Solve Thin-Lens Equation: Prove u=v, Find Focal Length Change

[sci0x]

Question
i)
Prove that if an object is at a distance of twice the focal length from a converging lens, the image is the same size as the object. Show this using an algebraic method.

Possible Answer
Must prove u = v
2F = U
2/u + 2/v = u
u + v = 2u
v=u

ii)
A small hand-held camera has a convex lens with a focal length of 5cm. The camera is arranged so that by moving the lens in and out, objects from 50cm away to "infinity" (very remote objects like mountains and scenery) can be brought to an exact focus on the film. How many centimeters will the lens have to be moved as the focus is changed from the closest to the farthest object?

Possible Answer:
Well, F = 5cm
U = 50cm
No I can't do this, could someone explain it please.

 

[link2001]

The solutions for both of these questions involve the thin lens equation. For (i) using the object distance and focal lenth will allow you to solve for the image distance. Then using a similar triangle argument you can show that the two heights are the same.

For (ii), find the image distances for each of the the two distances (50cm and infinity). Since objects at 50cm are the nearest objects that can be imaged, the flim must be located this far away from the lens... Hopefully this will start you along the right track.

 

[kyle8921]

To solve this problem, we can use the thin lens equation: 1/f = 1/u + 1/v, where f is the focal length, u is the object distance, and v is the image distance. We know that for the closest object, u = 50cm and for the farthest object, u = infinity. We can solve for v in both cases:

For the closest object: 1/5 = 1/50 + 1/v
1/v = 1/5 - 1/50
1/v = 9/50
v = 50/9 cm

For the farthest object: 1/5 = 0 + 1/v
1/v = 1/5
v = 5 cm

Now, to find the difference between these two values, we can simply subtract them: 50/9 - 5 = 45/9 = 5 cm. This means that the lens will have to be moved 5 cm as the focus is changed from the closest to the farthest object. This makes sense, as the closer the object is to the lens, the farther it will have to be moved to achieve the same focus on the film.