Question

In a single slit Fraunhofer diffraction pattern obtained at normal incidence, at the angular position of the second diffraction minimum the phase difference (in radian) between the waves from the opposite edges of the slit is

• A. \(\pi\)
• B. zero
• C. \(2\pi\)
• D. \(\frac{\pi}{2}\)

Hint:

For single slit diffraction minima, use \(a\sin\theta = n\lambda\) and convert path difference into phase difference using \( \Delta \phi = \frac{2\pi}{\lambda}\Delta x \).

Answer 1

The Correct Option is C

Step 1: Condition for minima in single slit diffraction.
For single slit Fraunhofer diffraction, minima occur when path difference between waves from opposite edges of the slit is:
\[ a \sin\theta = n\lambda \]

Step 2: Identify second diffraction minimum.
For the second minimum:
\[ n = 2 \]
So, the path difference is:
\[ \Delta x = 2\lambda \]

Step 3: Relation between phase difference and path difference.
Phase difference is related to path difference by:
\[ \Delta \phi = \frac{2\pi}{\lambda}\Delta x \]

Step 4: Substitute path difference.
\[ \Delta \phi = \frac{2\pi}{\lambda}(2\lambda) \]
\[ \Delta \phi = 4\pi \]

Step 5: Phase difference in effective phase cycle.
A phase difference of \(4\pi\) is equivalent to \(2\pi\) in phase cycle representation. Hence, among the given options, the corresponding phase difference is \(2\pi\).

Step 6: Final conclusion.
\[ \boxed{2\pi} \] Hence, correct answer is option (C).