## 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

1. anonymous

@arindameducationusc @irishboy123

2. IrishBoy123

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

i think this allows you to effectively ignore that minus sign https://gyazo.com/ab233816634c5df31c11d0d53ce7fd8b

4. anonymous

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

@irishboy123

6. IrishBoy123

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

Can you show me how?? @irishboy123

8. IrishBoy123

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

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