on a screen, virtual images cannot because the light doesn't actually converge. They have the unique trait of appearing inside the mirror.
  • Always upright
  • Found behind the mirror in the mirror's operational space

In the given exercise, the convex mirror creates virtual images. These images seem to be on the opposite side of the mirror from where the object actually is, reflecting the divergent nature of the rays. The image location is also reflected as a negative distance in calculations, marking its virtuality.
Plane Mirror
A plane mirror has a flat reflective surface. They're the kind of mirrors you find in bathrooms or on walls. With plane mirrors, an image is produced through a simpler reflection process:

- The distance between the object and the mirror equals the distance between the image and the mirror. - Plane mirrors always produce virtual images, which appear reversed horizontally.

In the problem, the image created by the convex mirror interacts with the plane mirror. It enables the formation of another image, which, interestingly, acts as a new object for further reflection by the convex mirror. This interplay between the types of mirrors helps in understanding complex image formations.

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Most popular questions from this chapter

To determine whether a frog can judge distance by means of the amount its lens must move to focus on an object, researchers covered one eye with an opaque material. An insect was placed in front of the frog, and the distance that the frog snapped its tongue out to catch the insect was measured with high-speed video. The experiment was repeated with a contact lens over the eye to determine whether the frog could correctly judge the distance under these conditions. If such an experiment is performed twice, once with a lens of power -9-D and once with a lens of power -15-D, in which case does the frog have to focus at a shorter distance, and why? (a) With the -9-D lens; because the lenses are diverging, the lens with the longer focal length creates an image that is closer to the frog. (b) With the -15-D lens; because the lenses are diverging, the lens with the shorter focal length creates an image that is closer to the frog. (c) With the -9-D lens; because the lenses are converging, the lens with the longer focal length creates a larger real image. (d) With the -15-D lens; because the lenses are converging, the lens with the shorter focal length creates a larger real image.

A lensmaker wants to make a magnifying glass from glass that has an index of refraction \(n\) = 1.55 and a focal length of 20.0 cm. If the two surfaces of the lens are to have equal radii, what should that radius be?

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 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 cm tall is placed 30.0 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 erect or inverted? (\(Note\): The results obtained here are not strictly accurate because the lens is embedded in fluids having refractive indexes different from that of air.)

An insect 3.75 mm tall is placed 22.5 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude 13.0 cm, and the index of refraction of the lens material is 1.70. (a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted? (b) Repeat part (a) if the lens is reversed.

Given that frogs are nearsighted in air, which statement is most likely to be true about their vision in water? (a) They are even more nearsighted; because water has a higher index of refraction than air, a frog's ability to focus light increases in water. (b) They are less nearsighted, because the cornea is less effective at refracting light in water than in air. (c) Their vision is no different, because only structures that are internal to the eye can affect the eye's ability to focus. (d) The images projected on the retina are no longer inverted, because the eye in water functions as a diverging lens rather than a converging lens.

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