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Where must you place an object in front of a concave mirror with radius \(R\) so that the image is erect and \(2 \frac{1}{2}\) times the size of the object? Where is the image?

Short Answer

Expert verified
Place the object at \(\frac{2R}{3.75}\) in front of the mirror. The image is \(\frac{5R}{3.75}\) on the same side.

Step by step solution

01

Understand the Mirror Formula

For any mirror, the mirror formula is given by \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\), where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance. For a concave mirror, \(f\) is negative and \(f = \frac{R}{2}\).
02

Determine the Focal Length

Since the radius of curvature \(R\) is given, the focal length \(f\) is \(-\frac{R}{2}\). This accounts for the negative sign because we are dealing with a concave mirror.
03

Set Up the Magnification Equation

The magnification \(m\) is given by the formula \(m = -\frac{v}{u}\). For an erect image (meaning the image is virtual and on the same side as the object), \(m\) should be positive and given as 2.5. So, \(m = 2.5 = -\frac{v}{u}\), solving gives \(v = -2.5u\).
04

Substitute v in the Mirror Formula

Substitute \(v = -2.5u\) into the mirror formula \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\) to find \(u\):\[\frac{1}{-\frac{R}{2}} = \frac{1}{-2.5u} + \frac{1}{u}\].
05

Simplify and Solve for u

Simplify the equation: \(-\frac{2}{R} = \frac{1}{-2.5u} + \frac{1}{u}\).Find a common denominator and solve:\(-\frac{2}{R} = \frac{u - 2.5u}{2.5u^2}\) => \(-\frac{2}{R} = \frac{-1.5u}{2.5u^2}\)=> \(-\frac{2}{R} = -\frac{1.5}{2.5u}\).
06

Solve for the Object Distance u

Cross multiply to solve for \(u\): \(\frac{2R}{1.5} = 2.5u\)=> \(u = \frac{2R}{3.75}\).This is the distance where the object must be placed.
07

Calculate the Image Distance v

Use \(v = -2.5u\) to find \(v\): \(v = -2.5\times\frac{2R}{3.75}\) => \(v = -\frac{5R}{3.75}\).The negative sign indicates that the image is virtual and located on the same side of the mirror as the object.

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Key Concepts

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

Mirror Formula
The mirror formula is a crucial equation used in optics to relate the object distance (\(u\)), the image distance (\(v\)), and the focal length (\(f\)) of a mirror. It is expressed as:
  • \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
This equation helps predict where the image of an object will form when placed in front of a mirror.
It's especially helpful when working with concave mirrors, which curve inward like a spoon.
For concave mirrors, the focal length (\(f\)) is typically negative, indicating that the focus is virtual, or on the same side as the object.
The focal length can be found using the radius of curvature (\(R\)), where \(f = \frac{R}{2}\).
Focal Length
The focal length is a key property of concave mirrors that defines how they converge light.
It is the distance from the mirror's surface to the point where parallel rays of light meet after reflection.
For a concave mirror, the focal length is calculated from the mirror's radius of curvature:
  • \(f = \frac{R}{2}\)
Because a concave mirror focuses light to one side, the focal length is assigned a negative value to indicate that the focal point is behind the mirror's surface.
Understanding this concept helps in determining the image characteristics such as position and size.
A concave mirror with a smaller radius will have a shorter focal length, reflecting light more sharply than a mirror with a larger radius.
Magnification
Magnification measures how much larger or smaller an image is compared to the object itself.
For mirrors, it is calculated with the formula:
  • \(m = -\frac{v}{u}\)
Where \(v\) is the image distance and \(u\) is the object distance.
In the scenario of an erect image from a concave mirror, the magnification factor (\(m\)) should be positive.
A positive magnification indicates that the image is upright and virtual.
For instance, a magnification of 2.5 means the size of the image is 2.5 times larger than the object's size.
The minus sign in the formula is typically used to denote the inversion of images, but for erect images which are virtual, the virtual image appears on the same side as the object.
Virtual Image
A virtual image is an image formed by rays that appear to diverge from a point in space.
Unlike real images, which can be projected onto a screen, virtual images cannot be captured directly and must be seen by looking into the mirror or lens.
In the case of concave mirrors, virtual images are formed when the object is placed within the focal length of the mirror.
These images are upright and appear larger than the actual object, depending on their magnification.
Virtual images are crucial in applications like makeup mirrors or shaving mirrors, where users need an enlarged, unobstructed view of themselves.
Understanding where and how virtual images form helps in placing objects correctly to create desired image characteristics.

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Most popular questions from this chapter

What should be the index of refraction of a transparent sphere in order for paraxial rays from an infinitely distant object to be brought to a focus at the vertex of the surface opposite the point of incidence?

A Glass Red. Both ends of a glass rod with index of refraction \(1.60\) are ground and polished to convex hemispherical surfaces. The radius of curvature at the left end is \(6.00 \mathrm{~cm}\), and the radius of curvature at the right end is \(12.0 \mathrm{~cm}\). The length of the rod between vertices is \(40.0 \mathrm{~cm}\). The object for the surface at the left end is an arrow that lies \(23.0 \mathrm{~cm}\) to the left of the vertex of this surface. The arrow is \(1.50 \mathrm{~mm}\) tall and at right angles to the axis. (a) What constitutes the object for the surface at the right end of the rod? (b) What is the object distance for this surface? (c) Is the object for this surface real or virtual? (d) What is the position of the final image? (e) Is the final image real or virtual? Is it erect or inverted with respect to the original object? (f) What is the height of the final image?

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen \(30.0 \mathrm{~cm}\) to the right of the lens. A diverging lens is now placed \(15.0 \mathrm{~cm}\) to the right of the converging lens, and it is found that the screen must be mowed \(19.2 \mathrm{~cm}\) farther to the right to obtain a sharp image. What is the focal length of the diverging lens?

A transparent rod \(50.0 \mathrm{~cm}\) long and with a refractive index of \(1.60\) is cut flat at the right end and rounded to a hemispherical surfuce with a \(15.0-\mathrm{cm}\) radius at the left end. An object is placed on the axis of the rod \(12.0 \mathrm{~cm}\) to the left of the vertex of the hemispherical end. (a) What is the position of the final image? (b) What is its magnification?

A double-convex thin lens has surfaces with equal radii of curvature of magnitude \(2.50 \mathrm{~cm}\). Looking through this lens, you observe that it forms an image of a very distant tree at a distance of \(1.87 \mathrm{~cm}\) from the lens. What is the index of refraction of the lens?

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