Optics, camera lenses and object height/size

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Homework Help Overview

The discussion revolves around a problem in optics involving a camera lens with a focal length of 6 cm, used to photograph a person who is 180 cm tall, positioned 3 m away from the camera. Participants are exploring how to calculate the size of the image on film and the necessary distance behind the lens for the film placement.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of the thin-lens equation and magnification formulas. There are attempts to solve for the image distance and magnification, with some participants questioning the use of units and signs in their calculations.

Discussion Status

Some participants have provided calculations and identified issues with unit consistency. There is an ongoing exploration of different methods, including ray diagrams, to visualize the problem. Guidance has been offered regarding the magnification ratio and the need for consistent units.

Contextual Notes

Participants note the importance of converting units for the object distance from meters to centimeters to align with the focal length. There is also a mention of uncertainty regarding the signs used in the equations.

map7s
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A camera with a focal length f of 6 cm is used to take the picture of a person who is 180 cm tall, standing at a distance 3 m in front of the camera.
How big will the image of that person be on film?
How far behind the lens (assumed to be very thin) must the film be placed?


I tried using the thin-lens equation and the equation of magnification, but I can't seem to be getting the right answer.

(1/d o)+(1/d i)=(1/f)
m=-(d i)/(d o)

Using the first equation, I plugged in the numbers for f and d o and solved for d i. I know that there is something special with the signs, but I tried them every way possible, and I still can't get the right answer.
 
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Show more of your calculation, and what answer it came to.
 
In order to solve for d i : (1/6)-(-1/3)=1/d i -> d i =2
m=2/-3
 
map7s said:
In order to solve for d i : (1/6)-(-1/3)=1/d i -> d i =2
m=2/-3

Be careful of your units. f is in centimeters, and the object distance is in meters! They need to be consistent for the equation to work.
I'm not sure why you have -1/3. Why did you pick negative?
 
Last edited:
okay, that was what was messing me up - the units. But is there anything special that you have to do for part a of the problem? b/c I tried using the equation for magnification. I got part b right, so I took that answer and divided by 300. (m = - (d i)/(d o)
 
If you know how to draw a ray diagram you could try doing that. The numbers in the question are quite round so you should be able to get quite an accurate answer from drawing a ray diagram.
 
I only know how to draw a ray diagram without using numbers (like the image appears on the opposite side of the lens inverted and smaller, but I wouldn't be able to tell how much smaller).
 
Magnification is also expressed as M = (image height)/(object height).

So you've already found the magnification factor. Now apply the above ratio to find the image height, since you also know the height of the person in the picture.
 

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