Magnification of an Image formed by a converging lens

In summary, the problem involves using a converging lens with a focal length of 14.2 cm to inspect a coin and determining the magnification given a virtual image formed 30.1 cm away. The solution involves using the equations (1/Si) + (1/So) = (1/f) and M = (hi/ho) = (-Si/So), taking into account the signs of the distances. The correct answer for the magnification is 1.12.
  • #1
fordy314
6
0

Homework Statement


A converging lens with a focal length of 14.2 cm is used to inspect a coin. The lens forms a virtual image 30.1 cm away. What is the magnification?


Homework Equations


(1/Si) + (1/So) = (1/f)
M = (hi/ho) = (-Si/So)


The Attempt at a Solution


1/So = 1/f - 1/Si

1/(1/So) = So = 26.88cm

M = (-Si/So) = -30.1/26.88 = 1.12


I'm pretty sure that I have the solution and answer correct, but I just wanted to double check. This homework is online and the site is telling me that I have the answer wrong and I just wanted to make sure that I wasn't missing anything before I bother my teacher about it. As of right now I think it's just an error in the code of the site, but I'm just double checking. Thanks.
 
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  • #2
fordy314 said:

Homework Statement


A converging lens with a focal length of 14.2 cm is used to inspect a coin. The lens forms a virtual image 30.1 cm away. What is the magnification?


Have you accounted for the fact that it's a virtual image?
 
  • #3
fordy314 said:

Homework Statement


A converging lens with a focal length of 14.2 cm is used to inspect a coin. The lens forms a virtual image 30.1 cm away. What is the magnification?


Homework Equations


(1/Si) + (1/So) = (1/f)
M = (hi/ho) = (-Si/So)


The Attempt at a Solution


1/So = 1/f - 1/Si

1/(1/So) = So = 26.88cm

M = (-Si/So) = -30.1/26.88 = 1.12


I'm pretty sure that I have the solution and answer correct, but I just wanted to double check. This homework is online and the site is telling me that I have the answer wrong and I just wanted to make sure that I wasn't missing anything before I bother my teacher about it. As of right now I think it's just an error in the code of the site, but I'm just double checking. Thanks.

This object [the coin] is placed inside the focus [giving a virtual image] so some/all of those distances may be negative.
I am never confident with that (1/Si) + (1/So) = (1/f) formula, so I always use a couple of alternates, and a ray diagram.
 
  • #4
Check your notes or your textbook. There are specific rules for deciding on the signs of the various quantities.
 
  • #5
TSny said:
Check your notes or your textbook. There are specific rules for deciding on the signs of the various quantities.

Ah, right, I forgot about the signs. Looks like the summer off has made me a little rusty. I just accounted for the signs and got the right answer. Thanks for your help.
 

1. What is the magnification of an image formed by a converging lens?

The magnification of an image formed by a converging lens is the ratio of the height of the image to the height of the object. It can be calculated by dividing the image distance by the object distance.

2. How does the focal length of the lens affect the magnification of the image?

The focal length of the lens directly affects the magnification of the image. A shorter focal length will result in a larger magnification, while a longer focal length will result in a smaller magnification.

3. Can the magnification of the image be greater than 1?

Yes, the magnification of the image can be greater than 1. This means that the image is larger than the object. This is typically the case for converging lenses.

4. How does the distance between the lens and the object affect the magnification of the image?

The distance between the lens and the object, also known as the object distance, is directly proportional to the magnification of the image. As the object distance increases, the magnification decreases and vice versa.

5. What is the difference between real and virtual images in terms of magnification?

A real image is formed when light rays actually converge at a point to form an image. In this case, the magnification is a positive value. A virtual image, on the other hand, is formed when light rays appear to be coming from a point behind the lens. In this case, the magnification is a negative value.

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