Solving a Virtual Image Problem Involving a Converging Lens

In summary, to determine the location of a virtual image formed by a converging lens, you can use the thin lens equation, 1/f = 1/do + 1/di, where f is the focal length of the lens, do is the object distance, and di is the image distance. A virtual image is an image that appears to be located behind the lens and cannot be projected onto a screen, while a real image can be projected onto a screen. To determine the magnification of a virtual image, you can use the equation M = -di/do, and it is possible for a virtual image to be larger than the object. The position of the object affects the location of the virtual image formed by a converging lens
  • #1
lpau001
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Homework Statement



A converging lens of focal length 0.246 m forms a virtual image of an object. The image appears to be .933 m from the lens on the same side as the object. What is the distance between the object and the lens?


Homework Equations


1/f = 1/di + 1/do

Since the image is virtual, it would be a negative distance right?

so 1/f = -1/di + 1/do


The Attempt at a Solution


I get do = (.246)(.993)/(.246+.993) = .197157 m
which is wrong..
 
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  • #2
The formula1/f=1/di - 1/do
 

1. How do I determine the location of a virtual image formed by a converging lens?

To determine the location of a virtual image formed by a converging lens, you can use the thin lens equation, 1/f = 1/do + 1/di, where f is the focal length of the lens, do is the object distance, and di is the image distance. Once you have these values, you can use the equation to solve for the image distance, which will give you the location of the virtual image.

2. What is the difference between a virtual image and a real image?

A virtual image is an image that appears to be located behind the lens, and cannot be projected onto a screen. It is only visible to the observer looking through the lens. A real image, on the other hand, is an image that can be projected onto a screen and is formed by the convergence of light rays at a specific point in space.

3. How do I determine the magnification of a virtual image formed by a converging lens?

To determine the magnification of a virtual image formed by a converging lens, you can use the magnification equation, M = -di/do, where di is the image distance and do is the object distance. This will give you a ratio that represents the size of the virtual image compared to the size of the object.

4. Can a virtual image be larger than the object?

Yes, a virtual image can be larger than the object. This can happen when the object is placed at a distance that is less than the focal length of the lens. In this case, the virtual image will appear magnified when viewed through the lens.

5. How does the position of the object affect the location of the virtual image formed by a converging lens?

The position of the object affects the location of the virtual image formed by a converging lens because it determines the object distance (do) in the thin lens equation. As the object distance changes, the image distance (di) and location of the virtual image will also change. When the object is placed closer to the lens, the virtual image will be located further away from the lens. When the object is placed further away, the virtual image will be located closer to the lens.

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