Converging Lens

[brickman7713]

I am lost on this. I know the equation 1/D_I + 1/D_O = 1/F but I don't know how it applies to this. Thanks

 

A 1.9 cm-tall object stands in front of a converging lens. It is desired that a virtual image 2.7 times larger than the object be formed by the lens. How far from the lens (in cm) must the object be placed to accomplish this task, if the final image is located 10 cm from the lens?

 

[studiot]

2 hours ago, brickman7713 said:

I am lost on this. I know the equation 1/D_I + 1/D_O = 1/F but I don't know how it applies to this. Thanks

 

A 1.9 cm-tall object stands in front of a converging lens. It is desired that a virtual image 2.7 times larger than the object be formed by the lens. How far from the lens (in cm) must the object be placed to accomplish this task, if the final image is located 10 cm from the lens?

 

This is where that still small voice of calm says

Draw a diagram.

That will help. 

As to your formula, you will also need to know what sign convention you are using to employ it.

Are you told the focal length?

[John Cuthber]

14 hours ago, studiot said:

Are you told the focal length?

Tacitly.

And you are, I think given more information than you need.

[brickman7713]

22 hours ago, studiot said:

This is where that still small voice of calm says

Draw a diagram.

That will help. 

As to your formula, you will also need to know what sign convention you are using to employ it.

Are you told the focal length?

Nope that is all I know from the question.

8 hours ago, John Cuthber said:

Tacitly.

And you are, I think given more information than you need.

Yeah I was wondering why I had three numbers and only needed two lol.

[studiot]

1 hour ago, brickman7713 said:

Nope that is all I know from the question.

So have you drawn a diagram?

If so, where is it?

In addition to the formula you quoted (this is known as the lens formula) you should also know this one

[math]\frac{{{\rm{size}}\;{\rm{of}}\;{\rm{image}}}}{{{\rm{size}}\;{\rm{of}}\;{\rm{object}}}}{\rm{ = }}\frac{{{\rm{distance}}\;{\rm{of}}\;{\rm{image}}}}{{{\rm{distance}}\;{\rm{of}}\;{\rm{object}}}}[/math]

 

I also said you need to understand the sign conventions.

I said this because you need to know whether the object and image distances and sizes are positive or negative.

and no, the problem is not overdetermined.

Edited April 27, 2019 by studiot

[John Cuthber]

18 hours ago, studiot said:

and no, the problem is not overdetermined.

What use do you anticipate making of the height of the object?

[studiot]

5 hours ago, John Cuthber said:

What use do you anticipate making of the height of the object?

Yes, thank you,  I see this duplication now.

You are right.