Could Paul Revere discern if there were one or two lanterns?

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SUMMARY

Paul Revere's ability to discern between one or two lanterns on the night of April 18, 1775, can be analyzed using the resolving power of the human eye. The equation θ_min = 1.22 * (wavelength / diameter) yields a resolving power of approximately 1.76E-4 radians, given a pupil diameter of 4mm and a lantern wavelength of 580 nm. Assuming the lanterns were 0.5 meters apart, calculations indicate that Revere would need to be within a specific distance to distinguish between the two signals. The discussion concludes that while the problem's wording is unclear, the approach to solving it is valid.

PREREQUISITES
  • Understanding of basic optics and resolving power
  • Familiarity with the equation θ_min = 1.22 * (wavelength / diameter)
  • Knowledge of human anatomy, specifically pupil diameter
  • Basic distance measurement concepts in physics
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  • Research the effects of distance on visual acuity
  • Explore the physics of light and wavelength in optics
  • Learn about the historical context of Paul Revere's midnight ride
  • Investigate similar optical resolution problems, such as satellite imaging capabilities
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Students of physics, historians interested in the American Revolution, and anyone studying optics and visual perception.

Plasmosis1
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"one if by land and two if by sea" is the famous saying of Paul Revere when he was a certain distance away from a signal sent from the people of old North Church in Boston on the night of April 18, 1775. if the average human pupil has a diameter of 4mm at night and the lanterns had a predominant wavelength of 580 nm how far away (in miles) was Paul Revere when he received the signal about the approaching British on that fateful night? assume he was at a minimum distance of .5 m and neglect any other influences- including the shortening of the wavelength in his eye (which does not change the answer). could he actually discern if there were one or two lanterns? Show work for why or why not.

Useful equation: thetamin= 1.22*wavelength/diameter

I don't believe that there is enough information for this problem. You have two variables, distance from the lanterns and the angle created between the lanterns. Any help would be appreciated.
 
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I think the equation gives the resolving power of the eye in radians. This comes out to be 1.76E-4 radians. I would think the lanterns would be 0.5 meters apart. Now you can determine how far away you can be to resolve the two lanterns., yes??
 
barryj said:
I think the equation gives the resolving power of the eye in radians. This comes out to be 1.76E-4 radians. I would think the lanterns would be 0.5 meters apart. Now you can determine how far away you can be to resolve the two lanterns., yes??

The problem's wording is terrible but I think you have the right approach to it nonetheless.
 
I think you have a similar question on this forum about can a spy satellite see a license plate.
 

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