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Chapter 26: Problem 128

Suppose the radius of curvature of the mirror is \(5.0 \mathrm{cm}\). (a) Find the object distance that gives an upright image with a magnification of \(1.5 .\) (b) Find the object distance that gives an inverted image with a magnification of -1.5 .

Short Answer

Expert verified
For both magnifications (1.5 and -1.5), the object distance is 1.875 cm.

Step by step solution

01

Understanding the Relationship

The magnification formula for mirrors is given by: \[ m = -\frac{q}{p} \] where \(m\) is the magnification, \(q\) is the image distance, and \(p\) is the object distance. Rearrange to find \(q\):\[ q = -mp \].
02

Identify Mirror Type

Since the radius of curvature \(R\) is given as \(5.0 \text{ cm}\), we can find the focal length \(f\) using the relation: \[ f = \frac{R}{2} = \frac{5.0}{2} = 2.5 \text{ cm} \]. The focal length indicates that this is a concave mirror (since \(f > 0\)).
03

Using Mirror Equation

Use the mirror equation: \[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \]. Substitute \(q = -1.5p\) for the upright image:\[ \frac{1}{2.5} = \frac{1}{p} - \frac{1}{1.5p} \].
04

Solve for Object Distance (Upright Image)

Simplify and solve for \(p\) from the equation in the previous step:\[ \frac{1}{2.5} = \frac{1}{p} - \frac{1}{1.5p} \]\[ \frac{1}{2.5} = \frac{1.5 - 1}{1.5p} \]\[ \frac{1}{2.5} = \frac{0.5}{1.5p} \]\[ p = \frac{0.5 \times 2.5 \times 1.5}{1} = 1.875 \text{ cm} \]. The object distance for an upright image with magnification 1.5 is \(1.875 \text{ cm}\).
05

Substitute for Inverted Image

Now consider \(q = 1.5p\) for the inverted image with magnification -1.5:\[ \frac{1}{2.5} = \frac{1}{p} + \frac{1}{1.5p} \].
06

Solve for Object Distance (Inverted Image)

Simplify and solve for \(p\) from the equation:\[ \frac{1}{2.5} = \frac{1 + 1/1.5}{p} \]\[ \frac{1}{2.5} = \frac{2.5}{1.5p} \]\[ p = \frac{2.5 \times 1.5}{2} = 1.875 \text{ cm} \]. The object distance for an inverted image with magnification -1.5 is also \(1.875 \text{ cm}\).

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Key Concepts

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

Concave Mirror
A concave mirror, also known as a converging mirror, is shaped like the inner surface of a sphere. These mirrors have the ability to converge light rays that are incident upon them, focusing them at a single point known as the focal point.

Here are some important features of concave mirrors:
  • They can produce both real and virtual images, depending on the position of the object.
  • When the object is placed between the focal point and the mirror, the image appears larger and upright.
  • If the object is beyond the focal point, the image is smaller, inverted, and real.
Concave mirrors are commonly used in applications where light needs to be focused, such as in telescopes and flashlights.
Magnification Formula
The magnification formula is a mathematical formula used to relate the size of an image to the size of the object and their respective distances from a mirror. It is expressed as:\[ m = -\frac{q}{p} \]where:
  • \( m \) is the magnification value.
  • \( q \) signifies the image distance from the mirror.
  • \( p \) represents the object distance from the mirror.
Positive magnification signifies an upright image, while negative magnification indicates an inverted image.

Magnification shows how much larger or smaller the image is compared to the object. A value greater than one means the image is larger than the object, whereas a value less than one means it is smaller.
Radius of Curvature
The radius of curvature \( R \) is the distance between the center of curvature and the mirror's surface. For spherical mirrors, the radius of curvature is twice the focal length (\( f \)). It is directly related to how strongly the mirror converges or diverges light.The radius of curvature helps determine the mirror's focal length using the relation:\[ f = \frac{R}{2} \]A concave mirror with a small radius of curvature will have a shorter focal length, indicating it is highly curved and will more sharply bend incoming light rays.
Focal Length
Focal length (\( f \)) is the distance between the focal point and the mirror. It is a crucial concept in understanding how concave mirrors work and is calculated from the radius of curvature using:\[ f = \frac{R}{2} \]For a concave mirror:
  • When the object is placed at the focal length, the rays parallel to the principal axis converge at a point on the opposite side.
  • It affects the size and nature (real or virtual) of the image formed by the mirror.
Understanding the focal length helps in predicting the behavior and characteristics of images created by concave mirrors in various situations.

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Most popular questions from this chapter

Two identical containers are filled with different transparent liquids. The container with liquid A appears to have a greater depth than the container with liquid B. Which liquid has the greater index of refraction? Explain.

A convex mirror with a focal length of \(-85 \mathrm{cm}\) is used to give a truck driver a view behind the vehicle. (a) If a person who is \(1.7 \mathrm{m}\) tall stands \(2.2 \mathrm{m}\) from the mirror, where is the person's image located? (b) Is the image upright or inverted? (c) What is the size of the image?

Light travels a distance of \(0.960 \mathrm{m}\) in \(4.00 \mathrm{ns}\) in a given substance. What is the index of refraction of this substance?

The focal length of a lens is inversely proportional to the quantity \((n-1),\) where \(n\) is the index of refraction of the lens material. The value of \(n,\) however, depends on the wavelength of the light that passes through the lens. For example, one type of flint glass has an index of refraction of \(n_{\mathrm{r}}=1.572\) for red light and \(n_{\mathrm{v}}=1.605\) in violet light. Now, suppose a white object is placed \(24.00 \mathrm{cm}\) in front of a lens made from this type of glass. If the red light reflected from this object produces a sharp image \(55.00 \mathrm{cm}\) from the lens, where will the violet image be found?

An object is a distance \(2 f\) from a convex lens. (a) Use a ray diagram to find the approximate location of the image. (b) Is the image upright or inverted? (c) Is the image real or virtual? Explain.
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