Physics *Important question* One concave & one convex lens both having focal lengths 4cm are placed very closed to each other. Their combined focal length is: a) 2 cm b) 4cm c) 0 d) 1cm e) Infinity
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http://www.schoolphysics.co.uk/age16-19/Optics/Refraction/text/Lenses_two_thin/index.html
if there is some distance 'd' between them, then we can use this formula Effective focal length \(EFD = \dfrac{f_1\times f_2}{f_1+f_2-d}\)
in our case, f1 = 4, f2 = -4 d=0
The first link gives you the derivation too :)
let me solve it :)
-16/0
yeah, but i don't see \(\large -\infty \) as one of the choice.
e is infinity
but its definitely not a finite number, so i'd go with +infinity
So E is answer .
Right ? @hartnn
yes, it is :)
Thank you so much ^_^ @hartnn Chocolate cake for you :P
yaay! nomnomnom :D welcome ^_^
^_^
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