Angular width of central maxima in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength `6000Å`. When the slit is illuminated by light of another wavelength, the angular width decreases by 30%. The wavelength of this light will be

Options

A. `3500Å`
B. `4200Å`
C. `4700Å`
D. `6000Å`

Answer

To solve the problem, we need to determine the new wavelength of light when the angular width of the central maxima in the Fraunhofer diffraction pattern decreases by 30%.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The angular width of the central maxima (β) in a single-slit diffraction pattern is given by the formula:
\[ \beta \propto \frac{\lambda}{d} \]
where \( \lambda \) is the wavelength of light and \( d \) is the width of the slit. Since \( d \) is constant, we can say:
\[ \beta \propto \lambda \]

2. **Initial Wavelength**:
The initial wavelength \( \lambda_1 \) is given as \( 6000 \, \text{Å} \).

3. **Change in Angular Width**:
When the wavelength changes, the angular width decreases by 30%. This means the new angular width \( \beta_2 \) can be expressed as:
\[ \beta_2 = \beta_1 - 0.3 \beta_1 = 0.7 \beta_1 \]

4. **Setting Up the Proportionality**:
Since \( \beta \) is proportional to \( \lambda \), we can write:
\[ \frac{\beta_1}{\beta_2} = \frac{\lambda_1}{\lambda_2} \]
Substituting \( \beta_2 = 0.7 \beta_1 \):
\[ \frac{\beta_1}{0.7 \beta_1} = \frac{\lambda_1}{\lambda_2} \]
This simplifies to:
\[ \frac{1}{0.7} = \frac{\lambda_1}{\lambda_2} \]

5. **Calculating the New Wavelength**:
Rearranging the equation gives:
\[ \lambda_2 = \lambda_1 \times 0.7 \]
Substituting \( \lambda_1 = 6000 \, \text{Å} \):
\[ \lambda_2 = 6000 \times 0.7 = 4200 \, \text{Å} \]

### Final Answer:
The wavelength of the new light is \( 4200 \, \text{Å} \).