1. Condition for bright fringe in YDSE

For a bright fringe at angle $\theta$, the path difference must satisfy

$d\sin\theta = n\lambda$

where:

• $d = 0.1\text{ mm} = 10^{-4}\text{ m}$
• $n$ is an integer
• $\lambda$ is the wavelength

2. Given angle

The angle is given as $\dfrac{1}{40}$ (in radians), so for small angles,

$\sin\theta \approx \theta = \frac{1}{40}$

Thus,

$d\sin\theta = 10^{-4}\times \frac{1}{40} = 2.5\times 10^{-6}\text{ m}$

So the bright fringe condition becomes

$n\lambda = 2.5\times 10^{-6}\text{ m} = 2500\text{ nm}$

Hence,

$\lambda = \frac{2500}{n}\text{ nm}$

3. Find visible wavelengths

Since $\lambda$ must lie in visible range $380\text{ nm}$ to $740\text{ nm}$,

$380 \le \frac{2500}{n} \le 740$

This gives possible integer values of $n$:

• $n=4 \Rightarrow \lambda = 625\text{ nm}$
• $n=5 \Rightarrow \lambda = 500\text{ nm}$
• $n=6 \Rightarrow \lambda \approx 416.7\text{ nm}$

These are the visible wavelengths possible.

4. Check options

We need two visible wavelengths that both produce a bright fringe at the same angle, i.e. both must satisfy

$n\lambda = 2500\text{ nm}$

• Option A: $400, 500$ nm
  $2500/400 = 6.25$ not integer, so $400$ nm is not valid.

• Option B: $625, 500$ nm
  $2500/625 = 4$, $2500/500 = 5$; both integers, so valid.

• Option C: $380, 500$ nm
  $2500/380 \approx 6.58$ not integer, so invalid.

• Option D: $380, 525$ nm
  $2500/380 \approx 6.58$, $2500/525 \approx 4.76$; neither is integer, so invalid.

5. Final answer

The correct pair is

$\boxed{625\text{ nm},\ 500\text{ nm}}$

So the correct option is B.