Solving a Camera Lens Problem: Height of Image on Film

AI Thread Summary
The discussion focuses on calculating the height of a woman's image on film using two different camera lenses with focal lengths of 35.0 mm and 150.0 mm. Using the thin lens equation and magnification formula, the image height for the 35.0-mm lens is determined to be 0.0063 m (6.3 mm), while for the 150.0-mm lens, it is 0.027 m (27 mm). The calculations emphasize that a longer focal length results in a larger image size. The relationship between object height, focal length, and distance from the lens is crucial for these calculations. This approach provides a clear understanding of how lens specifications affect image projection.
jribbe1
Messages
1
Reaction score
0
i need some help with this problem anyone want to give it a shot? thanks

A camera is supplied with two interchangeable lenses, whose focal lengths are 35.0 and 150.0 mm. A woman whose height is 1.80 m stands 10.00 m in front of the camera. What is the height (including sign) of her image on the film, as produced by
(a) the 35.0-mm lens and
m
(b) the 150.0-mm lens?
m
 
Physics news on Phys.org
thin lens equation

Why don't you give it a shot? You'll need the lens equation:
\frac{1}{f} = \frac{1}{o} + \frac{1}{i}

And you'll also need the magnification:
M = -\frac{i}{o}

To understand what these mean, and how to use the sign conventions, read your text!
 


Sure, I can help with this problem! First, we need to understand the relationship between the focal length of a lens and the image size. The focal length is the distance between the lens and the film when the object is in focus. The longer the focal length, the larger the image will appear on the film.

To solve this problem, we can use the formula:
Image height = Object height x (focal length of lens / distance between object and lens)

(a) For the 35.0-mm lens:
We know that the object height is 1.80 m and the distance between the object and lens is 10.00 m. We also know the focal length of the lens is 35.0 mm, which is equivalent to 0.035 m.
Plugging these values into the formula, we get:
Image height = 1.80 m x (0.035 m / 10.00 m) = 0.0063 m
Therefore, the height of the woman's image on the film produced by the 35.0-mm lens is 0.0063 m or 6.3 mm.

(b) For the 150.0-mm lens:
Using the same formula, we get:
Image height = 1.80 m x (0.150 m / 10.00 m) = 0.027 m
Therefore, the height of the woman's image on the film produced by the 150.0-mm lens is 0.027 m or 27 mm.

I hope this helps! Let me know if you have any further questions.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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...

Similar threads

Back
Top