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Determine Focal Length

1. Homework Statement
A person with presbyopia can see distant objects clearly, but is unable to accomadate to see near objects cuz of hardening lens. one person cannot see objects nearer than 1.5m.
a)the focal length of the eye glasses needed to see objects as close as 25cm.


2. Homework Equations

the answer is 30cm

3. The Attempt at a Solution

i know you gotta use the formula 1/f=1/do + 1/di
(i just dont understand the ?)
 

Answers and Replies

mgb_phys
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The glasses must create an image of the object at a distance of more than 1.5m
The persons eye's can then image this normally.
If you have an object distance of 25cm and an image distance of 1.5m what is the focal length?
 
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:S still dont get it , (the focal length is suppose to be 30)
 
so basically the distant of object is 25cm, and the distant of image should be 150cm, since he cant see objects less than 150cm , right?
 
mgb_phys
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Correct, ps be careful about the signs of object/image distance, it's a virtual image.
 
oh alright i got the answeR:D
 
but how do i know the image is virtual?
 
mgb_phys
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Because the lens is between the object and the eye - you must create an image behind the object. You can't create a real image here ( the rays would have to go backwards) so it is a virtual image.
 
thanks
 
heres a similar question im stuck on,
a tourist wishes to photpgraph 6.0m giraffe from a distance of 50m.
what focal length is required if the image is to fill a 24mm slide?
Ho=6.0m , Do=50m
 
mgb_phys
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What is the magnification he wants ?
Whats the realtionship between magnification and focal length?
 
my mistake - A tourist wishes to photpgraph a 6.0m giraffe from a distance of 50m.
what focal length of lens is required if the image is to fill a 24mm slide?
 
(we didnt really learn how to do these problems so thats why im having some trouble)
 
mgb_phys
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Simplest way is to draw a very simple diagram ( actually all of physics is just a matter of drawing the right diagram!)

Draw a 6.0 m high giraffe (you dont have to draw the giraffe, and arrow will do!) 50m away from a lens, now draw a straight line from the top of the giraffe through the centre of the lens, continue the straight line to the top of the image ( 24mm high)
now you have two similair triangles - all you need to know is the distance from the lens to image.
 
i stil dont get it, i just dont understand whta it means by 24mm slide
 
mgb_phys
Science Advisor
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Ah of course you don't!
For those of us over 20, photographs used to be taken on film - a film frame is 36mm wide by 24mm high. Slide film gave you a positive image, you cut a single film frame out, put it in a little plastic mount and projected it with a slide projector ( you can probably still see them in museums!)

What the question is saying is that you have an object 6.0m high, 50.0m in front of the lens and want to make an image 24mm high, an unknown distance behind the lens.
Since a ray going through the centre of a lens goes in a stright line, all you have to do is draw a ray from the top of the object through the centre of the lens to the top of the image and you have all the dimensions you need.

See this image for details http://upload.wikimedia.org/wikipedia/en/thumb/7/71/Lens3.svg/550px-Lens3.svg.png [Broken]

You can then work out the focal length from the object and image distance.
 
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hi... i am new here, but i try to give my answer...but i dont know if its right or wrong.. just try.
I assume that 24 mm is the height of the image of the object (girraphe) so its h'.. with h=6m and h'=24mm.. we can get the magnification... m=1/250. and now with m=-d1/d0... "d1" is the distance of the image to lens and "d0" is the distance between the object to the lens.. with m known, we can calculate the relation between d1 and d0... and with d0 is known (50 meter), we can calculate d1 and eventually we can find the focal length...
i dont know if its correct or not.. but my results is, the focal length needed is 50.2 meter.
 

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