Optics, resolution problem sheet

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SUMMARY

The discussion focuses on solving an optics problem involving the calculation of the size of letters 'Help!' needed for visibility from a distance of 250 km using a telescope with a 2.5 m diameter mirror and a focal ratio of f/4. The Rayleigh criterion is applied to determine the minimum angle of resolution, calculated as theta min = 1.22 x (500 nm / 2.5 m) = 0.0968 radians. Consequently, the required size of the letters is determined to be 24.2 m for visibility, while considering the physical size of the diffraction-limited image, the letters must be 10 m in size if limited by the charge-coupled device's physical size.

PREREQUISITES
  • Understanding of Rayleigh's Criterion in optics
  • Knowledge of diffraction limits and angular resolution
  • Familiarity with basic telescope specifications (diameter, focal ratio)
  • Ability to perform calculations involving wavelengths and distances
NEXT STEPS
  • Study the principles of diffraction and its impact on optical resolution
  • Learn about charge-coupled devices (CCDs) and their role in imaging systems
  • Explore advanced optics calculations involving focal lengths and angular magnification
  • Research practical applications of Rayleigh's Criterion in modern telescopes
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Students and professionals in optics, astronomy enthusiasts, and anyone involved in optical engineering or imaging technology will benefit from this discussion.

hayze2728
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Hey guys, I've been set this problem sheet for optics and I haven't got a clue how to complete it, the course was really rushed and all of us have been left confused, so any help would be greatly appreciated.

1. A man is stranded. He needs assistance and his mobile battery is dead. How large does he need to write the letters 'Help!' in the saad for the team back at CTU to read from the images taken by the spy satellite 250km above him if its mirror is 2.5m in diameter, has a focal ratio of f/4, uses a filter centred at 500nm and works at the diffraction limit? (6 marks)

Consider the physical size of the diffraction limited image at the focus assuming the angular magnification of the image is unity(?). If the telescope was limited by the physical size of each element of the charge-coupled device at the telescope focus, how much larger would he have to write the letters? (2 marks)




2. I'm assuming Rayleigh''s Criterion: theta min = 1.22 (lambda/D) is one to use.



3. Like I said we've tried many things and would just like to know how to tackle the problem.

Thanks!
 
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To answer the first part of your question, you need to use the Rayleigh criterion. This states that the minimum angle of resolution (theta min) is equal to 1.22 times the wavelength of the light divided by the diameter of the telescope objective (D). Using this equation, you can calculate the minimum angle of resolution for the given telescope: theta min = 1.22 x (500 nm / 2.5 m) = 0.0968 radians. To calculate the size of the letters 'Help!' that the man needs to write for them to be seen from 250 km away, you need to multiply the distance to the telescope and the minimum angle of resolution: size of letters = 250 km x 0.0968 radians = 24.2 m. For the second part of the question, you need to consider the physical size of the diffraction limited image at the focus, assuming the angular magnification of the image is unity. In other words, you need to calculate the physical size of the image created by the telescope. To do this, you need to use the equation physical size = focal length x (angular size / angular magnification). Using this equation, you can calculate the physical size of the image: physical size = 10 m x (0.0968 radians / 1) = 10 m. Therefore, if the telescope was limited by the physical size of each element of the charge-coupled device at the telescope focus, the man would have to write the letters 'Help!' at a size of 10 m.
 

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