Document Type

Article

Publication Date

2025

DOI

10.1119/5.0216438

Publication Title

The Physics Teacher

Volume

63

Issue

3

Pages

A215

Abstract

Question 1: Why is it easier to see through rain than fog?

Start thinking about this by imagining a fixed volume (V) of water being dispersed into, say, N identical droplets of diameter d. Surface area and volume considerations should lead to the answer in terms of V and d.

Solution to Question 1: N = VI(πd³/6) = 6V/πd³ ≈ 2V/d³.

The cross-sectional area A of each drop is πd²/4 ≈ 3d²/4, so the total area blocked off (assuming no overlapping drops—so this is an upper bound) is NA ≈ 1.5V/d, so the area blocked off is inversely proportional to the diameter of the drops for fixed volume of water.

Question 2: (a) How “long” (in meters) might such a rain shower or fogbank be?

Rights

© Copyright 2026 AIP Publishing LLC

This article may be downloaded for personal use only. Any other use requires prior permission of the author and AIP Publishing. This article appeared in 

Adam, J. (2025). A question of transparency: Solutions for Fermi questions, March 2025. The Physics Teacher, 63(3), A215. 

and may be found at: https://doi.org/10.1119/5.0216438

Original Publication Citation

Adam, J. (2025). A question of transparency: Solutions for Fermi questions, March 2025. The Physics Teacher, 63(3), A215. https://doi.org/10.1119/5.0216438

ORCID

0000-0001-5537-2889 (Adam)

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