Lenses - magnification, position, focal length help

In summary, the lens produces an image of magnification that is +0.667 the size of the object. The lense position and focal length can be found using the magnification equation and the thin lens equation, respectively. A sketch of the situation is shown, demonstrating the axis of symmetry and the principle axis of the lens, its optical centre, and the positions of the object and image.
  • #1
alicia113
103
0
Lenses - magnification, position, focal length help!

ok so here is the quesiton..

An object is loacted 3.5 cm from the optical centre of a lens. The lense produces an image of magnification that is +0.667 the size of the object

a) is this lens a diverging or converging lens? explain

b) Find the position of the image and the focal length of the lens using the magnification equation and the thin lens equation

c) make a diagram of the situation, showing the axis of symmetry and the prinicple axis of the lens, its optical centre itrs principle focal points and the positions of the object and image



ok so you said it was a diverging lens and i have NO IDEA how to do b... i am going back and back to my notes... but i just don't understand :confused: ... someone please help me ! this is due tomorrow too :cry:
 
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  • #2


alicia113 said:
ok so here is the quesiton..

An object is loacted 3.5 cm from the optical centre of a lens. The lense produces an image of magnification that is +0.667 the size of the object

a) is this lens a diverging or converging lens? explain

b) Find the position of the image and the focal length of the lens using the magnification equation and the thin lens equation

c) make a diagram of the situation, showing the axis of symmetry and the prinicple axis of the lens, its optical centre itrs principle focal points and the positions of the object and image



ok so you said it was a diverging lens and i have NO IDEA how to do b... i am going back and back to my notes... but i just don't understand :confused: ... someone please help me ! this is due tomorrow too :cry:

Welcome to the PF.

Per the PF rules, you need to show some effort before we can be of tutorial help.

What do you mean "you said it was diverging"? Who is you?

A good place to start is to list the lens equations, and show how you calculate magnification. Then try to make the sketch they are asking for -- that should get you pretty far along... Please show us your work.
 
  • #3


So magnification is


M= hi/ho = -di/do
 
  • #4


So hi is +3.5cm. But what's the ho?

M=3.5/? = +0.667

So it would end up being 3.5/0.667 = 5.24? Am I correct?
 
  • #5


Good so far. Sketch?
 
  • #6


Is there an equation for focal length.?
 
  • #7


But I thought the magnification was +0.667?
 
  • #8


alicia113 said:
Is there an equation for focal length.?

Yes, the lens equation.
 
  • #9


Is it

1/Di=1/f - 1/do
 
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  • #10
alicia113 said:
Is it

1/Di=1/f - 1/do

But for that I am not sure where the numbers fit in. I do not have focal length. All I have is magnification and where the object is located
 
  • #11
alicia113 said:
But for that I am not sure where the numbers fit in. I do not have focal length. All I have is magnification and where the object is located


I have 1/Di = 1/f - 1/3.5cm


I am missing the focal length


Wait that's the thin lens equation wow ok hold on.
 
  • #12


No I need the thin lens equation .. I am so confused right now !
 
  • #14


Ok so I have my focal length as +4.3 magnification as 5.24 and distance of image is -18.3
 
  • #15


How do I find position of image
 

FAQ: Lenses - magnification, position, focal length help

What is the difference between magnification and position in lenses?

Magnification refers to the ability of a lens to make objects appear larger or smaller, while position refers to where the lens is placed in relation to the object being viewed. Magnification is affected by the focal length of the lens, while position is affected by the distance between the lens and the object.

How does the focal length of a lens impact its magnification?

The focal length of a lens is directly related to its magnification. A shorter focal length will result in a wider field of view and a lower magnification, while a longer focal length will result in a narrower field of view and a higher magnification.

Can changing the position of a lens affect its focal length?

No, the position of a lens does not affect its focal length. The focal length of a lens is determined by its physical shape and cannot be changed by moving the lens.

How can I calculate the magnification of a lens?

The magnification of a lens can be calculated by dividing the image size by the object size. The image size is the size of the object as it appears through the lens, and the object size is the actual size of the object. This calculation can also be expressed as the ratio of the image distance to the object distance.

What is the relationship between magnification and image distance?

The relationship between magnification and image distance is inversely proportional. This means that as the magnification increases, the image distance decreases, and vice versa. This can be seen in the formula for magnification (m = -i/o), where i is the image distance and o is the object distance.

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