If the scattering intensity of a liquid is 8 | vedclass

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(C) According to Rayleigh's law of scattering, the scattering intensity $I$ is inversely proportional to the fourth power of the wavelength $\lambda$:

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) According to Rayleigh's law of scattering, the scattering intensity $I$ is inversely proportional to the fourth power of the wavelength $\lambda$:$I \propto \frac{1}{\lambda^{4}}$Given $I_{1} = 8 	ext{ units}$ at $\lambda_{1} = 500 	ext{ nm}$ and we need to find $I_{2}$ at $\lambda_{2} = 400 	ext{ nm}$.Using the ratio formula:$\frac{I_{2}}{I_{1}} = \left(\frac{\lambda_{1}}{\lambda_{2}}ight)^{4}$$\frac{I_{2}}{8} = \left(\frac{500}{400}ight)^{4} = (1.25)^{4}$$(1.25)^{4} = 2.4414 \approx 2.5$$I_{2} = 8 	imes 2.4414 \approx 19.53 	ext{ units}$Rounding to the nearest provided option, $I_{2} \approx 20 	ext{ units}$.

If the scattering intensity of a liquid is $8 \text{ units}$ at a wavelength of $500 \text{ nm}$, then the scattering intensity at a wavelength of $400 \text{ nm}$ will be approximately: (in $\text{ units}$)

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