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

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Homework Help Overview

The discussion revolves around the historical context of Paul Revere's signal involving lanterns and the physics of human vision, specifically focusing on whether Revere could discern between one or two lanterns based on distance and resolving power of the eye.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of an equation related to the resolving power of the eye and question the adequacy of the information provided in the problem. There is a focus on the distance from the lanterns and the angle created between them.

Discussion Status

Some participants have calculated the resolving power of the eye and suggested a distance for the lanterns, while others express concerns about the problem's wording and the information available. There is an ongoing exploration of how to determine the distance at which Revere could resolve the two lanterns.

Contextual Notes

Participants note the ambiguity in the problem's wording and the potential lack of sufficient information to reach a definitive conclusion.

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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