How Does Magnification Affect Object Distance in Lens Optics?

[Dx]

Hi!

I have 2 problems that I need some help with, please.

1) How far from a 50mm focal length lens, such as used in many 35mm cameras, must an object be positioned if it is so to form a real image magnified by a factor of 3?

how do i solve for this? I have tried to solve using lens equation but not the answer I would expect, my answer is 52mm. Is this correct if not how did you go about solving for this?


2) A biconvex lens is formed by using a piece of plastic(n=170). the radius of the front surface is 20cm and the radius of the back surface is 30cm. what is the focal length of the lens?

I know that r/2 = f but should i add both radius's and use that /2? I need some help?
Thanks!
Dx

 

[arcnets]

Hi Dx,

1) Sorry I can't reproduce your answer. What equation did you use, and what values did you plug in?
Anyway I think it's a strange question since in most cases a camera image will be much smaller than the object.

2) I think r/2=f can't be the correct equation since the refractive index n doesn't appear in it. Plus I doubt that the value n = 170 is correct. You would rather expect n to be in the range between n=1 and n=2. As for a biconvex lens, you should add the inverses of both focal lengths to get the inverse focal length, i.e. 1/f = 1/f1 + 1/f2.

 

[Dx]

1) Sorry I can't reproduce your answer. What equation did you use, and what values did you plug in?
Anyway I think it's a strange question since in most cases a camera image will be much smaller than the object.

My equation used was 1/do + 1/di = 1/f, since my f lenth is 1/50mm i tried to solve for do. Unless i am incorrect i believe that do and f are the same distance, right? that's how i solved for it using substitution. But the factor of 3 forlateral magnification, right?
m=+3, so di=m*do = 20cm. Sorry I don't have all my notes but i am trying to remember all my steps, i substituted the do answer to get 1/do - 1/f to get my answer. I am starting to think I am really confused here. Yes! Its a very strange question i think not enough info i though at first but my teacher sasy there's enough to solve. I am going to keep trying anyways...thanks!

2) I think r/2=f can't be the correct equation since the refractive index n doesn't appear in it. Plus I doubt that the value n = 170 is correct. You would rather expect n to be in the range between n=1 and n=2. As for a biconvex lens, you should add the inverses of both focal lengths to get the inverse focal length, i.e. 1/f = 1/f1 + 1/f2.

I think for the particular problem its refractive index n = 1.7 but your correct its normally not. i remeber reading that to get the f = r/2 but i need the inverses of the f to solve for this your saying, sir? I think the formula i found is it!
1/f = (1.7-=1)(1/20+1/30) = 17cm
is 17cm correct?

I would appreciate anymore of your help, I am still lost?
Thanks!
Dx

 

[arcnets]

Originally posted by Dx
My equation used was 1/do + 1/di = 1/f, since my f lenth is 1/50mm i tried to solve for do.

That is IMO correct.

i believe that do and f are the same distance

I think this is wrong. Because if you place the object in the focus, there will be no image. I think you should rather use di = 3*do, since you want a magnification of 3.

i need the inverses of the f to solve for this

Yes you need a proper formula for calculating f1 and f2. The value of n must appear in this formula. Then you use the formula I gave above. Remember this is only correct, if lenses are thin.

 

[Meninger]

1) How far from a 50mm focal length lens, such as used in many 35mm cameras, must an object be positioned if it is so to form a real image magnified by a factor of 3?

I am guessing that it is convex since a there is no way to get a magnified image with a concave.

m=-di/do

di=-3do

1/f=1/do - 1/3do

1/f=3/3do - 1/3do

1/f=2/3do

2f/3=do

2(50mm)/3=do

33.33mm=do, here is your answer