Solving a Physics Problem: Inverted Image & Lens Separation

AI Thread Summary
The physics problem involves determining the lens type and distances for an inverted image formed by a lens, with a known object-image separation of 0.40m and a magnification of 1/2. A converging lens is identified as necessary to produce the image. The initial step is to use the magnification equation to establish relationships between the distances from the object to the lens (p) and from the lens to the image (q). Once these distances are calculated, the focal length can be determined using the lens formula. The discussion emphasizes the importance of applying the correct equations to solve the problem effectively.
dlaszlo88
Messages
9
Reaction score
0
I have this Physics Problem which I am not to sure where to start it from. The problem is:

A real inverted image I of an object O is formed by a certain lens. The object-image speration is d = 0.40m, measured along the central axis of the lens. The image is just half the size of the object. There is a figure to this problem and its an object O larger than image I. I is a smaller and inverted, the lens has to go in between the object and image.


They ask you what kind of lens must be used to produce this image? I know that it should be a converging lens.
Then they ask How far from the object must the lens be placed and what is the focal length of the lens.

I think that the intial step is to look at the lateral magnification which is 1/2 But from there I am stuck. In order to find part C I know it will be easy just the second part is giving me a hard time. Any help will be greatly appreciated.
 
Physics news on Phys.org
Is this problem from the Halliday / Resnick / Walker book?
 
Yes that is my textbook
 
well, the problem gives you the distance from image to object.

the problem also gives you the magnification (sign and all!).

use the appropriate equation for magnification...

and now you can reduce the problem to a simple algebraic equation to find the distances from the object to the lens and the image to the lens.

(remember that "p" is the distance from the object to the lens and "q" is the distance from the image to the lens.)

and once you have those two distances, you can find the focal length, no problem, using the correct equation. :wink:

...you should be able to do this! :cool:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top