In a Young's double slit experiment slit separation 0.1 mm, one observes a bright fringe at angle $$\frac{1}{40}$$ rad by using light of wavelength $$\lambda_1$$. When the light of wavelength $$\lambda_2$$ is used a bright fringe is seen at the same angle in the same set up. Given that $$\lambda_1$$ and $$\lambda_2$$ are in visible range (380 nm to 740 nm), their values are:

For a bright (constructive) fringe in Young’s double-slit experiment we start from the condition

$$d\,\sin\theta = m\lambda,$$

where $$d$$ is the slit separation, $$\theta$$ the angular position of the fringe, $$\lambda$$ the wavelength of light and $$m$$ an integer (the order of the fringe).

We are told that the same angular position $$\theta = \dfrac{1}{40}\,{\rm rad}$$ gives bright fringes for two different wavelengths $$\lambda_1$$ and $$\lambda_2$$. Hence for the two cases we have

$$d\,\sin\theta = m_1\lambda_1 \qquad\text{and}\qquad d\,\sin\theta = m_2\lambda_2.$$

Because the left-hand side is the same in both equations, we can write

$$m_1\lambda_1 = m_2\lambda_2.$$ Now we substitute the numerical values that are common to both situations. The slit separation is

$$d = 0.1\,{\rm mm} = 0.1 \times 10^{-3}\,{\rm m} = 1.0 \times 10^{-4}\,{\rm m}.$$

To work conveniently with nanometres, we convert $$d$$ to nanometres. We recall that $$1\,{\rm m}=10^{9}\,{\rm nm},$$ therefore

$$d = 1.0 \times 10^{-4}\,{\rm m} = 1.0 \times 10^{-4}\times 10^{9}\,{\rm nm}=1.0 \times 10^{5}\,{\rm nm}.$$

The given angle is small, so $$\sin\theta \approx \theta,$$ and we evaluate

$$\sin\theta = \frac{1}{40} = 0.025.$$

Putting these numbers together, the product $$d\,\sin\theta$$ that must equal each $$m\lambda$$ is

$$d\,\sin\theta = (1.0 \times 10^{5}\,{\rm nm})\,(0.025) = 2.5 \times 10^{3}\,{\rm nm} = 2500\,{\rm nm}.$$

Thus both wavelengths must divide the value $$2500\,{\rm nm}$$ exactly so that the corresponding orders $$m_1$$ and $$m_2$$ are integers:

$$m_1 = \frac{2500}{\lambda_1},\qquad m_2 = \frac{2500}{\lambda_2}.$$

We now inspect each option to see which pair of wavelengths makes both fractions integral and still keeps the wavelengths in the visible range (380 nm - 740 nm).

Option A:  $$\lambda_1 = 400\,{\rm nm},\ \lambda_2 = 500\,{\rm nm}$$ gives $$\frac{2500}{400}=6.25\ (\text{not integer}),$$ so this fails.
Option B:  $$\lambda_1 = 380\,{\rm nm},\ \lambda_2 = 525\,{\rm nm}$$ gives $$\frac{2500}{380}=6.579,\ \frac{2500}{525}=4.762 \ (\text{neither integer}),$$ so this fails.
Option C:  $$\lambda_1 = 625\,{\rm nm},\ \lambda_2 = 500\,{\rm nm}$$ gives $$\frac{2500}{625}=4,\qquad \frac{2500}{500}=5,$$ both perfect integers, so this works.
Option D:  $$\lambda_1 = 380\,{\rm nm},\ \lambda_2 = 500\,{\rm nm}$$ gives $$\frac{2500}{380}=6.579\ (\text{not integer}),$$ so this fails.

Only Option C satisfies the integer requirement while keeping both wavelengths within the visible spectrum. For completeness we can note the corresponding fringe orders:

$$m_1 = 4\ \text{for}\ \lambda_1 = 625\,{\rm nm},\qquad m_2 = 5\ \text{for}\ \lambda_2 = 500\,{\rm nm},$$

and indeed $$m_1\lambda_1 = 4\times625 = 2500\,{\rm nm} = 5\times500 = m_2\lambda_2,$$ as required.

Hence, the correct answer is Option C.