Find Largest Focal Length Lens for Photo of Poster - 2.79mm

Click For Summary
SUMMARY

The largest focal length lens that can be used to photograph a poster measuring 0.8 m wide and 1.2 m tall from a maximum distance of 3.1 m is 2.79 mm. The calculations involved using the lens formula 1/f = 1/do + 1/di and the magnification equation M = -di/do. The initial approach of using area ratios was incorrect; the correct method involves using the ratios of the sides of the poster and film. The final focal length was confirmed to be -2.79 mm after proper calculations.

PREREQUISITES
  • Understanding of lens formulas, specifically 1/f = 1/do + 1/di
  • Knowledge of magnification equations, particularly M = -di/do
  • Basic geometry for calculating ratios of dimensions
  • Familiarity with unit conversions, especially between meters and millimeters
NEXT STEPS
  • Study the principles of optics and lens design
  • Learn about different types of camera lenses and their focal lengths
  • Explore practical applications of the lens formula in photography
  • Investigate the impact of distance on image quality and magnification
USEFUL FOR

Photography students, optical engineers, and anyone interested in understanding lens specifications and calculations for capturing images effectively.

deenuh20
Messages
50
Reaction score
0

Homework Statement



You take a picture of a poster on the wall of a room and maximum distance you can get from the picture is 3.1 m. The poster is 0.8 m wide and 1.2 m tall and you want it to fit in the 24 x 36 mm frame of the film in your camera.

What is the largest focal length lens that you can use? ____mm

Homework Equations



1/f = 1/do + 1/di
M=-di/do



The Attempt at a Solution



I have absolutely no idea how to go about this. What I tried, for starters, was find the ratio of the areas of the poster (0.96m^2) and the film (8.64E-4 m^2). I thought that by finding the ratio of the film to the poster would give me magnification which was equal to 9E-4. Then, using this number, I used the equation M=-di/do, plugged in 9E-4 for M and 3.1 for do and tried solving di and got -0.00279m. Then, I plugged this in the equation 1/f=1/do +1/di to solve for f and I got: 1/f=(1/3.1)+ -(1/.00279) = (-358.1 m) and took the inverse to be -0.00279 m. Then, converting to mm, it was -2.79mm.
 
Physics news on Phys.org
Don't use the ratios of the areas. Use the ratios of the sides.
 
Thank you very much. I finally got it!
 

Similar threads

Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K