Optics Question

  • Thread starter dzza
  • Start date
  • #1
14
0
Hey, so i've got a HW for an optics lab and haven't studied optics for a while, so I could use a little help.

The problem is a lensmaker's equation question involving a system of optical devices. An object at a distance of 30cm to the left of a converging lens of focal lenght +30cm. The converging lens is then 10 cm away from a divering lens of known focal length.

Now, I know the general procedure for solving this sort of problem, using the image of the light through the first lens as the object for the second lens. If I could just find the image distance from the first lens I'd be in good shape. But, isn't the object on the focal plane? and if it is, don't all rays of light incident on the first lens from the object emerge parallel to optical axis? In that case, no image is formed.

From the lensmaker's equation, i get 1/30 + 1/s = 1/30, where s is the image distance, and this has no solution. My question is do I have a sign messed up for the focal length or object distance?
 

Answers and Replies

  • #2
andrevdh
Homework Helper
2,128
116
No.You are in the right track. Remeber mathematically that the inverse of avery large number is very small. That is if one make the denominater larger and larger the inverse appoaches zero. So the image will be formed at infinity. This means that the object dinstance for the divergiing lens will be [tex]+\infty[/tex].
 

Related Threads on Optics Question

  • Last Post
Replies
1
Views
861
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
961
D
  • Last Post
Replies
2
Views
874
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
2
Views
705
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
9
Views
2K
Top