# Need help with contraconcave mirrors

1. Jun 27, 2012

### utkarsh5

need help with contraconcave mirrors....

i saw the video of mirage illusion on the inter-net .it consists of two concave mirrors placed in front of each other such that focus point on each mirror is on the pole of the other mirror.one of the mirrors has a hole in between.i decided to make one for myself(as i can't get it here where i live).i calculated that the aperture of the mirrors should be equal to ?7 times their focal length(f*2.64).are my calculations correct?also,what size should i make the hole?i am asking this since i can't take risk of ordering wrong mirrors,since i can't get them here,so they will be costly.please answer.thanks in advance!

2. Jun 28, 2012

### Simon Bridge

Re: need help with contraconcave mirrors....

Welcome to PF.
The hole needs to be big enough for you to see the object. It is not a projection image - you need to be able to see a bit of the mirror for it to work.

I cannot tell if your calculations are correct if I don't know your calculations.

Compare what you plan with the following:
http://www.wfu.edu/physics/demolabs/demos/6/6a/6A2035.html

3. Jul 4, 2012

### utkarsh5

Re: need help with contraconcave mirrors....

thank you for your answer!i am going to go with my calculations,and use the corresponding focal lengths(what's science if there is no risk?)as for the size of the hole,i am definitely following your advice.again,thanks!

4. Jul 4, 2012

### utkarsh5

Re: need help with contraconcave mirrors....

as for my calculations,i have uploaded a rough diagram as an attachment.i used the mirrors as as part of two spheres,intersecting each other.since focal length is half the radius of curvature,the line segment joining the two centers is trisected,with each part equal to focal length.between the mirrors,f is bisected again (just an imaginary bisection,to ease the calculation).applying pythagoras' theorem,we get half of aperture = root7 / 2 times f.so, aperture is root7 times f.

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