/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 A small postage stamp is placed ... [FREE SOLUTION] | 91Ó°ÊÓ

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A small postage stamp is placed in front of a concave mirror (radius \(=R)\) so that the image distance equals the object distance. (a) In terms of \(R,\) what is the object distance? (b) What is the magnification of the mirror? (c) State whether the image is upright or inverted relative to the object. Draw a ray diagram to guide your thinking.

Short Answer

Expert verified
(a) The object distance is \(R\). (b) The magnification is \(-1\). (c) The image is inverted.

Step by step solution

01

Understand the mirror equation

The mirror equation relates object distance \(d_o\), image distance \(d_i\), and the focal length \(f\) of the mirror: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). For a concave mirror, the focal length \(f\) is half the radius of curvature, \(f = \frac{R}{2}\). Since the image distance equals the object distance, we let \(d_o = d_i\).
02

Solve for object distance using symmetry

Substitute \(d_o = d_i\) into the mirror equation: \(\frac{1}{f} = \frac{2}{d_o}\). Since \(f = \frac{R}{2}\), this becomes \(\frac{2}{R} = \frac{2}{d_o}\). Solving for \(d_o\), we obtain \(d_o = R\). Hence, the object distance is equal to the radius of curvature.
03

Calculate the magnification

The magnification \(m\) is given by the formula \(m = -\frac{d_i}{d_o}\). Since \(d_i = d_o\), substituting gives \(m = -1\). This means the image is of the same size as the object but inverted.
04

Image orientation and ray diagram conclusion

The negative magnification indicates that the image is inverted relative to the object. In a ray diagram for this scenario, rays diverging from the object will converge to form an inverted image on the same location but on the opposite side of the principal axis.

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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 formula used in optics to relate different distances involved with a concave mirror. It connects the object distance ( d_o ), the image distance ( d_i ), and the focal length ( f ) of the mirror through the equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] The focal length is particularly important as it is half the radius of curvature ( R ) for concave mirrors, expressed as f = \frac{R}{2} . This makes it easier to connect the radius of the mirror to its focal abilities.
In many scenarios, you might encounter situations where the distances ( d_o and d_i ) have a relationship, such as equality. Solving this equation can provide direct insight into the nature and conditions required for a specific image formation.
Object Distance
The term "object distance" ( d_o ) refers to the distance between the object being observed and the mirror. It is one of the key variables in the mirror equation. Given that the image distance ( d_i ) is equal to the object distance in some cases, you might substitute d_o = d_i into the mirror equation to solve for d_o .
For a concave mirror scenario where the object distance equals the image distance, the calculations simplify the equation to: \[ \frac{1}{f} = \frac{2}{d_o} \] Since the focal length is half of the radius ( f = \frac{R}{2} ), the equation transforms to \frac{2}{R} = \frac{2}{d_o} . Solving this results in d_o = R , indicating that the object is placed at the radius of curvature from the mirror. This placement is crucial for special cases of image formation.
Magnification
Magnification in mirror calculations is a measure of how much larger or smaller the image is compared to the object. It is given by the formula: \[ m = -\frac{d_i}{d_o} \] This ratio indicates by what factor the image size is changed by the mirror. A negative sign in the magnification indicates the orientation of the image.
In situations where the object distance is set equal to the image distance, such as in our problem, it results in m = -1 . This tells us that the image is the same size as the object, but inverted. Negative magnification commonly indicates that the image is upside down relative to the original object.
Image Orientation
Image orientation refers to the direction (upright or inverted) in which an image appears after reflection from a mirror. With concave mirrors, this orientation is directly linked with the sign of the magnification.
In the problem, since the magnification is negative ( m = -1 ), the image is inverted. This means the image appears upside down relative to the object.
Well-drawn ray diagrams can help visualize this inversion. They generally show that rays reflecting off a concave mirror converge at a point opposite to the object's location, flipping the image upside down along the principal axis.

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Most popular questions from this chapter

The outside mirror on the passenger side of a car is convex and has a focal length of \(-7.0 \mathrm{m}\). Relative to this mirror, a truck traveling in the rear has an object distance of \(11 \mathrm{m}\). Find \((\mathrm{a})\) the image distance of the truck and (b) the magnification of the mirror.

When an object is located very far away from a convex mirror, the image of the object is \(18 \mathrm{cm}\) behind the mirror. Using a ray diagram drawn to scale, determine where the image is located when the object is placed \(9.0 \mathrm{cm}\) in front of the mirror. Note that the mirror must be drawn to scale also. In your drawing, assume that the height of the object is \(3.0 \mathrm{cm}\).

Suppose that you are walking perpendicularly with a velocity of \(+0.90 \mathrm{m} / \mathrm{s}\) toward a stationary plane mirror. What is the velocity of your image relative to you? The direction in which you walk is the positive direction.

An object is placed in front of a convex mirror. Draw the convex mirror (radius of curvature \(=15 \mathrm{cm}\) ) to scale, and place the object \(25 \mathrm{cm}\) in front of it. Make the object height \(4 \mathrm{cm} .\) Using a ray diagram, locate the image and measure its height. Now move the object closer to the mirror, so the object distance is \(5 \mathrm{cm}\). Again, locate its image using a ray diagram. As the object moves closer to the mirror, (a) does the magnitude of the image distance become larger or smaller, and (b) does the magnitude of the image height become larger or smaller? (c) What is the ratio of the image height when the object distance is \(5 \mathrm{cm}\) to its height when the object distance is \(25 \mathrm{cm} ?\) Give your answer to one significant figure.

A drop of water on a countertop reflects light from a flower held \(3.0 \mathrm{cm}\) directly above it. The flower's diameter is \(2.0 \mathrm{cm},\) and the diameter of the flower's image is \(0.10 \mathrm{cm} .\) What is the focal length of the water drop, assuming that it may be treated as a convex spherical mirror?

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