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anonymous

  • one year ago

A lens of focal length "f" is cut into two equal halves then the focal length will be.. (A)f. (B)f/2 (C)2f. (D)f/√2

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  1. anonymous
    • one year ago
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    @arindameducationusc @irishboy123

  2. IrishBoy123
    • one year ago
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    actually this is more helpful http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenmak.html seeing the sign convention explained pretty much allows you to ignore that negative sign so for the original lens you have \(\frac{1}{f} = K (\frac{1}{R} + \frac {1}{R})\) wjere K is just a proxy for the refractive indexes and R assumes its symmetrical: \(f = \frac{R}{2K}\) then if we slice it in half you have \(\frac{1}{f^*} = K (\frac{1}{R} + \frac {1}{\infty})\), same K as same materials, R = \(\infty\) for flat edge, so \(f^* = \frac{R}{K} = 2f\) would check that before proceeding

  3. IrishBoy123
    • one year ago
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    i think this allows you to effectively ignore that minus sign https://gyazo.com/ab233816634c5df31c11d0d53ce7fd8b

  4. anonymous
    • one year ago
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    Would you be kind enough to tell me whether it can be done from this formula or not??? 1/f =1/p + 1/q

  5. anonymous
    • one year ago
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    @irishboy123

  6. IrishBoy123
    • one year ago
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    of course you get the same result what is the formula, i used one from a legit website so i think it is reliable

  7. anonymous
    • one year ago
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    Can you show me how?? @irishboy123

  8. IrishBoy123
    • one year ago
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    i think you mean this https://gyazo.com/5dc884453c3bdefb5ec0da3141e13451 if so you could do pretty much the same thing \(\large \frac{1}{f}=\frac{1}{p}+\frac{1}{q} = \frac{1}{R}+\frac{1}{R} = \frac{2}{R}, f = \frac{R}{2}\) \(\large \frac{1}{f^*}=\frac{1}{p}+\frac{1}{q} = \frac{1}{R}+\frac{1}{\infty} = \frac{1}{R}, f^* = \frac{R}{1}\) \(\large f^* = 2f\)

  9. anonymous
    • one year ago
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    Why you have written "R" in place of p and q because its not in my syllabus... @irishboy123

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