The interference fringes for sodium light `(lambda = 5890 Å)` in a double slit experiment have an angular width of `0.2^(@)`. For what wavelength will width be `10%` greater.

Explanation

To solve the problem step by step, we need to find the new wavelength for which the angular width of the interference fringes is 10% greater than the original width.

### Step 1: Understand the relationship between angular width and wavelength
The angular width (θ) of the interference fringes in a double slit experiment is directly proportional to the wavelength (λ) of the light used. This can be expressed as:
\[ \theta \propto \lambda \]
This means that if the angular width changes, the wavelength will also change proportionally.

### Step 2: Define the original parameters
We are given:
- Original wavelength, \( \lambda_1 = 5890 \, \text{Å} \)
- Original angular width, \( \theta_1 = 0.2^\circ \)

### Step 3: Calculate the new angular width
We need to find the new angular width (θ₂) which is 10% greater than the original width.
\[ \theta_2 = \theta_1 + 0.1 \times \theta_1 \]
Calculating this:
\[ \theta_2 = 0.2 + 0.1 \times 0.2 = 0.2 + 0.02 = 0.22^\circ \]

### Step 4: Set up the proportional relationship
Using the proportional relationship between the angular widths and wavelengths:
\[ \frac{\theta_1}{\theta_2} = \frac{\lambda_1}{\lambda_2} \]
Substituting the known values:
\[ \frac{0.2}{0.22} = \frac{5890}{\lambda_2} \]

### Step 5: Solve for the new wavelength (λ₂)
Cross-multiplying gives:
\[ 0.2 \cdot \lambda_2 = 0.22 \cdot 5890 \]
Now, solving for λ₂:
\[ \lambda_2 = \frac{0.22 \cdot 5890}{0.2} \]
Calculating the right side:
\[ \lambda_2 = \frac{1290.8}{0.2} = 6454 \, \text{Å} \]

### Step 6: Final answer
Thus, the new wavelength for which the angular width will be 10% greater is:
\[ \lambda_2 \approx 6479 \, \text{Å} \]