XPost: uk.d-i-y   
   From: tw_usenet@dionic.net   
      
   On 22/04/15 15:48, Phil Hobbs wrote:   
   > On 04/21/2015 06:15 PM, john james wrote:   
   >>   
   >>   
   >> "Tim Watts"  wrote in message   
   >> news:0mmi0c-j0m.ln1@squidward.dionic.net...   
   >>> On 21/04/15 19:51, Cursitor Doom wrote:   
   >>>> On Tue, 21 Apr 2015 19:37:31 +0100, Tim Watts wrote:   
   >>>>   
   >>>>> On the subject of glasses, for a single focus lens for a short sighted   
   >>>>> person, how are intermediate lenses sepcified (ones you might use for   
   >>>>> reading plus computer work at up to say 1m)?   
   >>>>>   
   >>>>> Is the far sight dioptre power changed (by a factor or actually   
   >>>>> prescribed by the optician) or is the far sight dioptre given to the   
   >>>>> lens maker along with a request for "intermediate"?   
   >>>>   
   >>>> People have varying requirements when it comes to intermediates, so you   
   >>>> can't just specify according to some preset formula. It requires a   
   >>>> separate sight test at the specific distance - one meter in your   
   >>>> case; it   
   >>>> was 24" in mine. Assuming you're short sighted, the prescription   
   >>>> will be   
   >>>> less negative. So for example if your distance script is -3.25 for each   
   >>>> eye your intermediate script will be -2.75 or thereabouts give or   
   >>>> take a   
   >>>> bit. But you need to test at the appropriate distance to ascertain what   
   >>>> the 'give or take bit' is.   
   >>>>   
   >>>   
   >>> So in principle, if you have your far prescription, it ought to be   
   >>> possible to calculate an intermediate based on "I want to focus upto   
   >>> 1m instead of infinity"?   
   >>   
   >> Yes, and there is a newsgroup that covers that stuff, I've added it.   
   >   
   > Yup.  Optical power in dioptres is the reciprocal of the focal length in   
   > metres.  This is useful because of the lens formula:   
   >   
   > 1/d_o + 1/d_i = 1/f,   
   >   
   > where d_o is the object distance, d_i is the image distance, and f is   
   > the focal length.   
   >   
   > In terms of optical power P=1/f, this is   
   >   
   > 1/d_o + 1/d_i = P.   
   >   
   > Optical power just adds, i.e. for thin lenses, if you have two of them   
   > adjacent, P1 and P2, then P = P1 + P2.  To change focus from infinity to   
   > 1 m, you could just hold a 1 m focal length lens in front of your   
   > distance glasses, right?  So now you know how to compute the required   
   > correction: since the powers add, just add +1 dioptre to your infinity   
   > prescription.   
   >   
   > More formally, changing from d_o = infinity to d_o = 1m requires the   
   > addition of power   
   >   
   > delta P = (1/new d_o) - 1/old_d_o)   
   >   
   > i.e. delta P = 1 dioptre   
   >   
   > So as I say, just take your distance prescription and add 1 dioptre--if   
   > you're -2.75 for infinity, -1.75 will make a focus at 1 m, -0.75 -> 0.5   
   > m, etc.  (When you get to really close distances, the position of the   
   > lenses starts to matter, so you have to be more precise, but the above   
   > thin-lenses-in-contact approximation is better than good enough for   
   > eyeglasses.)   
   >   
   > Cheers   
   >   
   > Phil Hobbs   
   >   
      
   That is really cool - thank you Phil.   
      
   So presumably this applies equally when the eye lens is focussed at its   
   other extreme (ie near)?   
      
   So if -3.50 dioptre (my far sight power) lets me view in focus down to   
   0.5m, then if I add 1.25 for reading (as the optician specified), then I   
   should be able to focus down to:   
      
   1.25 = 1/X - 1/0.5   
   1.25+2 = 1/X   
   X = 1/2.25   
   X=0.44m   
      
   Something seems not right there - what did I do wrong?   
      
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