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-   -   Grade 11 lenses question (http://www.physicsforums.com/showthread.php?t=367915)

usermanual Jan7-10 08:59 PM

grade 11 lenses question
 
1. The problem statement, all variables and given/known data
a candle is placed 36cm from a screen. where between the candle and the screen should a converging lens with a focal length of 8.0cm be placed to produce a sharp image on the screen?


2. Relevant equations
1/focal length = 1/distance of object from lens + 1/distance of image created from the lens
1/f= 1/Do+1/di
magnification= height (image)/height (object)
magnification= -distance (image)/distance (object)

3. The attempt at a solution
36cm = d0
f= 8cm
di=?
1/di=1/8-1/36
= 1/di 1(x9)/8(x9)-1(x2)/36(x2)
1/di=9/72-2/72
1/di=7/72
then i cross multiplied so 1 times 72 and 7 times di
then 72/7=7d/7 divided each side by 7 to get rid of 7
then i got di to be 10.28

but in the back of the book it says the answer 10.28 is wrong

the right answer in the back of the book is 12cm or 24cm

thx you guys and gals for spending the time and reading this thread and hopefully you guys and gals can answer it

ehild Jan7-10 09:09 PM

Re: grade 11 lenses question
 
The candle is the object, the image appears on the screen. The distance between the object and the image is 36 cm.

In the formula 1/f =1/do +1/di, do means the distance of the object from the lens, and di means the distance of the image from the lens.
Neither of them is 36 cm.

Draw a picture, and find out what relation is between di and do.

ehild

usermanual Jan7-10 09:23 PM

Re: grade 11 lenses question
 
Quote:

Quote by ehild (Post 2521529)
The candle is the object, the image appears on the screen. The distance between the object and the image is 36 cm.

In the formula 1/f =1/do +1/di, do means the distance of the object from the lens, and di means the distance of the image from the lens.
Neither of them is 36 cm.

Draw a picture, and find out what relation is between di and do.

ehild

ehild thx for the quick reply
btw you kno how you ended your post with "ehild"
lol i thought you were calling me a child lmao
ok but on topic ehild would i do 36cm divdied by 2 so 18 and then i just plug in 18 as the value of do O.O
cuz when i did that i got 14.4 as the value of di
andother question is that the question is asking where should the lens be placed between the candle and the screen so after i find di do i just subtract 36-di which will equal the distance the lens is at?

ehild Jan7-10 09:41 PM

Re: grade 11 lenses question
 
You can not just plug-in a number as you like. This would be very childish :). Try to make a drawing. Candle, object, lens in between, and show di and do.

ehild (this is from my real name: E. Hild)

ehild

usermanual Jan7-10 09:55 PM

Re: grade 11 lenses question
 
um E. Hild

i have drawn the picture and i still dont get it =(

candle _____________________________________screen
<---------------------------->
36cm
btw is this a trick question, or does it have something to do with the word "sharp" in the question ^.^ and im also wondering isnt there like infinite possibilities cuz the distance of object is proportional to distance of the image.

another method i tried is that since in the back of the book it says 24cm and 12 cm is the answer so when i subbed in 12cm in di to solve for do i got 24cm and when i subbed in 24 i got 12cm. i kno that 24+12= 36 right but doesnt that mean i can use other 2 value so x+y= 36 ??

srry i may sound (type in this case) a bit confused

also im wondering how i can use the focus 8cm in this question

ehild Jan7-10 10:33 PM

Re: grade 11 lenses question
 
Place the lens somewhere in the drawing, (see attachment) and show which distance is di and which is do.

ehild

usermanual Jan7-10 10:38 PM

Re: grade 11 lenses question
 
Quote:

Quote by ehild (Post 2521643)
Place the lens somewhere in the drawing, (see attachment) and show which distance is di and which is do.

ehild

ok so "Do" is from the candle to the lens and "Di" is from the lens to the screen

ehild Jan7-10 10:44 PM

Re: grade 11 lenses question
 
Well, all right. See attachment. do+di=36 cm. If do=x, di = 36-x. Plug in these expressions into your equation.

[tex]\frac{1}{x}+\frac{1}{36-x}=\frac{1}{8}[/tex]

Cross multiply, you get a quadratic equation for x. Solve

ehild

usermanual Jan7-10 11:07 PM

Re: grade 11 lenses question
 
Quote:

Quote by ehild (Post 2521661)
Well, all right. See attachment. do+di=36 cm. If do=x, di = 36-x. Plug in these expressions into your equation.

[tex]\frac{1}{x}+\frac{1}{36-x}=\frac{1}{8}[/tex]

Cross multiply, you get a quadratic equation for x. Solve

ehild

ehild how would i cross multiply when there is an addition sign [tex]\frac{1}{x}+\frac{1}{36-x}[/tex] isnt it true that i cant break up this up
i tired the quadratic form " x2 -36x -8 =0 just wondering if this is the correct quadratic form.

srry i kno im being a nuisance

ehild Jan7-10 11:29 PM

Re: grade 11 lenses question
 
36-x+x=1/8 (36-x)x

36*8=36x-x^2

x^2-36x+288=0

x1=12
x2=24

so you can place the lens either 12 cm far from the candle or 24 cm from it to get a sharp image on the screen.

ehild

usermanual Jan8-10 12:08 AM

Re: grade 11 lenses question
 
Quote:

Quote by ehild (Post 2521716)
36-x+x=1/8 (36-x)x


x1=12
x2=24

ehild


actually E.Hild i have 1 last question. you know for that answer right for x=12and x=24 cuz 12+24 dont equal -36
shouldn't it be x= -12 and x=-24 because -12*-24= 288 and -12 + (-24)= -36
but if this is the case, then why would the distance be negative.

ehild Jan8-10 12:10 AM

Re: grade 11 lenses question
 
Quote:

Quote by usermanual (Post 2521739)
i dont understand how you went from [tex]\frac{1}{x}+\frac{1}{36-x}[/tex] to 36-x+x=1/8 (36-x)x

im srry math and physics arent my strongest subject

I see. Multiply both sides of the equation by x(36-x).

[tex]x(36-x)(\frac{1}{x}+\frac{1}{36-x})=\frac{1}{8}x(36-x)[/tex]

resolve:

[tex]\frac{x(36-x)}{x}+\frac{x(36-x)}{36-x}=\frac{1}{8}(36x-x^2)[/tex]

simplify: [tex](36-x)+x=\frac{1}{8}(36x-x^2)[/tex]

simplify further and multiply by 8

[tex]36*8=36x-x^2[/tex]

Collect everything at one side:

x^2-36x+288=0

ehild

usermanual Jan8-10 12:19 AM

Re: grade 11 lenses question
 
Quote:

Quote by ehild (Post 2521755)
I see. Multiply both sides of the equation by x(36-x).

[tex]x(36-x)(\frac{1}{x}+\frac{1}{36-x})=\frac{1}{8}x(36-x)[/tex]

resolve:

[tex]\frac{x(36-x)}{x}+\frac{x(36-x)}{36-x}=\frac{1}{8}(36x-x^2)[/tex]

simplify: [tex](36-x)+x=\frac{1}{8}(36x-x^2)[/tex]

simplify further and multiply by 8

[tex]36*8=36x-x^2[/tex]

Collect everything at one side:

x^2-36x+288=0

ehild

ehild thx you so much i get everything except for 1 thing
. you know for that answer right for x=12and x=24 because 12+24 doesn't equal -36
shouldn't it be x= -12 and x=-24 because -12*-24= 288 and -12 + (-24)= -36
but if this is the case, then why would the distance be negative.
i promise this is the last question =)

and thank you for all your time and patience *hug* =)

ehild Jan8-10 12:28 AM

Re: grade 11 lenses question
 
Why should it be equal to -36????

di+do=36. You can see it in the picture.

do can be either 12 or 24.
If do=12, di =24.
If do=24, di=12.

ehild

usermanual Jan8-10 12:31 AM

Re: grade 11 lenses question
 
Quote:

Quote by ehild (Post 2521763)
Why should it be equal to -36????

di+do=36. You can see it in the picture.

do can be either 12 or 24.
If do=12, di =24.
If do=24, di=12.

ehild

cuz ehild i was thinking that for x^2-36x+288=0 since a=1 b=-36 c=288 so dont that mean that i have to find 2 values that add to equal -36 and multiply to give us 288?

ehild Jan8-10 01:41 AM

Re: grade 11 lenses question
 
Quote:

Quote by usermanual (Post 2521766)
cuz ehild i was thinking that for x^2-36x+288=0 since a=1 b=-36 c=288 so dont that mean that i have to find 2 values that add to equal -36 and multiply to give us 288?

The sum of the two roots is -b/a, and their product is c/a.

ehild


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