The source that illuminates the double-slit in 'double-slit interference experiment' emits two distinct monochromatic waves of wavelength 500 nm and 600 nm, each of them producing its own pattern on the screen. At the central point of the pattern when path difference is zero, maxima of both the patterns coincide and the resulting interference pattern is most distinct at the region of zero path difference. But as one moves out of this central region, the two fringe systems are gradually out of step such that maximum due to one wavelength coincides with the minimum due to the other and the combined fringe system becomes completely indistinct. This may happen when path difference in nm is:

In the double-slit interference experiment with two wavelengths, λ₁ = 500 nm and λ₂ = 600 nm, we need to find the path difference Δ where the maximum of one pattern coincides with the minimum of the other, causing indistinct fringes.

For a maximum (bright fringe) at wavelength λ₁, the path difference must satisfy Δ = nλ₁, where n is an integer. For a minimum (dark fringe) at wavelength λ₂, the path difference must satisfy Δ = (m + ½)λ₂, where m is an integer. Setting these equal gives:

$$n \lambda_1 = \left(m + \frac{1}{2}\right) \lambda_2$$

Substituting λ₁ = 500 nm and λ₂ = 600 nm:

$$n \times 500 = \left(m + \frac{1}{2}\right) \times 600$$

Simplify the equation:

$$500n = 600m + 300$$

Divide both sides by 100:

$$5n = 6m + 3$$

Rearrange to:

$$5n - 6m = 3$$

Solve for integer solutions. Express n in terms of m:

$$n = \frac{6m + 3}{5}$$

So, (6m + 3) must be divisible by 5. Thus, 6m + 3 ≡ 0 mod 5. Simplify modulo 5:

$$6 \equiv 1 \pmod{5}, \quad 3 \equiv 3 \pmod{5} \implies 1 \cdot m + 3 \equiv 0 \pmod{5} \implies m \equiv 2 \pmod{5}$$

Therefore, m = 5k + 2 for integer k ≥ 0. Substitute back:

$$n = \frac{6(5k + 2) + 3}{5} = \frac{30k + 12 + 3}{5} = \frac{30k + 15}{5} = 6k + 3$$

The path difference Δ is:

$$\Delta = n \lambda_1 = (6k + 3) \times 500$$

For k = 0 (smallest positive Δ):

$$\Delta = (6 \times 0 + 3) \times 500 = 3 \times 500 = 1500 \text{ nm}$$

Verify with λ₂:

$$\Delta = \left(m + \frac{1}{2}\right) \lambda_2 = \left(5 \times 0 + 2 + \frac{1}{2}\right) \times 600 = \left(2.5\right) \times 600 = 1500 \text{ nm}$$

At Δ = 1500 nm:

• For λ₁ = 500 nm: Δ = 1500 = 3 × 500 → maximum (n=3).
• For λ₂ = 600 nm: Δ = 1500 = 2.5 × 600 = (2 + ½) × 600 → minimum (m=2).

Thus, a maximum of λ₁ coincides with a minimum of λ₂, satisfying the condition. Checking other k values gives larger Δ (e.g., k=1 yields 4500 nm, not in options). Considering the reverse case (maximum of λ₂ and minimum of λ₁) leads to no integer solutions, as shown by:

$$n \lambda_2 = \left(m + \frac{1}{2}\right) \lambda_1 \implies 600n = 500\left(m + \frac{1}{2}\right) \implies 12n = 10m + 5 \implies 12n - 10m = 5$$

The left side is even, right side odd, so no integer solutions exist. Therefore, the only valid solution in the options is Δ = 1500 nm.

Hence, the correct answer is Option D (1500 nm).