Wavelength of the incident light is changed from 600 nm to 1200 nm. The intensity of the scattered light in Rayleigh scattering will become times the initial scattered intensity.

Question

Wavelength of the incident light is changed from 600 nm to 1200 nm. The intensity of the scattered light in Rayleigh scattering will become times the initial scattered intensity. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

According to Rayleigh's law of scattering, I propto (1/λ4) ∴ ( I 2/ I 1)=((λ1/λ2))4=((600 × 10-9/1200 × 10-9))4 =(1/24)=0.0625 ≈ 0.06

Q. Wavelength of the incident light is changed from $600 nm$ to $1200 nm$ . The intensity of the scattered light in Rayleigh scattering will become _____ times the initial scattered intensity.

According to Rayleigh's law of scattering,
$I \propto \frac{1}{λ ^{4}}$
$∴ \frac{I _{2}}{I _{1}} = (\frac{λ _{1}}{λ _{2}})^{4} = (\frac{600 \times 1 0 ^{- 9}}{1200 \times 1 0 ^{- 9}})^{4}$
$= \frac{1}{2 ^{4}} = 0.0625 \approx 0.06$

Q. Wavelength of the incident light is changed from 600nm to 1200nm. The intensity of the scattered light in Rayleigh scattering will become _____ times the initial scattered intensity.

According to Rayleigh's law of scattering, I∝λ41​ ∴I1​I2​​=(λ2​λ1​​)4=(1200×10−9600×10−9​)4 =241​=0.0625≈0.06

Q. Wavelength of the incident light is changed from $600 nm$ to $1200 nm$ . The intensity of the scattered light in Rayleigh scattering will become _____ times the initial scattered intensity.

According to Rayleigh's law of scattering,
$I \propto \frac{1}{λ ^{4}}$
$∴ \frac{I _{2}}{I _{1}} = (\frac{λ _{1}}{λ _{2}})^{4} = (\frac{600 \times 1 0 ^{- 9}}{1200 \times 1 0 ^{- 9}})^{4}$
$= \frac{1}{2 ^{4}} = 0.0625 \approx 0.06$