Question from simple lens diagram (camera)

AI Thread Summary
The discussion revolves around calculating the image distance (v) for a simple camera with a converging lens of focal length 50.0 mm, used to photograph a tree 8.00 m away. Initially, a negative image distance was calculated, indicating an incorrect assumption about the image's position, but this was corrected to 50.3 mm. The linear magnification (m) was then discussed, with the correct formula being m = -v/u, resulting in a value of -0.0063, indicating the image is inverted. It was clarified that the height of the tree is not necessary for calculating linear magnification, as it focuses on the ratio of image distance to object distance. The conversation emphasizes the importance of understanding image orientation and distance in lens calculations.
ZedCar
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Homework Statement


A simple camera has a converging lens of focal length 50.0 mm.

It is used to photograph a tree 3.00 m high.

The tree is 8.00 m away from the lens.

Calculate the distance v between the lens and the film, giving your answer in mm to three significant figures.


Homework Equations



1/f = 1/u + 1/v

The Attempt at a Solution



1/50 = 1/8000 + 1/(-v)
v = -50.3 mm

Is this correct? Thank you! :smile:
 
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Why did you reason that the image distance was negative?
 
Hmmm, not sure why I done that actually...

Have corrected to:

1/f = 1/u + 1/v

1/50 = 1/8000 + 1/v

v = 50.3 mm
 
That looks right. The negative image distance you got before implies that the film needed to be outside the camera to capture the image.
 
Thanks Squeezebox!

When I want to calculate the linear magnification of the image, the following equation is used:

m = v/u

Do I use the same v and u values from above? (both in metres)

i.e.

0.0503 / 8 = 0.0063
 
ZedCar said:
Thanks Squeezebox!

When I want to calculate the linear magnification of the image, the following equation is used:

m = v/u

Do I use the same v and u values from above? (both in metres)

i.e.

0.0503 / 8 = 0.0063

Now you have the opposite problem. There's a negative sign in this one.

m= -v/u

The image gets flipped upside down. This happens with your eyes too; your brain flips the image back right side up.
 
Ah, so it'd be

-0.0503 / 8 = -0.0063

So is the height of the tree, stated in the question as being 3.00 m, not required to obtain the linear magnification?
 
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