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Chapter 5: Problem 79

An LED 0.60 cm tall is on the central axis 30.0 \(\mathrm{cm}\) in front of a convex spherical mirror. If the radius of curvature of the mirror is \(12.0 \mathrm{cm}\) determine the location of the image, describe it, and draw a ray diagram. How big is the image?

Short Answer

Expert verified
The image is 5.0 cm behind the mirror, upright, virtual, and 0.10 cm tall.

Step by step solution

01

Understand the Mirror Equation

To find the image location in a spherical mirror, use the mirror equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. For a convex mirror, the focal length \( f \) is negative and is half the radius of curvature: \( f = -\frac{R}{2} \). Since the radius of curvature \( R \) is 12.0 cm, \( f = -6.0 \text{ cm} \).
02

Solve for Image Distance

Substitute the known values into the mirror equation to find the image distance \( d_i \):\[ \frac{1}{-6.0} = \frac{1}{30.0} + \frac{1}{d_i} \]Simplify this to find \( d_i \):\[ \frac{1}{d_i} = \frac{1}{-6.0} - \frac{1}{30.0} \]Calculate \( \frac{1}{d_i} \):\[ \frac{1}{d_i} = -\frac{5}{30} - \frac{1}{30} = -\frac{6}{30} = -\frac{1}{5} \]Thus, \( d_i = -5.0 \text{ cm} \), meaning the image is 5.0 cm behind the mirror.
03

Calculate the Magnification

The magnification \( m \) of a mirror is given by \( m = -\frac{d_i}{d_o} \). Substitute the values:\[ m = -\left( \frac{-5.0}{30.0} \right) = \frac{5.0}{30.0} = \frac{1}{6} \].Therefore, the magnification is \( \frac{1}{6} \). The height of the image \( h_i \) is the magnification times the height of the object \( h_o \):\[ h_i = m \times h_o = \frac{1}{6} \times 0.60 \text{ cm} = 0.10 \text{ cm} \].
04

Describe the Image

The image is virtual (since \( d_i \) is negative), upright (as indicated by the positive magnification), reduced in size (\( \frac{1}{6} \) of the original size), and located behind the mirror at 5.0 cm.

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Key Concepts

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

Mirror Equation
The mirror equation is a fundamental tool used to find the relationship between the object distance, image distance, and the focal length of a mirror. It is given by: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Where:
  • \( f \) is the focal length of the mirror
  • \( d_o \) is the distance from the object to the mirror
  • \( d_i \) is the distance from the image to the mirror
For convex mirrors, which spread out light rays, the focal length \( f \) is always negative. This negativity indicates that the focal point is located behind the mirror, emphasizing the mirror's nature of creating virtual images.
Focal Length
The focal length is a key property of any mirror, defining how strongly it converges or diverges light. It is specifically half the radius of curvature of the mirror surface: \[ f = \frac{R}{2} \]In the case of a convex mirror, the focal length is negative, reflecting the fact that the focal point is not in the same direction as the incident light rays. For our problem with a radius of curvature of 12.0 cm, the focal length computes to \[ f = -6.0 \text{ cm} \]This negative focal length plays a crucial role in determining image properties such as image distance and size.
Image Distance
The image distance \( d_i \) is defined as the distance from the mirror to the location of the image formed by the mirror. In cases involving convex mirrors, image distance often results in negative values, reflecting the fact that the images are virtual. By inserting the given values into the mirror equation:\[ \frac{1}{-6.0} = \frac{1}{30.0} + \frac{1}{d_i} \]Rearranging and solving gives:\[ d_i = -5.0 \text{ cm} \]This result indicates the image appears 5.0 cm behind the mirror surface, verifying it is virtual because convex mirrors typically form virtual images, meaning they can鈥檛 be projected onto a screen like real images.
Magnification
Magnification determines how much larger or smaller an image appears compared to the original object. The magnification \( m \) is calculated by:\[ m = -\frac{d_i}{d_o} \]For this example:\[ m = -\left( \frac{-5.0}{30.0} \right) = \frac{1}{6} \]A positive magnification indicates the image is upright compared to the object, while a value of \( \frac{1}{6} \) shows the image is reduced to six times smaller than the object. Applying this to the object's height:\[ h_i = m \times h_o = \frac{1}{6} \times 0.60 \text{ cm} = 0.10 \text{ cm} \]Thus, the image retains that reduced height of 0.10 cm, reinforcing characteristics typical of convex mirrors.

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Most popular questions from this chapter

A stepped-index multimode glass fiber has indices of 1.481 and 1.461. Its core diameter is \(100 \mu \mathrm{m}\). Determine the fiber's acceptance angle when immersed in air.

A device used to measure the radius of curvature of the cornea of the eye is called a keratometer. This is useful information when fitting contact lenses. In effect, an illuminated object is placed a known distance from the eye, and the image reflected off the cornea is observed. The instrument allows the operator to measure the size of that virtual image. If the magnification is found to be \(0.037 \times\) when the object distance is set at \(100 \mathrm{mm}\), what is the radius of curvature?

Consider the case of two positive thin lenses, \(L_{1}\) and \(L_{2},\) separated by \(5 \mathrm{cm} .\) Their diameters are 6 and \(4 \mathrm{cm},\) respectively, and their focal lengths are \(f_{1}=9 \mathrm{cm}\) and \(f_{2}=3 \mathrm{cm} .\) If a diaphragm with a hole \(1 \mathrm{cm}\) in diameter is located between them, \(2 \mathrm{cm}\) from \(L_{2},\) find (a) the aperture stop and (b) the locations and sizes of the pupils for an axial point, \(S, 12 \mathrm{cm}\) in front of (to the left of) \(L_{1}\).

Given a fused silica fiber with an attenuation of \(0.2 \mathrm{dB} / \mathrm{km},\) how far can a signal travel along it before the power level drops by half?

In an amusement park a large upright convex spherical mirror is facing a plane mirror \(10.0 \mathrm{m}\) away. A girl \(1.0 \mathrm{m}\) tall standing midway between the two sees herself twice as tall in the plane mirror as in the spherical one. In other words, the angle subtended at the observer by the image in the plane mirror is twice the angle subtended by the image in the spherical mirror. What is the focal length of the latter?
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