Angular magnification of lens

[grouper]

Homework Statement

A small insect is placed 6.55 cm from a 7.00 cm-focal-length lens. Calculate the position of the image. Calculate the angular magnification.

Homework Equations

1/f=1/do+1/di

M=angular magnification=θ'/θ=angle of image/angle of object=N/f where N=near point

The Attempt at a Solution

Finding the image distance is easy using the lens equation and is -101.9 cm. I don't think this problem deals with the near point (although I may be wrong), so the angles need to be known for both the object and image, but I don't know how to do this if the height of the object is unknown. Perhaps I am missing the significance of something (probably something to do with the near point). Thanks for any help.

 

[technician]

I calculated the image distance and got the same as you, 102cm (I have rounded off!)
If this image is being viewed through the lens then it is not at the near point ( I take the near point to be 25cm) so, like you, I am not certain what the question wants.
The next obvious thing I did was calculate the linear magnification 102/6.55 = 15.6.
Don't know if this is any help to you.

 

[grouper]

Yes, the image distance is ≈-100 cm, I got that part of the question correct. I already tried the linear magnification of 15.6, but that was incorrect. Hopefully someone can help me out before Sunday when it's due. Thanks for trying though. I've got another one nobody has responded to if you'd like to take a look: https://www.physicsforums.com/showthread.php?t=549189

 

[technician]

One other thing strikes me about this question: the object (small insect) is practically at the focal point of the lens (6.55 cm, 7cm). If it was at the focal point the image would be at infinity and the angular magnification would be 25/7 (D/f) = 3.6.
I know this is not what is being asked but I like to get what I can from the question.
It would be interesting to have a chat with the person who made up the question !