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Formation of Images by lenses!

  • Thread starter lenaosu
  • Start date
  • #1
2
0
1. An object is placed in front of a converging lens in such a position that the lens (f = 11.0 cm) creates a real image located 27.0 cm from the lens. Then, with the object remaining in place, the lens is replaced with another converging lens (f = 19.0 cm). A new, real image is formed. What is the image distance of this new image?


2. I tried using 1/f= 1/d0-1/di to solve for this. But it doesn't seem to be working.
Any suggestions.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
1
Hi lenaosu,

The thin lens formula is

[tex]
\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}
[/tex]

Depending on the problem, any of these variables might turn out to be negative numbers. What numbers are you using, and what do you get?
 
  • #3
2
0
i used the thin lens equations twice.
1.) 1/f-1/di=1/do
1/11-1/27=1/di
di=18.6cm

2.) 1/f-1/do=1/di
1/19-1/18.6=1/di
di= -883.5cm

My calculations for my second image doesn't seem correct. My image should be to the left of my lens and not to the righ tof my lens.
 
  • #4
alphysicist
Homework Helper
2,238
1
Since the object distance is less than the focal length for the second lens, it's image should be virtual, and so I think the fact that the final image length is negative is right.

However, you should not round your result from step 1 to three digits. Keep all the digits you can, and then round the final answer. Here it makes a relatively large difference.
 

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