Hardest Lens Question EK 982<--wrong answer

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Spiker

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Ok be I am pretty sure the EK book is wrong on this one.

this is question 982

there are 2 convergent lens place 28 cm appart. First lens f=24cm second lens f=9cm. a object is placed 6 cm to the left of first lense so it looks like this

object---6cm---L1----28cm-----L2

Anyways so I work on the first lens first

1/f=1/di+1/do
1/24=1/6+1/do
do=-8
so it looks like this

Image (upright) --2cm--object-6cm---L1----28cm-----L2

the image is upright and and virtual and to the left

so the new do would be 8+28=36 and...

1/9=1/36+1/di
di=12

Image (upright) --2cm--object-6cm---L1----28cm---L2---12cm---image (inverted!)

Since the image is on the other side it is inverted

maginfication of the l1=-(-8/6)=4/3 l2=-(12/36)=-1/3 which means the total magnification -4/9 however all the answers are positive WTF....

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I don't have this book, but maybe they were just asking for the magnification and didn't care about whether the object was inverted or not? The magnitude of the amplification is 4/9, so your calculation is correct. I mean, the negative/positive sign is really only there to tell us if the object is inverted or upright and doesn't really say much about the magnification itself.
 
Yea however they made specific point that magnification is negative when it is inverted and they had a similar problem earlier where the magnification is negative...
 
Edit: nvm, didn't see the - sign in front of the do.
 
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Ok be I am pretty sure the EK book is wrong on this one.

this is question 982

there are 2 convergent lens place 28 cm appart. First lens f=24cm second lens f=9cm. a object is placed 6 cm to the left of first lense so it looks like this

object---6cm---L1----28cm-----L2

Anyways so I work on the first lens first

1/f=1/di+1/do
1/24=1/6+1/do
do=-8
so it looks like this

Image (upright) --2cm--object-6cm---L1----28cm-----L2

the image is upright and and virtual and to the left

so the new do would be 8+28=36 and...

1/9=1/24+1/di <---- (EDIT BY DOODLES: so why didnt you use 36 instead of 24 here?)
di=12

Image (upright) --2cm--object-6cm---L1----28cm---L2---12cm---image (inverted!)

Since the image is on the other side it is inverted

maginfication of the l1=-(-8/6)=4/3 l2=-(12/36)=-1/3 which means the total magnification -4/9 however all the answers are positive WTF....

see quote

EDIT:... woops, you did use 36 to get 12, you just wrote 24 by accident.. my bad....
 
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Don't have the book, but if you have two converging lenses, the final image will be upright.

That is because the first lens inverts it.

Then the second lens uses the inverted image as the object, and the image it produces is upright.

EDIT: Nevermind, the first image should be upright and virtual because it is placed within the focal point of the first lens. Is this correct????

Was 4/9 a correct answer? I did this problem another way, and came up with the same answer as you.
 
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Don't have the book, but if you have two converging lenses, the final image will be upright.

That is because the first lens inverts it.

Then the second lens uses the inverted image as the object, and the image it produces is upright.

EDIT: Nevermind, the first image should be upright and virtual because it is placed within the focal point of the first lens. Is this correct????

Was 4/9 a correct answer? I did this problem another way, and came up with the same answer as you.

yes first image is within the focal so second is inverted therefore the total magnification should be negative 4/9
 
Yea, my last post was at 3 a.m., now I actually did the math completely objective from looking at whatever you did and got the same answer. Its probably a magnitude answer for this question, so sign must not matter, but even if you DONT do the math. At first its within focal, so its positive magnification (upright, virtual) then its outside the focal point (inverted, real)... so in the end its inverted real and MUST be negative no matter the math....
 
I just did this problem and got -4/9

My image distance for the for the first lens was -8cm; object distance was 6cm.

The image distance for the second lens was 12 cm; object distance was 36cm.

For total magnification I used

m1*m2 = (-si1/so2)(-si2/so2) = (si1si2)/(so1so2)

Should be an inverted, reduced, real image.
 
EK has an online forum where problem are explained and errata are clarified. I use it all the time.
 
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