1. In Young’s double slit experiment, the position of the $n$th bright fringe is $y_n = n\beta = n\frac{\lambda D}{d}$ where $\lambda$ is the wavelength.

2. Since the central maxima coincide, fringe counting for both patterns starts from the same origin.

3. Given:

• Known wavelength: $\lambda_1 = 590\,\text{nm}$
• Third bright fringe of known light coincides with fourth bright fringe of unknown light.

So their positions are equal: $3\frac{\lambda_1 D}{d} = 4\frac{\lambda_2 D}{d}$

4. Cancel the common factors $\frac{D}{d}$: $3\lambda_1 = 4\lambda_2$

5. Substitute $\lambda_1 = 590\,\text{nm}$: $3(590) = 4\lambda_2$ $1770 = 4\lambda_2$ $\lambda_2 = \frac{1770}{4} = 442.5\,\text{nm}$

6. Therefore, the unknown wavelength is $\boxed{442.5\,\text{nm}}$

7. Option check:

• A: $885.0\,\text{nm}$ — incorrect
• B: $442.5\,\text{nm}$ — correct
• C: $776.8\,\text{nm}$ — incorrect
• D: $393.4\,\text{nm}$ — incorrect

Hence the correct option is $\boxed{\text{B}}$.

Comparison with stored answer: stored correct answer is B, which matches the derived result.