Question from simple lens diagram (camera)

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SUMMARY

The discussion focuses on calculating the image distance (v) for a simple camera using a converging lens with a focal length of 50.0 mm and an object distance (u) of 8.00 m. The correct calculation yields an image distance of 50.3 mm, confirming that the film must be positioned inside the camera. Additionally, the linear magnification (m) is calculated using the formula m = -v/u, resulting in a magnification of -0.0063, indicating that the image is inverted. The height of the tree is not necessary for determining linear magnification.

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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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