Question

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

Answer 1

@arindameducationusc @irishboy123

Answer 2

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

Answer 3

i think this allows you to effectively ignore that minus sign

https://gyazo.com/ab233816634c5df31c11d0d53ce7fd8b

Answer 4

Would you be kind enough to tell me whether it can be done from this formula or not???
1/f =1/p + 1/q

Answer 5

@irishboy123

Answer 6

of course

you get the same result

what is the formula, i used one from a legit website so i think it is reliable

Answer 7

Can you show me how?? @irishboy123

Answer 8

i think you mean this
https://gyazo.com/5dc884453c3bdefb5ec0da3141e13451

if so you could do pretty much the same thing
\(\large \frac{1}{f}=\frac{1}{p}+\frac{1}{q} = \frac{1}{R}+\frac{1}{R} = \frac{2}{R}, f = \frac{R}{2}\)

\(\large \frac{1}{f^*}=\frac{1}{p}+\frac{1}{q} = \frac{1}{R}+\frac{1}{\infty} = \frac{1}{R}, f^* = \frac{R}{1}\)

\(\large f^* = 2f\)

Answer 9

Why you have written "R" in place of p and q because its not in my syllabus... @irishboy123