Location final image diveraging and converging lens

[crazyog]

Homework Statement

A 1.5 cm tall object is placed 0.50 m to the left of a diverging lens with a focal length of 0.20 m. A converging lens with a focal length of 0.17 m is located 0.08 m to the right of the diverging lens.
a)What is the location of the final image with respect to the object? answer = final image is lcoated 1.3 m to the right of the object
b)What is the height and orientation with respect to the orginal object of the final image?
answer = 1.4 cm, inverted

Homework Equations

1/q+1/p =1/f
M=hi/h =-q/p

The Attempt at a Solution

Ok, well i need to figure out part a before I can do part b...so
I did 1/p1+1/q2=1/f
1/0.5+1/q=-1/0.2 (negative focal length for diverging)
solve for q = -0.14285
I'm a little confused, when I draw my picture to help me visualize where the image is...i don't know whether or not to add the 0.08 or subtract it ...
I thought I would add it (negative q means the image is in front of the lens (virtual))
so i did...
1/0.22285 +1/q2=1/0.17
1/q2 = 1.39 but if i inverse it then it become 0.71 and that is not the answer...hm

I can't seem to figure this out, please help! thanks in advance!

 

[alphysicist]

Hi crazyog,

Homework Statement

A 1.5 cm tall object is placed 0.50 m to the left of a diverging lens with a focal length of 0.20 m. A converging lens with a focal length of 0.17 m is located 0.08 m to the right of the diverging lens.
a)What is the location of the final image with respect to the object? answer = final image is lcoated 1.3 m to the right of the object
b)What is the height and orientation with respect to the orginal object of the final image?
answer = 1.4 cm, inverted

Homework Equations

1/q+1/p =1/f
M=hi/h =-q/p

The Attempt at a Solution

Ok, well i need to figure out part a before I can do part b...so
I did 1/p1+1/q2=1/f
1/0.5+1/q=-1/0.2 (negative focal length for diverging)
solve for q = -0.14285
I'm a little confused, when I draw my picture to help me visualize where the image is...i don't know whether or not to add the 0.08 or subtract it ...
I thought I would add it (negative q means the image is in front of the lens (virtual))
so i did...
1/0.22285 +1/q2=1/0.17
1/q2 = 1.39 but if i inverse it then it become 0.71 and that is not the answer...hm

I can't seem to figure this out, please help! thanks in advance!

I think everything you've done so far looks fine, but you still have one more step to go. You have found the image distance for the second lens, which tells you where the final image is relative to the second lens. But the question asks you to find out how far the final image is from the object.

 

[crazyog]

Oh ok I think I get it, so since it is the distance from the object to the image I want to add 0.71683 + 0.08+0.5 = 1.3 and the image is to the right
thanks!

ok now for part b...
I know M= -q/p so would I do -0.7168/.22285 = -3.216
and we know the orginal height M=hi/h
-3.216=hi/1.5
but the answer is not 1.4...what am i missing?

 

[crazyog]

Oh I just realized when I find the magnification due to the first lens, and then the second lens, the product of the two gave me the final magnification and I used this with the orginal height to find it to be 1.4 ...
why does the product give me the final magnification? and inverted?

 

[crazyog]

Ok I just realized again haha...my first M is postive and 2nd negative

M1=-2/7
M2=-3.216
-0.918=hi/1.5
hi=-1.4 so inverted
ok I get the answer! but I don't understand why it is the product why not add together?

 

[alphysicist]

Oh I just realized when I find the magnification due to the first lens, and then the second lens, the product of the two gave me the final magnification and I used this with the orginal height to find it to be 1.4 ...
why does the product give me the final magnification? and inverted?

The product gives the overall magnification, because of the way the system of lens works. For example, just to use easy numbers, let's say your first lens had a magnification of -2, and the second lens had a magnification of -3. So the image of the first lens is twice as large as its object and inverted, and the image of the second lens is three times larger than its object and inverted. But the image of the first lens is the object for the second lens, so overall the final image is 6 times larger than the (original) object; also, since its inverted twice, the final image is upright.

or in equation form:

<br /> \begin{align}<br /> m_{\rm total}&amp;=m_1 m_2\nonumber\\<br /> m_{\rm total} &amp;=(-2) (-3) = +6\nonumber<br /> \end{align}<br />

 

[crazyog]

Oh ok that makes a lot more sense! Thank you so much! You have been a great help :) !

 

[alphysicist]

Sure, glad to help!