#1

Hey

There's gotta be someone on UG who can do this, cmon, restore my faith in the Pit Goers, please.

I need the following formula reagganged so it's in the form 'u=...'

(1/u)+(1/v)=(1/f)

EDIT: I realise I typed 'rearranging' incorrectly, sorry

There's gotta be someone on UG who can do this, cmon, restore my faith in the Pit Goers, please.

I need the following formula reagganged so it's in the form 'u=...'

(1/u)+(1/v)=(1/f)

EDIT: I realise I typed 'rearranging' incorrectly, sorry

#2

I guess this has to do with focal lengths. Can't you just work out 1/U and then find the inverse of it?

#3

Yes it doesn, of all the things we could do physics coursework on we end up doing it on feckin' lenses. I'm not sure how to go about working out inverses so I'm kinda knackered there

#4

(1/u)+(1/v)=(1/f)

1+u/v=u/f

1=u/f-u/v

Erm, got confused now. Sorry.

1+u/v=u/f

1=u/f-u/v

Erm, got confused now. Sorry.

#5

u=f-v ???

Just a guess

Just a guess

#6

1/u + 1/v=1/f

1/u=1/f - 1/v

1/((1/f) - (1/v))= u

I think. I never actually bother, I just sub all the numbers in. Rearranging is long and unnecessary.

1/u=1/f - 1/v

1/((1/f) - (1/v))= u

I think. I never actually bother, I just sub all the numbers in. Rearranging is long and unnecessary.

#7

Inverse of 1/u is 1/1/u lol. Can you use a calculator?

#8

uhhhhhhhhhh

(-1/f)+(1/v)=(1/u)?

(-1/f)+(1/v)=(1/u)?

#9

if you want to try it daft's way

1/u = 1/f-1/v. work out 1/f-1/v to give you your answer, which I'll call "y".

1/u = y

then just divide 1 by y. to get u.

(EDIT: this assumes you have values for f and v)

1/u = 1/f-1/v. work out 1/f-1/v to give you your answer, which I'll call "y".

1/u = y

then just divide 1 by y. to get u.

(EDIT: this assumes you have values for f and v)

#10

Inverse of 1/u is 1/1/u lol. Can you use a calculator?

How did I not see that? Thanks.

Thanks to everyone else who responded too.

#11

(1/u) + (1/v) = (1/f)

(1/u) = (1/f) - (1/v)

u = f - v

(1/u) = (1/f) - (1/v)

u = f - v

#12

^ He has it. I was gonna say multiply everything buy 'ufv', but it's simple.

#13

^ He has it. I was gonna say multiply everything buy 'ufv', but it's simple.

u=f-v ???

#14

he actually doesn't have it.

*Last edited by vantage4 at Jan 29, 2007,*

#15

^ He has it. I was gonna say multiply everything buy 'ufv', but it's simple.

No he doesn't. That's wrong.

It's 1/((1/f - 1/v)) = u

#16

No he doesn't. That's wrong.

It's 1/((1/f - 1/v)) = u

i don't believe you're right either.

#17

garg.

burningcowsrule: what values for u, f and v do you already have? cos that'd help. a heck of a lot.

burningcowsrule: what values for u, f and v do you already have? cos that'd help. a heck of a lot.

#18

^He wants to rearrange the equation, not solve it.

I am right.

1/u + 1/v = 1/f

1/f - 1/v = 1/u

u(1/f - 1/v)= 1

u = 1/(1/f - 1/v)

i don't believe you're right either.

I am right.

1/u + 1/v = 1/f

1/f - 1/v = 1/u

u(1/f - 1/v)= 1

u = 1/(1/f - 1/v)

#19

neater way:

1/u + 1/v = 1/f

1/u = 1/f - 1/v

1/u = (f-v)/(fv)

u= (fv)/(f-v).

And note: 1/u = 1/f - 1/v does not imply u = f - v, for example if u=v=4 and f=2, we have:

1/4 = 1/2 - 1/4, which is true, but 4 =/= 2 - 4.

Also, multiplying everything by ufv at the start would work nicely.

1/u + 1/v = 1/f

1/u = 1/f - 1/v

1/u = (f-v)/(fv)

u= (fv)/(f-v).

And note: 1/u = 1/f - 1/v does not imply u = f - v, for example if u=v=4 and f=2, we have:

1/4 = 1/2 - 1/4, which is true, but 4 =/= 2 - 4.

Also, multiplying everything by ufv at the start would work nicely.

#20

i actually figured this out earlier today as chance would have it;

{"."="multiply"}

(u.v.f/u)+(u.v.f/v)=(u.v.f/f)

v.f+u.f=u.v

i think

{"."="multiply"}

(u.v.f/u)+(u.v.f/v)=(u.v.f/f)

v.f+u.f=u.v

i think

#21

neater way:

1/u + 1/v = 1/f

1/u = 1/f - 1/v

1/u = (f-v)/(fv)

u= (fv)/(f-v).

And note: 1/u = 1/f - 1/v does not imply u = f - v, for example if u=v=4 and f=2, we have:

1/4 = 1/2 - 1/4, which is true, but 4 =/= 2 - 4.

Also, multiplying everything by ufv at the start would work nicely.

DING!

This guy is right, u = fv/(v-f)

#22

^He wants to rearrange the equation, not solve it.

ok, cool.