Question

Two beams of visible light (400–700 nm) interfere at a point. The optical path difference is 5000 nm. Which wavelength interferes constructively?

• A. 416.67 nm
• B. 555.55 nm
• C. 625 nm
• D. 666.66 nm

Hint:

Use $\Delta = n\lambda$ for constructive interference and check allowed wavelength range.

Answer 1

The Correct Option is A, B, C

Step 1: Condition for constructive interference.
$\Delta = n\lambda$ where $\Delta=5000\ \mathrm{nm}$.

Step 2: Compute wavelengths.
Possible wavelengths: $\lambda = \frac{5000}{n}$. Check which lies between 400–700 nm. $n=7 $\Rightarrow$ \lambda \approx 714$ nm (not allowed) $n=8 $\Rightarrow$ \lambda = 625$ nm (option C) $n=9 $\Rightarrow$ \lambda = 555.55$ nm $n=12 $\Rightarrow$ 416.67$ nm $n=15 $\Rightarrow$ 333$ nm (not visible) But the correct interference condition must match EXACT geometry: Only 666.66 = 5000/7.5 corresponds to given list. But the closest visible constructive match in options is 666.66 nm.

Step 3: Conclusion.
Option (D) best satisfies the condition within visible range.