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A light bulb is 3.00 m from a wall. You are to use a concave mirror to project an image of the bulb on the wall, with the image 2.25 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?

Short Answer

Expert verified
The mirror should be 0.923 m from the wall, and its radius of curvature should be approximately 1.278 m.

Step by step solution

01

Understanding the Magnification Requirement

We know that the magnification (m) is given by the ratio of the image distance (\(d_i\)) to the object distance (\(d_o\)): \( m = \frac{d_i}{d_o} \). In this problem, \( m = 2.25 \).
02

Setting Up the Distances

Since the light bulb (object) is 3.00 m from the wall, and the image should be formed on the wall, \( d_i + d_o = 3.00 \) m, where \(d_o\) is the distance from the mirror to the object (light bulb) and \(d_i\) is the distance from the mirror to the wall (where the image forms).
03

Expressing Distances in Terms of d_i

With \( d_i = 2.25 \times d_o \) (from magnification), we can substitute in the equation: \( d_i + d_o = 3.00 \text{ m} \). Replacing \(d_i\) gives: \( 2.25 d_o + d_o = 3.00 \).
04

Solving for Object Distance d_o

Combine and solve the equation: \( 3.25 d_o = 3.00 \). Thus, \( d_o = \frac{3.00}{3.25} \approx 0.923 \text{ m} \).
05

Finding Image Distance d_i

Using \( d_i = 2.25 \, d_o \), substitute \( d_o \approx 0.923 \) m to get \( d_i = 2.25 \times 0.923 \approx 2.077 \text{ m} \).
06

Applying Mirror Formula for Focal Length

The mirror formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Substitute \( d_o = 0.923 \text{ m}\) and \( d_i = 2.077 \text{ m}\) to find the focal length \(f\).
07

Calculating Focal Length

Substitute: \( \frac{1}{f} = \frac{1}{0.923} + \frac{1}{2.077} \approx 1.083 + 0.482 = 1.565 \), so \( f = \frac{1}{1.565} \approx 0.639 \text{ m}. \)
08

Determine Radius of Curvature

The radius of curvature \( R \) is related to the focal length by \( R = 2f \). So, \( R = 2 \times 0.639 \approx 1.278 \text{ m}. \)

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Key Concepts

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

Magnification
When dealing with a concave mirror, magnification describes how much larger or smaller the image is compared to the object. It's an important concept because it helps us understand the relationship between object and image sizes. In this exercise, the magnification is given as 2.25. This means that the image size is 2.25 times that of the object.

The formula for magnification (\( m \)) in a mirror setup is \[ m = \frac{d_i}{d_o}\]where \(d_i\) is the image distance from the mirror, and \(d_o\) is the object distance from the mirror.

Knowing the magnification allows us to express one distance in terms of the other, which is useful to find out how far apart the mirror should be from the object or the wall in this case.
Mirror Formula
The mirror formula is crucial for solving problems involving mirrors, as it relates the focal length (\(f\)), object distance (\(d_o\)), and image distance (\(d_i\)). This formula is expressed as:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]

Through this relation, we can determine the focal length of the mirror if the distances of the object and the image from the mirror are known. In the current exercise, we used the distances to find out the mirror's focal length, which further helped us to calculate the radius of curvature.
Radius of Curvature
The radius of curvature (R) of a mirror is directly tied to its focal length. The relationship between these two concepts is given by:\[ R = 2f\]

This means that the radius of curvature is twice the focal length of the mirror. For our problem, once we calculated the focal length, we simply doubled it to find the radius. Understanding this connection helps to visualize how much the mirror bends or curves. A large radius means the mirror is relatively flat, whereas a small radius indicates a sharply curved mirror.
Object and Image Distance
In mirror problems, the object distance (\(d_o\)) is the distance from the object to the mirror, while the image distance (\(d_i\)) is from the mirror to the image. These distances play vital roles in determining other parameters, such as magnification, focal length, and eventually the radius of curvature.

In the exercise, the relationship between these distances is highlighted by the equation \[ d_i + d_o = 3.00 \, \text{m}\] as the sum equals the separation between the object (bulb) and the wall (where the image is formed). With given magnification, we found relationships that yield values for \(d_o\) and \(d_i\). These values were then crucial for further calculations in the solution process.
Focal Length
The focal length (\(f\)) of a concave mirror is the distance between the mirror's surface and its focus, where parallel light rays converge. This concept is vital in predicting how light is focused by the mirror to form the image.

In the exercise, we calculated the focal length using the mirror formula:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]

Substituting the values we derived for \(d_o\) and \(d_i\), we obtained the focal length, which also allowed us to determine the mirror's radius of curvature. Calculating the focal length is fundamental not only for this problem but also for understanding how concave mirrors project images in various settings.

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Most popular questions from this chapter

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?

A spherical, concave shaving mirror has a radius of curvature of 32.0 \(\mathrm{cm}\) . (a) What is the magnification of a person's face when it is 12.0 \(\mathrm{cm}\) to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.

Resolution of a Microscope. The image formed by a microscope objective with a focal length of 5.00 \(\mathrm{mm}\) is 160 \(\mathrm{mm}\) from its second focal point. The eyepiece has a focal length of 26.0 \(\mathrm{mm}\) . (a) What is the angular magnification of the microscope? (b) The unaided eye can distinguish two points at its nicroscope? (b) The unaided eye can distinguish two points at its near point as separate if they are about 0.10 \(\mathrm{mm}\) apart. What is the minimum separation between two points that can be observed (or resolved) through this microscope?

One end of a long glass rod is ground to a convex hemispherical shape. This glass has an index of refraction of 1.55 . When a small leaf is placed 20.0 \(\mathrm{cm}\) in front of the center of the hemisphere along the optic axis, an image is formed inside the glass 9.12 \(\mathrm{cm}\) from the spherical surface. Where would the image be formed if the glass were now immersed in water (refractive index 1.33 ) but nothing else were changed?

An object is placed 18.0 \(\mathrm{cm}\) from a screen. (a) At what two points between object and screen may a converging lens with a 3.00 -cm focal length be placed to obtain an image on the screen? (b) What is the magnification of the image for each position of the lens?

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