# Finding Image Distance with a Converging Lens: Magnifying Glass Problem

• FritoTaco
In summary, JaredMTg found a place where you can play with a lot of physics simulations, also with lenses.
FritoTaco

## Homework Statement

A magnifying glass has a converging lens of focal length 15.0 cm. At what distance from a nickel should you hold this lens to get an image with a magnification of +2.00?

## Homework Equations

1/f = 1/di + 1/do
m = -di/do or hi/ho

## The Attempt at a Solution

I tried to set it up like this:

1/15 = 1/di + 1/2do I don't know if that's even how it's set up or I tried to set up the magnification equation like

this: m = -di / 2do and cross multiply to get m2do = /di. I don't know how to set the problem up properly.

Hello Taco, welcome to PF

You surprise me writing 1/f = 1/di + 1/do on one line and then 1/15 = 1/di + 1/2do a little further. Where does the 2 come from ?
If that is the magnification, then why first m = -di/do and a little further m = -di / 2do ?

Thanks! I know the magnification is 2 in order to help find the distance of the object. I first thought 2 would be plugged straight into m, so 2 = -di/do. But I don't have di or do. I only have one other number which is 15 for the focal length. Do I need to manipulate the equations? (i.e. 1/f = 1/di + 1/do and m = -di/do) Is there something I'm not seeing because I feel like it should be easier then it looks.

2 = -di/do is good. So you can express di in terms of do.
the manipulation is simple: multiply left and right with do.
In mathematical terms: you have eliminated di. If you now substitute that in the other equation, you get one equation with one unknown (namely do), from which you should eb able to derive do.

There is one small problem: the minus sign. If you write "m = -di/do or hi/ho", do you have some coordinate system in mind ? Is the lens equation in the same system ?
Is the exercise really asking for a magnification +2 in the sense that it wants the image upright ? I think it is.
Can you draw and post a sketch of the situation ?

The real lens equation is similar. I have a screen shot of magnification and tin lens equation. My teacher said we can use di/do and hi/ho since its easier. I don't really know if the negative sign in di matters but I know it means virtual image and on the same side as the object, so yes, you are right about it being upright. Also the question says the image has a magnification of +2. Here is what i have done so far with what we've discussed

.

OK, textbook page is clear on what they mean (and you reproduced that correctly in your original post).
Your notes leave something to be desired, though: M > 0 leads us to expect one of the two of p and q to be negative. Let it be q (di).
So you have ##\ 2 = {-d_i\over d_o} \Rightarrow -d_i = 2 d_o## . Now for the lens formula: If I substitute ##\ d_i = - 2 d_o \ ## I sure don't get what you get !

In the mean time I've made a drawing and the situation is completetly clear ! I invite you to do so too, it really helps understandding, also for later exercises.

FritoTaco
Thank you BvG! I wouldn't have solved it without your help. The answer is 7.50cm and the book answer is also 7.50cm. I've made a red pen mark in what you've got that I also missed.

That looks very good! Up to you to make a drawing. It's really useful to develop that skill, it'll help you a lot in other exercises.

Oh yeah, I forgot about that, I think this is right but it's not drawn to scale.

BvU

Not to scale is not relevant. You used the proper phsyical principles
1. rays going through focal point come out parallel to the optical axis
2. rays coming in parallel go through focal point
3. rays going through center go through unaffected
A (perhaps) dashed line from foot of red arrow to tip of black would heve made it even clearer, but it's obvious that you understand it all.

And you still have one of the three rules left over for checking purposes !

gracy
Hey, glad I made your day as you did to me too! :) Yeah, the 3rd line going through the center is good but probably will use it on the test. Glad it's solved, cya around BvU!

JaredMTg found a place where you can play with a lot of physics simulations, also with lenses !
You can ask for the virtual image and shift the object to get a picture like

FritoTaco and gracy
Hey, thank you again! I will definitely try it out to prepare me for the exam.

## 1. What is a converging lens?

A converging lens is a type of lens that is thicker in the middle and thinner at the edges. It is also known as a convex lens and it causes light rays to converge or come together at a focal point after passing through it.

## 2. How do you calculate the focal length of a converging lens?

The focal length of a converging lens can be calculated using the formula 1/f = 1/u + 1/v, where f is the focal length, u is the distance of the object from the lens, and v is the distance of the image from the lens.

## 3. What is the difference between a real and a virtual image in a converging lens problem?

A real image is formed when the light rays actually converge and intersect at a point, whereas a virtual image is formed when the light rays appear to come from a point but do not actually converge. In a converging lens problem, a real image is formed when the object is placed beyond the focal point, and a virtual image is formed when the object is placed between the lens and the focal point.

## 4. 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 using the formula m = -v/u, where m is the magnification, u is the distance of the object from the lens, and v is the distance of the image from the lens.

## 5. How can the position of an object be adjusted to create a larger or smaller image with a converging lens?

To create a larger image, the object should be placed closer to the lens. To create a smaller image, the object should be placed farther away from the lens. The distance of the object from the lens affects the magnification of the image formed.

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