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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?

Short Answer

Expert verified
The filament should be placed 20cm in front of the mirror. The radius of curvature of the mirror should be 36.36cm.

Step by step solution

01

Determine the image and object dimensions

First, we need to keep in mind that both the dimension of the object and the image are given. The image is negative since it is a real image; therefore, we can use the magnification formula \( h'= -h \frac{i}{o} \), where \( h'= -24cm \), \( h= 6mm = 0.6cm \), \( i= -800cm \). Solving that we get the object distance, \( o \).
02

Determine the object distance

Substitute the known values into the equation. So, \( -24cm = -0.6cm \frac{-800cm}{o} \$\). By cross multiplying and solving for \( o \), we find that \( o = -20cm \). Notice that \( o \) is negative because the object is on the same side of the mirror as the incoming light.
03

Determine the radius of curvature

Use the mirror equation to solve for the radius of curvature. The equation is \( \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \), where f is the focal length. We found \( o \) in the previous step and \( i \) is given, so we can solve for \( f \). After we find the focal length, we use the fact that the radius \( R = 2f \) to find the radius of curvature.
04

Solve for the radius of curvature

Substitute the known values into the mirror equation: \( \frac{1}{f} = \frac{1}{-20} + \frac{1}{-800} \). Solve that to get \( f = -18.18cm \). Then, use \( R = 2f \) to find \( R = 36.36cm \). The radius is positive since the center of curvature is on the same side as the incoming light.

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Key Concepts

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

Magnification Formula
Understanding the magnification formula in a concave mirror scenario helps us determine the relation between the size of the object and the image. The formula is given by:\[ h' = -h \frac{i}{o} \]where:
  • \(h'\) is the height of the image.
  • \(h\) is the height of the object.
  • \(i\) is the image distance from the mirror.
  • \(o\) is the object distance from the mirror.
The negative sign reflects that the image is inverted relative to the object. In the exercise, we had \(h' = -24 \text{ cm}\) and \(h = 0.6 \text{ cm}\), showing an inverted image. We use these values along with the known image distance \(i = -800 \text{ cm}\) to solve for the object distance \(o\), ensuring we consider the sign conventions for real and virtual images.
Image Formation
Image formation in a concave mirror involves understanding how light rays reflect to form images. A concave mirror can create real or virtual images depending on the position of the object relative to the mirror's focal point.
In our exercise, the goal was to create a real, inverted image of a lamp filament. Real images are typically formed when reflected rays converge at a point. We used the given dimensions to predict that the screen would capture the image 8.00 meters from the mirror, indicating how image distance plays a vital role in placement for visibility.
An important aspect to remember is the use of sign conventions:
  • Distances are positive if located in front of the mirror and negative if behind.
  • A negative image height denotes an inverted image.
These conventions ensure we can predict both the kind of image (real or virtual) and its orientation.
Radius of Curvature
The radius of curvature for a concave mirror is crucial as it is directly tied to the mirror's focusing ability. It is defined as the distance from the mirror's surface to its center of curvature, where reflected rays align neatly.
In the task, we calculated the radius using the focal length, given by the relationship:\[ R = 2f \]This linking of the radius of curvature and focal length shows that any changes in one affect the mirror's overall behavior.
The radius we derived was 36.36 cm, positive here because the center of curvature lies on the same side as the incident light. This positive result confirms that users should expect the mirror setup to behave in alignment with traditional concave mirror physics.
Mirror Equation
The mirror equation is a foundational formula in optics that ties together object distance, image distance, and focal length:\[ \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \]This equation is vital in solving the task's exercise. By knowing any two of the parameter values, we can solve for the third.
Here's how it was applied:
  • First, we derived the object distance \(o\) using the magnification formula.
  • Then, we plugged it into the mirror equation along with the image distance \(i\) of -800 cm.
  • This allowed us to find the focal length \(f\).
  • Finally, with \(f\), we calculated the radius of curvature using \( R = 2f \).
Understanding and applying this formula allows students to approach various mirror problems with confidence, utilizing the connections between physical quantities.

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Most popular questions from this chapter

A candle \(4.85 \mathrm{~cm}\) tall is \(39.2 \mathrm{~cm}\) to the left of a plane mirror. Where is the image formed by the mirror, and what is the height of this image?

A mirror on the passenger side of your car is convex and has a radius of curvature with magnitude \(18.0 \mathrm{~cm} .\) (a) Another car is behind your car, \(9.00 \mathrm{~m}\) from the mirror, and this car is viewed in the mirror by your passenger. If this car is \(1.5 \mathrm{~m}\) tall, what is the height of the image? (b) The mirror has a warning attached that objects viewed in it are closer than they appear. Why is this so?

A person swimming \(0.80 \mathrm{~m}\) below the surface of the water in a swimming pool looks at the diving board that is directly overhead and sees the image of the board that is formed by refraction at the surface of the water. This image is a height of \(5.20 \mathrm{~m}\) above the swimmer. What is the actual height of the diving board above the surface of the water?

The left end of a long glass rod \(8.00 \mathrm{~cm}\) in diameter, with an index of refraction of 1.60 , is ground and polished to a convex hemispherical surface with a radius of \(4.00 \mathrm{~cm}\). An object in the form of an arrow \(1.50 \mathrm{~mm}\) tall, at right angles to the axis of the rod, is located on the axis \(24.0 \mathrm{~cm}\) to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

The science museum where you work is constructing a new display. You are given a glass rod that is surrounded by air and was ground on its left-hand end to form a hemispherical surface there. You must determine the radius of curvature of that surface and the index of refraction of the glass. Remembering the optics portion of your physics course, you place a small object to the left of the rod, on the rod's optic axis, at a distance \(s\) from the vertex of the hemispherical surface. You measure the distance \(s^{\prime}\) of the image from the vertex of the surface, with the image being to the right of the vertex. Your measurements are as follows: $$\begin{array}{l|rrrrrr} \boldsymbol{s}(\mathbf{c m}) & 22.5 & 25.0 & 30.0 & 35.0 & 40.0 & 45.0 \\ \hline s^{\prime}(\mathbf{c m}) & 271.6 & 148.3 & 89.4 & 71.1 & 60.8 & 53.2 \end{array}$$ Recalling that the object-image relationships for thin lenses and spherical mirrors include reciprocals of distances, you plot your data as \(1 / s^{\prime}\) versus \(1 / s .\) (a) Explain why your data points plotted this way lie close to a straight line. (b) Use the slope and \(y\) -intercept of the best-fit straight line to your data to calculate the index of refraction of the glass and the radius of curvature of the hemispherical surface of the rod. (c) Where is the image if the object distance is \(15.0 \mathrm{~cm} ?\)

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