Solving a Virtual Image Problem Involving a Converging Lens

Click For Summary
A converging lens with a focal length of 0.246 m creates a virtual image located 0.933 m from the lens, on the same side as the object. The appropriate formula for this scenario is 1/f = -1/di + 1/do, acknowledging that the virtual image distance (di) is negative. The user attempts to calculate the object distance (do) but arrives at an incorrect value of 0.197157 m. The discussion highlights the importance of correctly applying the lens formula for virtual images. Accurate calculations are crucial for solving lens problems effectively.
lpau001
Messages
25
Reaction score
0

Homework Statement



A converging lens of focal length 0.246 m forms a virtual image of an object. The image appears to be .933 m from the lens on the same side as the object. What is the distance between the object and the lens?


Homework Equations


1/f = 1/di + 1/do

Since the image is virtual, it would be a negative distance right?

so 1/f = -1/di + 1/do


The Attempt at a Solution


I get do = (.246)(.993)/(.246+.993) = .197157 m
which is wrong..
 
Physics news on Phys.org
The formula1/f=1/di - 1/do
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
988
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
9
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K