Question:
physics problem involving diverging lens?
emperor
2011-07-11 19:23:42 UTC
An object is 21 cm in front of a diverging lens that has a focal length of -9 cm. How far in front of the lens should the object be placed so that the size of its image is reduced by a factor of 3.5?




now some tools include:
the mirror equation which is: 1/f=1/do+1/di
and the magnification equation which is: m=di/d0(h0)

it seems the problem is asking at which d0 will the current d0's magnification be reduced by 3.5.
mathematically it would seem like i would need to calculate the current magnification.
then i could set up a proportion involving the given d0/calculated magnification = new do/ calculated m/3.5


it would probably be alot easier if i could somehow turn this into a calculus problem.

but please help because im stuck
Four answers:
Quark
2011-07-11 19:55:09 UTC
do = 21 cm

f = -9 cm

ho/hi = 3.5



di = do * f/do - f

= 21 * (-9)/(21 -(-9)) = -6.3 cm

M = di/do = 6.3/21 = 0.3

M' = 0.3/3.5 = 0.086



M' = f/(do - f )

-0.086 = -9/(do -(-9))

0.086 = 9/(do +9)

0.086(do + 9) = 9

0.086do +0.77 = 9

do =95.7 cm
Quicksilver
2011-07-11 20:11:01 UTC
First lets, find di in first case.

1/(-9) = 1/(-21) + 1/di

1/21 - 1/9 = 1/di

di = - 189/12 cm

m = di / do = (-189/12) / -21 = 9/12 = 3/4

New magnification = m/3.5 = 3/14 = di / do (These di and do are the new ones that we are going to find out )

1/f = 1/do + 1/di

Now we have two equations and two unknowns.

Substitute di from the first equation in the second.

di = 3 do/14

1/f = 1/do + 14/3do

1/(-9) = (3 + 14)/3 do

-1/9 = 14/3do

do = - (14 x 9)/3 = -42 cm

42 cm from the lens.
Erika
2016-10-23 01:51:55 UTC
A diverging ng lens, by way of itself, can in straightforward terms type a small digital photograph on the comparable area of the lens because of the fact the item. The rays will diverge on the properly suited hand area of the lens into infinity yet finding into the lens this is going to look as though they seem to converge on a smaller digital photograph on the comparable area of the lens because of the fact the item. In different words the image you spot can't be located on a viewing reveal screen. think of using a magnifying glass to amplify some thing. this is a digital photograph on the comparable area of the lens because of the fact the item. A paraxial ray is a hypothetical ray that propagates alongside the optical axis of the device, this is going to continually stay alongside the optical axis until eventually a lens is tilted.
anonymous
2014-04-30 16:07:17 UTC
in the best answer, where did the the +0.77 come from?


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