Step 1: Write the condition for coincidence of bright fringes.
In YDSE, the position of the $n$-th bright fringe for wavelength $\lambda$ is: \[ y_n = \frac{n \lambda D}{d} \] For two wavelengths $\lambda_1$ and $\lambda_2$, bright fringes coincide when: \[ n_1 \lambda_1 = n_2 \lambda_2 \]
Step 2: Substitute the given wavelengths.
Given: $\lambda_1 = 3750\,\text{\AA}$ and $\lambda_2 = 7500\,\text{\AA}$. Note that $\lambda_2 = 2\lambda_1$. So: \[ n_1 \times 3750 = n_2 \times 7500 \implies n_1 = 2n_2 \]
Step 3: Find the minimum order for coincidence.
For the minimum (first non-trivial) coincidence, take the smallest positive integers: $n_2 = 1$ and $n_1 = 2$. (The $n_1 = n_2 = 0$ case is the central maximum, which is not counted.)
Step 4: Calculate the position of coincidence.
Using $n_2 = 1$, $\lambda_2 = 7500\,\text{\AA} = 7500 \times 10^{-10}\,\text{m}$, $D = 4\,\text{m}$, $d = 3\,\text{mm} = 3 \times 10^{-3}\,\text{m}$: \[ y = \frac{n_2 \lambda_2 D}{d} = \frac{1 \times 7500 \times 10^{-10} \times 4}{3 \times 10^{-3}} \]
Step 5: Simplify the calculation.
\[ y = \frac{7500 \times 10^{-10} \times 4}{3 \times 10^{-3}} = \frac{30000 \times 10^{-10}}{3 \times 10^{-3}} = 10000 \times 10^{-7} = 10^{-3}\,\text{m} = 1\,\text{mm} \]
Step 6: State the final answer.
The minimum distance from the central bright fringe at which bright fringes of the two wavelengths coincide is: \[ \boxed{y = 1\,\text{mm}} \]