Q.Which of the following relations is correct for intensity of scattered light (I) and wavelength of light (λ)? (A) I ∝ 1/λ
(B) I ∝ 1/λ^2
(C) I ∝ 1/λ^3
(D) I ∝ 1/λ^4

Fog is made of countless tiny water droplets that scatter light in all directions (Tyndall scattering), destroying the sharp image-forming rays.

Fog consists of a suspension of very fine water droplets in air, comparable in size to the wavelength of visible light. When light passes through this suspension, it undergoes scattering (Tyndall effect) in random directions off the droplets, rather than travelling in the original straight, well-defined direction. Because rays from objects get scattered every which way instead of reaching the eye/camera in a coherent bundle, distinct images cannot form and visibility drops sharply.

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By Rayleigh's law, the intensity of scattered light I ∝ 1/λ⁴.

For scattering by particles much smaller than the wavelength (Rayleigh scattering), the scattered intensity is inversely proportional to the fourth power of the wavelength:

$I \propto \dfrac{1}{\lambda^4}.$

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The sky looks blue because air molecules scatter the shorter-wavelength (blue) component of sunlight much more strongly than the longer wavelengths.

Sunlight passing through the atmosphere is scattered by air molecules and fine dust particles, a process called Rayleigh scattering. The intensity of this scattering varies as $I \propto \dfrac{1}{\lambda^4}$, so shorter wavelengths (blue/violet end of the visible spectrum) are scattered far more than longer wavelengths (red end). Although violet is scattered even more than blue, the eye is more sensitive to blue and the sun's spectrum has less violet, so the overall effect we see when looking away from the sun is a blue sky. (By the same reasoning, at sunris …

Sunlight travels through a much thicker layer of atmosphere at sunrise/sunset; Rayleigh scattering removes most of the blue light from the direct beam, leaving the transmitted light red.

Rayleigh scattering intensity $\propto\dfrac{1}{\lambda^4}$, so shorter wavelengths (blue/violet) scatter far more strongly off air molecules than longer wavelengths (red/orange).

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The scattering of light is inversely proportional to the fourth power of its wavelength. Blue light has the shortest wavelength among red, blue, and yellow, so its scattering is maximum.

Scattering is the redirection of light in various directions by small particles in its path, rather than the light continuing to travel in a straight line.

When a beam of light passes through a medium containing fine particles (molecules, dust, water droplets) whose size is comparable to or smaller than the wavelength of light, the particles absorb the light momentarily and re-radiate it in all directions. This redirection of light is called scattering.

For particles much smaller than the wavelength of light (like air molecules), Lord Rayleigh showed that the intensity of scattered light varies as:

$I\propto \dfrac{1}{\lambda^4}$

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Red is chosen for danger signals not because the eye is most sensitive to it, but because red travels farthest through the atmosphere without scattering.

The human eye's photopic (daylight) sensitivity curve peaks around $\lambda \approx 555\,\text{nm}$, which lies in the yellow-green region - this is why yellow/green appear 'brightest' to us for the same power of light.

However, for a danger or warning signal, what matters most is not peak sensitivity but range and reliability of visibility, especially through fog, mist, dust or smoke. By Rayleigh's scattering law, the amount a wave is scattered by small atmospheric particles varies as

$I_{scattered} \propto \frac{1}{\lambda^4}$ …