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Chapter 34: Problem 66

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
The object should be placed at a distance of \(0.833R\) from the mirror on the same side light is coming from, and the image is located at a distance \(2.08R\) from the mirror on the same side as the light.

Step by step solution

01

Use the magnification equation to express \(v\) in terms of \(u\)

First, express the image distance \(v\) as a function of the object distance \(u\) using the magnification formula. As the magnification \(m = 2 \frac{1}{2}\), then \(v = 2.5u\).
02

Substitute \(v\) into the mirror equation

Plug the expression obtained from Step 1 into the mirror equation \(\frac{1}{v} + \frac{1}{u} = \frac{2}{R}\) to find \(u\) in terms of \(R\). Replace \(v\) with \(2.5u\). So, the equation becomes \(\frac{1}{2.5u} + \frac{1}{u} = \frac{2}{R}\). Simplifying gives: \(u = -0.833R\). A negative sign indicates that the object is placed on the same side as the light is coming from, which is usual in these problems.
03

Find the image location

Substitute \(u\) found in Step 2 back into \(v = 2.5u\) to find the image distance. Hence, \(v = 2.5(-0.833R) = -2.08R\). Again, a negative sign indicates that the image is formed on the same side as the light, which is expected for a virtual image.

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Key Concepts

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

Magnification Equation
Understanding the magnification equation is crucial when dealing with mirrors and lenses. It relates the height of the image to the height of the object and gives insight into the size of the image produced by the mirror. The formula for magnification (\(m\)) is given by:
\[ m = -\frac{v}{u} = \frac{h_{i}}{h_{o}} \]
where \(v\) is the image distance from the mirror, \(u\) is the object distance from the mirror, \(h_i\) is the height of the image, and \(h_o\) is the height of the object. The negative sign indicated in the magnification formula arises when using the sign conventions for mirrors; it tells us the orientation of the image. If the magnification is negative, the image is inverted relative to the object. In the given exercise, a positive magnification of 2.5 indicates an erect image, which is 2.5 times the size of the object. This particular value of magnification helps in solving for the relationship between the image and the object distances, highlighting the importance of understanding this equation in image formation.
Mirror Equation
The mirror equation plays a vital role in geometric optics, especially when determining the position and characteristics of the image formed by a mirror. The equation is typically written as:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Here, \(f\) is the focal length of the mirror, \(v\) is the distance from the mirror to the image, and \(u\) is the distance from the mirror to the object. The focal length (\(f\)) is related to the radius of curvature (\(R\)) of the mirror by the formula \(f = \frac{R}{2}\). In our exercise, we are given the radius of the mirror, which is used to find the focal length. By substituting the magnification equation's expression for \(v\) into the mirror equation, we can find the value of \(u\) in terms of \(R\). It is essential to use the correct sign conventions here; distances are positive if they are in the direction in which light is travelling and negative if they are in the opposite direction. Using this equation, we can correctly locate the position of the object, which is necessary to produce the required image.
Virtual Image
When dealing with concave mirrors, or any mirrors for that matter, it's significant to understand the concept of virtual images. A virtual image is formed when the reflected light rays from a mirror appear to diverge, causing the light rays to appear to come from a point behind the mirror. Unlike a real image which can be cast onto a screen, a virtual image cannot be projected as it doesn鈥檛 exist at a real point in space.
In the context of concave mirrors, virtual images are typically formed when the object is placed between the focal point and the mirror itself. These images are always upright, enlarged, and appear on the same side of the mirror as the object. As was calculated in the exercise, the image distance obtained a negative value (\(v = -2.08R\)), which is a strong indication of the image being virtual. The negative image distance confirms the image formation on the same side as the incoming light and that it is a virtual image. Students need to be mindful of these characteristics in order to accurately predict the nature of the image formed by a concave mirror.

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Most popular questions from this chapter

A convex spherical mirror with a focal length of magnitude \(24.0 \mathrm{~cm}\) is placed \(20.0 \mathrm{~cm}\) to the left of a plane mirror. An object \(0.250 \mathrm{~cm}\) tall is placed midway between the surface of the plane mirror and the vertex of the spherical mirror. The spherical mirror forms multiple images of the object. Where are the two images of the object formed by the spherical mirror that are closest to the spherical mirror, and how tall is each image?

(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?

A tank whose bottom is a mirror is filled with water to a depth of \(20.0 \mathrm{~cm}\). A small fish floats motionless \(7.0 \mathrm{~cm}\) under the surface of the water. (a) What is the apparent depth of the fish when viewed at normal incidence? (b) What is the apparent depth of the image of the fish when viewed at normal incidence?

Figure P34.99 shows a simple version of a zoom lens. The converging lens has focal length \(f_{1}\) and the diverging lens has focal length \(f_{2}=-\left|f_{2}\right| .\) The two lenses are separated by a variable distance \(d\) that is always less than \(f_{1}\). Also, the magnitude of the focal length of the diverging lens satisfies the inequality \(\left|f_{2}\right|>\left(f_{1}-d\right) .\) To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius \(r_{0}\) entering the converging lens. (a) Show that the radius of the ray bundle decreases to \(r_{0}^{\prime}=r_{0}\left(f_{1}-d\right) / f_{1}\) at the point that it enters the diverging lens. (b) Show that the final image \(I^{\prime}\) is formed a distance \(s_{2}^{\prime}=\left|f_{2}\right|\left(f_{1}-d\right) /\left(\left|f_{2}\right|-f_{1}+d\right)\) to the right of the diverging lens. (c) If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left of the diverging lens, they will eventually expand to the original radius \(r_{0}\) at some point \(Q .\) The distance from the final image \(I^{\prime}\) to the point \(Q\) is the effective focal length \(f\) of the lens combination; if the combination were replaced by a single lens of focal length \(f\) placed at \(Q\), parallel rays would still be brought to a focus at \(I^{\prime} .\) Show that the effective focal length is given by \(f=f_{1}\left|f_{2}\right| /\left(\left|f_{2}\right|-f_{1}+d\right) .\) (d) If \(f_{1}=12.0 \mathrm{~cm}, f_{2}=-18.0 \mathrm{~cm},\) and the separation \(d\) is adjustable between 0 and \(4.0 \mathrm{~cm}\), find the maximum and minimum focal lengths of the combination. What value of \(d\) gives \(f=30.0 \mathrm{~cm} ?\)

The focal length of a simple magnifier is \(8.00 \mathrm{~cm}\). Assume the magnifier is a thin lens placed very close to the eye. (a) How far in front of the magnifier should an object be placed if the image is formed at the observer's near point, \(25.0 \mathrm{~cm}\) in front of her eye? (b) If the object is \(1.00 \mathrm{~mm}\) high, what is the height of its image formed by the magnifier?
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