Location final image diveraging and converging lens

In summary, Homework Statement says 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 and a converging lens with a focal length of 0.17 m is located 0.08 m to the right of the diverging lens. The attempt at a solution found that q=-0.14285 and the final image is 1.3 m to the right of the object.
  • #1
crazyog
50
0

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!
 
Physics news on Phys.org
  • #2
Hi crazyog,

crazyog said:

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.
 
  • #3
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?
 
  • #4
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?
 
  • #5
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?
 
  • #6
crazyog said:
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:

[tex]
\begin{align}
m_{\rm total}&=m_1 m_2\nonumber\\
m_{\rm total} &=(-2) (-3) = +6\nonumber
\end{align}
[/tex]
 
  • #7
Oh ok that makes a lot more sense! Thank you so much! You have been a great help :) !
 
  • #8
Sure, glad to help!
 

1. How does the location of the final image differ for diverging and converging lenses?

For a diverging lens, the final image is always virtual and located on the same side as the object. This means that the image is upright and smaller than the object. In contrast, for a converging lens, the final image can be either real or virtual, depending on the object distance and focal length of the lens. The real image is inverted and located on the opposite side of the lens, while the virtual image is upright and located on the same side as the object.

2. What factors affect the location of the final image for a diverging lens?

The location of the final image for a diverging lens is primarily affected by the object distance and the focal length of the lens. As the object moves closer to the lens, the final image moves farther away from the lens. The focal length also plays a role, as a shorter focal length will result in a final image that is closer to the lens.

3. How does the location of the final image change when using multiple converging lenses?

When using multiple converging lenses, the final image will depend on the combined focal length of the lenses and the distance between them. If the lenses are close together, the final image will be located farther away from the lenses. On the other hand, if the lenses are far apart, the final image will be closer to the lenses.

4. Can the location of the final image be manipulated by changing the curvature of a lens?

Yes, the location of the final image can be manipulated by changing the curvature of a lens. For a converging lens, a more curved lens will result in a shorter focal length and a final image that is closer to the lens. For a diverging lens, a flatter lens will result in a longer focal length and a final image that is farther away from the lens.

5. How does the size of the object affect the location of the final image for a converging lens?

The size of the object does not directly affect the location of the final image for a converging lens. However, a larger object will result in a larger final image, and a smaller object will result in a smaller final image. This is because the size of the final image is directly proportional to the size of the object, as determined by the magnification equation.

Similar threads

Understanding a Converging Lens and Its Two Positions
    • Introductory Physics Homework Help
Replies
3
Views
1K
Image position in an optical system?
    • Introductory Physics Homework Help
Replies
5
Views
1K
Optics Problem with a Double Lens System
    • Introductory Physics Homework Help
Replies
1
Views
1K
Position of the object and image in this optics problem
    • Introductory Physics Homework Help
Replies
10
Views
1K
Calculating Chromatic Aberration
    • Introductory Physics Homework Help
Replies
2
Views
694
Imaging resolution for a microscope and eye
    • Introductory Physics Homework Help
Replies
5
Views
851
Inserting thick lenses into a thin lens system and deducing values
    • Introductory Physics Homework Help
Replies
1
Views
958
Lens with different refraction index on each side
    • Introductory Physics Homework Help
Replies
7
Views
1K
Image with an Plano-Convex lens
    • Introductory Physics Homework Help
Replies
7
Views
2K
Image distance of a Keplerian telescope
    • Introductory Physics Homework Help
Replies
7
Views
1K

Hot Threads

Recent Insights

Back
Top