## Related questions with answers

On the night of April 18, 1775, a signal was sent from the steeple of Old North Church in Boston to Paul Revere, who was $1.80 \mathrm{mi}$ away: "One if by land, two if by sea." At what minimum separation did the sexton have to set the lanterns for Revere to receive the correct message about the approaching British? Assume that the patriot's pupils had a diameter of $4.00 \mathrm{~mm}$ at night and that the lantern light had a predominant wavelength of $580 \mathrm{~nm}$.

Solution

VerifiedTo solve this problem we are going to use the formula for the resolving power of the circular aperture which says that the angle of the first minimum is given as

$\theta_{min}=\frac{1.22\lambda}{ D}$

On the other hand in the small-angle approximation, we have that $\theta_{min}=\frac{d}{L}$ where $d$ is the distance between the two latnerns. We have that

$\frac{1.22\lambda}{ D}=\frac{d}{L}$

We can express $d$ as

$d=\frac{1.22\lambda L}{ D}=\frac{1.22\times 580\times 10^{-9}\times 2.9\times 10^{3}}{4\times 10^{-3}}$

Finally, we obtain that the distance between the lanterns should be

$d=0.513\textrm{m}$

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