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A converging lens with a focal length of \(9.00 \mathrm{~cm}\) forms an image of a \(4.00-\mathrm{mm}\) -tall real object that is to the left of the lens. The image is \(1.30 \mathrm{~cm}\) tall and erect. Where are the object and image located? Is the image real or virtual?

Short Answer

Expert verified
The object is located at a distance of approximately -38 cm while the image is formed at approximately 20 cm on the same side as the object. This implies that the image is virtual and not real.

Step by step solution

01

Finding Object Distance

First, we use the lens equation \(1/f = 1/d_o + 1/d_i\), where f is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. We rearrange this to find \(d_o = 1/ (1/f - 1/d_i)\). The focal length (f) is given as 9 cm. The image distance (\(d_i\)) though not directly given can be implied to be negative because the image is erect, hence it is a virtual image and virtual images are formed on the same side as the object which is conventionally taken as the negative side.
02

Derive Image Distance from Magnification

To derive the image distance (\(d_i\)), we need to utilize the magnification formula which for a lens is given by \(M = -d_i/d_o = h_i/h_o\), where \(h_i\) and \(h_o\) are the image height and object height respectively. We are given \(h_i\) and \(h_o\). As \(h_i\) is positive, this implies that the magnification (M) is also positive. From the magnification formula, we now see that \(d_i = -d_o * M\). As M is positive, \(d_i\) is negative.
03

Use lens formula to determine object location

Substitute equation from Step 2 into equation from Step 1 to determine \(d_o\). This will give object distance which will be on the left of the lens as specified in the question.
04

Use lens formula to determine image location

The image distance (\(d_i\)) can now be found by substituting \(d_o\) into the lens formula. A negative \(d_i\) indicates that the image is on the same side of the lens as the object.
05

Conclude if image is real or virtual

Lastly, we conclude if the image is real or virtual based on the calculated \(d_i\). If image distance is negative, it means image is on the same side of the lens as the object and hence it's a virtual image (as suggested by the erect image) else if it's positive it would have been a real image.

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Key Concepts

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

Lens Formula
Understanding how a converging lens forms an image involves deciphering the lens formula: \( 1/f = 1/d_o + 1/d_i \), where \(f\) represents the lens's focal length, \(d_o\) the object distance, and \(d_i\) the image distance. This equation is a powerful tool that relates these three critical variables allowing for the calculation of any one when the other two are known. For instance, in an exercise where an erect and a larger image is formed by a converging lens, this formula helps to infer the nature of the image—whether it's real or virtual—and the respective positions of the object and image relative to the lens.

One crucial aspect of the lens formula is its sign convention. Real images (formed by converging lenses when objects are placed outside the focal point) produce a positive image distance, \(d_i\), because they are formed on the opposite side of the lens from the object. Conversely, a negative \(d_i\) signals a virtual image, which occurs when the object is within the focal length of a converging lens, resulting in an image on the same side as the object.
Object Distance
The object distance, \(d_o\), is simply how far the object is from the lens. It's a pivotal factor in determining the characteristics of the resulting image. Typically, in this context, the object distance will dictate the size and type of image that the lens will form. If an object is placed outside of a lens's focal length, a real, inverted image is produced. However, if an object is within the focal length, the image will be virtual and erect—just as depicted in the textbook example where an erect image suggests the object was placed close to the lens.

Understanding Object Distance with Real Examples

A real-world implication of object distance can be seen in photography. Photographers adjust the object distance to ensure they capture a clear image of their subject; too close, and the subject might not be in focus (akin to a virtual image), too far, and the subject will be too small in the final shot (similar to the effects of a real image at a large object distance).
Image Distance
Image distance, \(d_i\), refers to the distance between the lens and the image it forms. It is a direct outcome of the interplay between the object distance and the lens's focal length as they conform to the lens formula. When \(d_i\) is found to be negative, like in the given exercise, this indicates an image appearing on the same side of the lens as the object—key to identifying a virtual image.

An understanding of image distance can be particularly useful when setting up optical instruments, such as microscopes or projectors. Whether an image is sharp and accurate can depend on precise adjustments to the image distance, balancing the focal length and how far the actual object is from the lens.
Magnification
Magnification is a measure of how much larger or smaller the image is compared to the object, and is given by the formula \(M = -d_i/d_o = h_i/h_o\), where \(h_i\) is the image height and \(h_o\) is the object height. This equation provides us with a ratio that conveys the degree to which the lens magnifies the object. Using the textbook exercise as an example, a positive magnification value suggested by an erect image height tells us that the image distance is negative, hinting that the image formed is virtual and not real.

Magnification in Everyday Life

Magnification isn't just a theoretical concept; it applies to everyday scenarios such as using a magnifying glass to read small print, where a larger, virtual image is seen, or when looking at distant objects through binoculars, where the desired outcome is a closer, larger view of the object in question.
Real and Virtual Images
The nature of the images formed by lenses—real or virtual—is dictated by the object's position relative to the focal point of the lens. Real images are produced when the object is farther from the lens than the focal length, and these images are inverted and can be projected onto a screen. In contrast, virtual images, which are upright and cannot be projected, form when the object is within the focal length of the lens.

In the described exercise, the creation of an erect and larger image immediately informs us that the image is virtual since real images are always inverted. This is a critical concept in optics, as understanding whether an image is real or virtual influences multiple real-life applications, including the design and use of cameras, glasses, and telescopes, all of which rely on the precise manipulation of light to produce the desired type of image.

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Most popular questions from this chapter

It is your first day at work as a summer intern at an optics company. Your supervisor hands you a diverging lens and asks you to measure its focal length. You know that with a converging lens, you can measure the focal length by placing an object a distance \(s\) to the left of the lens, far enough from the lens for the image to be real, and viewing the image on a screen that is to the right of the lens. By adjusting the position of the screen until the image is in sharp focus, you can determine the image distance \(s^{\prime}\) and then use Eq. (34.16) to calculate the focal length \(f\) of the lens. But this procedure won't work with a diverging lens - by itself, a diverging lens produces only virtual images, which can't be projected onto a screen. Therefore, to determine the focal length of a diverging lens, you do the following: First you take a converging lens and measure that, for an object \(20.0 \mathrm{~cm}\) to the left of the lens, the image is \(29.7 \mathrm{~cm}\) to the right of the lens. You then place a diverging lens \(20.0 \mathrm{~cm}\) to the right of the converging lens and measure the final image to be \(42.8 \mathrm{~cm}\) to the right of the converging lens. Suspecting some inaccuracy in measurement, you repeat the lenscombination measurement with the same object distance for the converging lens but with the diverging lens \(25.0 \mathrm{~cm}\) to the right of the converging lens. You measure the final image to be \(31.6 \mathrm{~cm}\) to the right of the converging lens. (a) Use both lens-combination measurements to calculate the focal length of the diverging lens. Take as your best experimental value for the focal length the average of the two values. (b) Which position of the diverging lens, \(20.0 \mathrm{~cm}\) to the right or \(25.0 \mathrm{~cm}\) to the right of the converging lens, gives the tallest image?

You hold a spherical salad bowl \(60 \mathrm{~cm}\) in front of your face with the bottom of the bowl facing you. The bowl is made of polished metal with a \(35 \mathrm{~cm}\) radius of curvature. (a) Where is the image of your \(5.0-\mathrm{cm}\) -tall nose located? (b) What are the image's size, orientation, and nature (real or virtual)?

A speck of dirt is embedded \(3.50 \mathrm{~cm}\) below the surface of a sheet of ice \((n=1.309) .\) What is its apparent depth when viewed at normal incidence?

A light bulb is \(3.00 \mathrm{~m}\) from a wall. You are to use a concave mirror to project an image of the bulb on the wall, with the image 3.50 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?

A camera with a 90-mm-focal-length lens is focused on an object \(1.30 \mathrm{~m}\) from the lens. To refocus on an object \(6.50 \mathrm{~m}\) from the lens, by how much must the distance between the lens and the sensor be changed? To refocus on the more distant object, is the lens moved toward or away from the sensor?

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