For destructive interference to take place two monochromatic light waves of wavelength `lamda`, the path difference should be

Options

A. `((2n-1)lamda)/4`
B. `2n lamda`
C. `((2n+1)lamda)/2`
D. `n lamda`

Answer

To determine the path difference required for destructive interference of two monochromatic light waves of wavelength \( \lambda \), we can follow these steps:

### Step 1: Understand the Condition for Destructive Interference
Destructive interference occurs when the waves are out of phase, meaning that the crest of one wave coincides with the trough of another. This results in a reduction in intensity.

### Step 2: Relate Path Difference to Phase Difference
The phase difference \( \phi \) between two waves is related to the path difference \( \Delta x \) by the formula:
\[ \phi = k \Delta x \]
where \( k \) is the wave number given by:
\[ k = \frac{2\pi}{\lambda} \]

### Step 3: Determine the Condition for Minimum Intensity
For destructive interference, the phase difference must be an odd multiple of \( \pi \):
\[ \phi = (2n + 1)\pi \quad \text{for } n = 0, 1, 2, \ldots \]

### Step 4: Substitute the Phase Difference into the Path Difference Equation
Substituting the expression for \( \phi \) into the equation relating phase difference to path difference, we have:
\[ (2n + 1)\pi = \frac{2\pi}{\lambda} \Delta x \]

### Step 5: Solve for Path Difference
To find the path difference \( \Delta x \), we can rearrange the equation:
\[ \Delta x = \frac{(2n + 1)\pi \lambda}{2\pi} \]
This simplifies to:
\[ \Delta x = \frac{(2n + 1)\lambda}{2} \]

### Conclusion
Thus, for destructive interference to take place, the path difference \( \Delta x \) should be:
\[ \Delta x = \frac{(2n + 1)\lambda}{2} \]
where \( n \) is any non-negative integer (0, 1, 2, ...).

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