# Two Converging Lense Questions

• JM2107
In summary, the conversation discusses the possibility of obtaining a non-inverted image with a converging spherical lens and the corresponding distance between the object and the lens needed to achieve a magnification of one. One method to answer this question is to use graphical methods or the Law of lenses. It is not possible to obtain a non-inverted image using a single converging lens, though it is possible to obtain a magnification of -1. The reason for this is purely geometrical, and a virtual image can be created as long as the object is between the focal length point and the lens. However, there is no distance that would give a magnification of one in this scenario.

#### JM2107

For a lens of focal length f, what value of the distance between the object and the lens[D0] would give an image with a magnification of one?

Is it possible to obtain a non-inverted image with a converging spherical lens? Explain please?

Any response would be greatly appreciated, and I would like to thank anyone for their response to this thread in advance.

Hi JM2107,
to answer questions like this, you could use use 2 methods:
1) Draw 3 rays, remembering that...
... focal ray becomes parallel ray
... parallel ray becomes focal ray
... central ray is not refracted
2) Use the Law of lenses: 1/o + 1/i = 1/f
where
o = distance of object from lens
i = distance of image from lens
f = focal length

Got it?

It is not possible to obtain a non inverted image using a single converging lens. It is therefore not possible to obtain a magnification of 1, though it is possible to obtain a magnification of -1. (Inverted images have a negative magnification by convention).

The reason for this is purely geometrical. Arcnets outlined a standard graphical method of seeing why this is.

Originally posted by Claude Bile
It is not possible to obtain a non inverted image using a single converging lens.
R u sure? How about a virtual image? See here...

Last edited by a moderator:

Yes, it is possible to obtain a non-inverted image with a converging lens. As long as the object is between the focal length point and the lens it is possible. Of course this would be a virtual image and the image would be magnified.

m=-d(image)/d(object)

If m=1 than we have an equation where we can put d (object) = - d (image). Since 1/f = 1/d (object) + 1/d (image) and since d (object) = - d (image), therefore 1/f = 0

So there is no distance that would give a magnification of one.

## What are two converging lenses and how do they work?

Two converging lenses are two lenses that are placed close to each other and have the same focal length. They work by bending light rays, which causes the light to converge or come together at a specific point called the focal point. This creates a magnified image of the object placed in front of the lenses.

## What is the difference between a converging lens and a diverging lens?

A converging lens is thicker in the middle and thinner at the edges, while a diverging lens is thinner in the middle and thicker at the edges. Converging lenses bend light rays towards a central point, while diverging lenses spread out light rays.

## How do you calculate the magnification of an object using two converging lenses?

The magnification of an object can be calculated by dividing the image distance by the object distance. With two converging lenses, the image distance is the sum of the focal lengths of both lenses, and the object distance is the distance between the object and the first lens.

## Can two converging lenses be used to create a virtual image?

Yes, two converging lenses can be used to create a virtual image. This occurs when the object is placed between the first lens and the focal point of the second lens. The image created will be magnified and upright, but it will appear to be behind the second lens and cannot be projected onto a screen.

## What is the relationship between the distance between two converging lenses and the magnification of the image?

The distance between two converging lenses and the magnification of the image are inversely proportional. This means that as the distance between the lenses increases, the magnification of the image decreases. Conversely, as the distance between the lenses decreases, the magnification of the image increases.