In young double slit experiment , when two light waves form third minimum, they have

Options

A. Phase difference by `3pi`
B. Path difference of `3lamda`
C. Phase difference of `(5pi)/2`
D. Path difference of `(5lamda)/2`

Answer

To solve the problem regarding the Young's double slit experiment and the conditions for the third minimum, we can break it down into a series of steps.

### Step-by-Step Solution:

1. **Understanding the Young's Double Slit Experiment**:
In the Young's double slit experiment, two coherent light sources (S1 and S2) produce an interference pattern on a screen. The pattern consists of alternating bright and dark fringes.

2. **Identifying the Conditions for Minima**:
A dark fringe (minimum) occurs when the path difference between the two waves from the slits is an odd multiple of half the wavelength. The condition for minima can be expressed as:
\[ \Delta x = \left(m + \frac{1}{2}\right) \lambda \]
where \( m \) is an integer (0, 1, 2, ...).

3. **Finding the Path Difference for the Third Minimum**:
For the third minimum, we set \( m = 2 \) (since we start counting from 0):
\[ \Delta x = \left(2 + \frac{1}{2}\right) \lambda = \frac{5}{2} \lambda \]

4. **Relating Path Difference to Phase Difference**:
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by the formula:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting the path difference for the third minimum:
\[ \phi = \frac{2\pi}{\lambda} \left(\frac{5}{2} \lambda\right) = 5\pi \]

5. **Conclusion**:
Therefore, when the two light waves form the third minimum, the phase difference is \( 5\pi \) and the path difference is \( \frac{5\lambda}{2} \).

### Final Answer:
When two light waves form the third minimum in Young's double slit experiment, the path difference is \( \frac{5\lambda}{2} \).

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