Solving Optics Lens Problems: 2 Possibilities for Object Placement

[oneplusone]

Hello,

If you have an object, and a bi-convex lense, you are given the focal length, and you are given the magnification factor (M) to be 2, how are there two possibilities to place the object?

One question I was doing asked this, and said:

a) if the image is virtual, and M=2, find the distance.
b) if the image is real, and M=2 find the distance.

Basically, i found one answer easily, and noticed that the absolute value of the difference between my distance and the focal length, equaled the absolute value of the other distance and the focal length.

So is this always the case?

 

[Simon Bridge]

The short answer is because there are two ways to get a magnified image from a converging lens.

To understand this - draw the ray diagrams.
image height is 2units and object height is 1unit (M=2)
put image and object some reasonable distance apart and draw the rays that must connect them ... and you'll find where the lens has to be and what the focal length is.

Repeat for a virtual image.

 

[oneplusone]

So would this be correct analytically? :

------|--------------[C]---------------|-----------

If the distances between the vertical segments, and the focal length [C] are equal, then both will produce an image of the Same size, same distortion, except one will be real, and one will be virtual?

 

[Simon Bridge]

That makes no sense to me, sorry.

 

[sardar]

Hello,

Thank you for your question. When solving optics lens problems, there are two possibilities for object placement due to the nature of lenses and how they form images. The first possibility is when the image is real, meaning it can be projected onto a screen, and the second possibility is when the image is virtual, meaning it cannot be projected onto a screen and can only be seen through the lens.

In the given scenario, you have an object placed in front of a bi-convex lens with a known focal length and a magnification factor of 2. In order to find the distance at which the object must be placed from the lens, we need to use the thin lens equation: 1/f = 1/di + 1/do, where f is the focal length, di is the image distance, and do is the object distance.

For the first possibility, when the image is virtual, the image distance is negative (since it is on the same side as the object) and the object distance must be greater than the focal length. This means that the object is placed between the lens and its focal point. Solving for do, we get do = -f/2, which gives us the distance at which the object must be placed in order to produce a virtual image with a magnification factor of 2.

For the second possibility, when the image is real, the image distance is positive (since it is on the opposite side of the object) and the object distance must be less than the focal length. This means that the object is placed beyond the focal point. Solving for do, we get do = 2f, which gives us the distance at which the object must be placed in order to produce a real image with a magnification factor of 2.

In both cases, we can see that the absolute value of the difference between the object distance and the focal length is equal to the absolute value of the image distance and the focal length. This is because of the nature of the thin lens equation and the fact that the image and object distances are inversely related.

In conclusion, there are two possibilities for object placement when solving optics lens problems, depending on whether the image is real or virtual. The distance at which the object must be placed can be determined using the thin lens equation, and in both cases, the absolute value of the difference between the object distance and the focal length will be equal to the absolute value of the image distance and