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Lenses - drawing ray diagram

  1. Mar 21, 2005 #1
    Hi i have a question about a lens,im not sure if this is the right place to ask but il ask anyways...please help im really stuck!!

    An object is placed 24cm infront of a concave mirror of focal length 18cm. Where is the image formed and what is its magnification?Include an accurately dranw ray diagram??
  2. jcsd
  3. Mar 21, 2005 #2
    \frac{1}{f} = \frac{1}{u} + \frac{1}{v}

    where f is the focal lenth of the lense,v is the object distace and u the image distance.

    [note: allways good to say what you have allready tried!]

    that didn`t look quite right did it .... (latex is harder that i 1st thought!!)

    1/f = 1/u + 1/v
    Last edited: Mar 22, 2005
  4. Mar 21, 2005 #3


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    Zanazzi: Your slash before the ending tex tag should be forward not backwards
  5. Mar 21, 2005 #4
    This is an exercise that belongs in some homework section of the forums. From a mental ray diagram, I think the image is upright, shrunken and is a virtual image on the same side of the lens as the object.

    It is really quite simple. Use the lens equation:

    [tex] \frac{1}{s_0} + \frac{1}{s_i} = \frac{1}{f} [/tex]

    The only things you must be careful about are the signs (+,-) of your quantities. Do we count a concave lens as negative or positive? Look that up in your notes (and read about the lens equation).
    Last edited by a moderator: Mar 21, 2005
  6. Mar 22, 2005 #5

    oh right,thanks,that was really helpfull...so would the equation turn out like this:

    1/f = 1/u +1/V
    1/18 = 1/u + 1/24
    u = 1/72

    so is this the image distance?? or should the lengths have been in Metres?? and also how do i work out the magnification??
  7. Mar 22, 2005 #6
    thanks for your help,the equation you have given is the same as this one:
    1/f = 1/u + 1/v

    is that correct?? i dont know if the concave lens is positive or negative,it doesnt say in the question??
  8. Mar 22, 2005 #7
    Your question stated the distance in cm so the asnwer you worked out is also in cm!

    When it comes to magnification, what`s happened? The image has changed scale, so it`s a question about proportions! have a think and if you can work it out ...
  9. Mar 22, 2005 #8
    I`ve always used some thing calle da sign convention (there are diffrent ones but they all give the same answers!)

    For the "real is positive" convention:

    both f and R are positive for concave mirrors/lenses (real focus)
    both f and R are negative for convex mirrors/lenses (virtual focus)
    u is positive for real objects and negative for virtual objects
    v is positive for real images and negative for virtual images

    this becomes very important when you start putting more than one lense in a system!
  10. Mar 22, 2005 #9
    ok so i think if the object distance is 24cm and the image distance is 0.01cm then its magnification is 2400???..im nots ure i quite understand that??
  11. Mar 22, 2005 #10
    Whats "R"???...oh so if i had the same question but the lens was convex then i would use the same equation as before but the focal length would be negative??
  12. Mar 22, 2005 #11
    Maybe the language I used confused you, sorry
    If the object is 24 cm away from the lense and the image formed is 72cam away then, the magnification is the relationship between the two distances,
    [tex] - \frac {72}{24} = -3 [/tex]

    you can also do it if you know the heights of the image and object

    basically [tex] Mag = \frac{image height}{object height}= - \frac {v}{u} [/tex]
    Last edited: Mar 22, 2005
  13. Mar 22, 2005 #12
    You got it!

    R is the radius of a mirror, sorry (again) I should have explained all the terms
  14. Mar 23, 2005 #13
    thats cool but i dont get where you got the 72 from in the equation -72/24??..isnt the number ontop of the fraction meant to be 1/72??
  15. Mar 23, 2005 #14
    so when the lens is convex does the magnification formula change or does it stay as "image distance/object distance???"...thanks, what your telling me really is making sense to me.
  16. Mar 23, 2005 #15
    Ok first I must readdress an errror I made! I gave you the wrong sign convention! Ooops

    So ... for the Real is positive convention ;

    f is positive for a convex lense (real focus)
    f is negative for a concave lense (virtual focus)
    u is positive for real objects and negative for virtual objects
    v is positive for real images and negative for virtual images

    so your lense equation should look like this
    [tex] - \frac {1}{18} = \frac{1}{24} + \frac {1}{v} [/tex]

    giving you an answer of v = -72 cm (not 1/72)

    the minus sign indicates that the image formed is virtual (as it would be for ALL concave lense images!)

    the magnification therefore goes ike this;
    [tex] Mag = - \frac {-72}{24} = +3 [/tex]

    therefore you have an image 3 times bigger than the object

    i`ve attached a ray diagram to help

    keep practising it will become natural soon as you`ve allready got a good understanding of optics!

    A my idol once said " Optics are either very easy or very hard" - Richard Feynman
  17. Mar 23, 2005 #16
    Thanks,that all great..the only problem i have is when i try to use that equation i cant get -72??..i keep gettin 1/72 or -7/72???..how do you end up with just -72??...and finally when i do my ray diagram it should look like the one you have attatched except that my lengths will be different,is that right??
  18. Mar 23, 2005 #17
    Also you have written v = -72......but i thought we were working out u??
    sorry,ive confused myself again.
  19. Mar 23, 2005 #18
    [tex] \frac{1}{72} = \frac{1}{v} [/tex]
    multiply both sides by v
    [tex] 1 = \frac {v}{72}[/tex]
    multiply both sides by 72
    [tex] 72=v [/tex]


    don`t worry too much about what letters to use as long as you remember

    [tex] \frac{1}{focal length} = \frac {1}{object distance} + \frac {1}{image distance} [/tex]

    Maybe using the notation crosson gave may be a little less confusing, it doesn`t matter what the letters are as long as the equation is applied correctly!
    Last edited: Mar 23, 2005
  20. Mar 23, 2005 #19
    as long as the drawing is in proportion it will be fine. The one i've done is to scale, 1:10, the focus I've placed on the opposite side to the object to remind me it`s 'negative' (from the sign convention) and the image is on the same side showing that it is virtual :smile:
  21. Mar 23, 2005 #20

    thanks,i think ive got the understanding of that now and should be able to do a similar question if it was on a convex lens...can i ask you another question??its a pretty long question and i dont know how to tackle it??
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