On decreasing the wavelength of incident light from 8000 Å to 4000 Å. The intensity of the scattered light in Rayleigh scattering will become _________ time the initial scattered intensity.

Options

A. 2
B. 4
C. 16
D. 8

Answer

To solve the problem, we need to determine how the intensity of scattered light changes when the wavelength of the incident light is decreased from 8000 Å to 4000 Å, using Rayleigh scattering principles.

### Step-by-Step Solution:

1. **Understanding Rayleigh Scattering**:
- According to Rayleigh scattering, the intensity \( I \) of the scattered light is inversely proportional to the fourth power of the wavelength \( \lambda \). This can be expressed mathematically as:
\[ I \propto \frac{1}{\lambda^4} \]

2. **Setting Up the Ratios**:
- We can express the relationship between the initial intensity \( I_1 \) (at wavelength \( \lambda_1 = 8000 \, \text{Å} \)) and the final intensity \( I_2 \) (at wavelength \( \lambda_2 = 4000 \, \text{Å} \)) using the proportionality:
\[ \frac{I_1}{I_2} = \frac{\lambda_2^4}{\lambda_1^4} \]

3. **Substituting the Wavelength Values**:
- Substitute \( \lambda_1 = 8000 \, \text{Å} \) and \( \lambda_2 = 4000 \, \text{Å} \):
\[ \frac{I_1}{I_2} = \frac{(4000)^4}{(8000)^4} \]

4. **Calculating the Ratio**:
- Simplifying the ratio:
\[ \frac{I_1}{I_2} = \frac{4000^4}{8000^4} = \left(\frac{4000}{8000}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16} \]

5. **Finding the Final Intensity**:
- Rearranging the equation gives:
\[ I_2 = 16 \cdot I_1 \]
- This shows that the final intensity \( I_2 \) is 16 times the initial intensity \( I_1 \).

6. **Conclusion**:
- Therefore, the intensity of the scattered light when the wavelength is decreased from 8000 Å to 4000 Å will become **16 times** the initial scattered intensity.

### Final Answer:
The intensity of the scattered light will become **16 times** the initial scattered intensity.