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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) You want to use a lens with a focal length of \(35.0 \mathrm{~cm}\) to produce a real image of an object, with the height of the image twice the height of the object. What kind of lens do you need, and where should the object be placed? (b) Suppose you want a virtual image of the same object, with the same magnification-what kind of lens do you need, and where should the object be placed?

Two thin lenses with a focal length of magnitude \(12.0 \mathrm{~cm},\) the first diverging and the second converging, are located \(9.00 \mathrm{~cm}\) apart. An object \(2.50 \mathrm{~mm}\) tall is placed \(20.0 \mathrm{~cm}\) to the left of the first (diverging) lens. (a) How far from this first lens is the final image formed? (b) Is the final image real or virtual? (c) What is the height of the final image? Is it erect or inverted? (Hint: See the preceding two exercises.)

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 smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns \((\mu \mathrm{m})\) is typical near the center of the eye. We shall model the eye as a sphere \(2.50 \mathrm{~cm}\) in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells \(5.0 \mu \mathrm{m}\) in diameter. (a) What is the smallest object you can resolve at a near point of \(25 \mathrm{~cm} ?\) (b) What angle is subtended by this object at the eye? Express your answer in units of minutes \(\left(1^{\circ}=60 \mathrm{~min}\right),\) and compare it with the typical experimental value of about \(1.0 \mathrm{~min} .\) (Note: There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.)

A concave mirror has a radius of curvature of \(34.0 \mathrm{~cm}\). (a) What is its focal length? (b) If the mirror is immersed in water (refractive index 1.33 ), what is its focal length?

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