A parallel beam of light of wavelength λ is incident normally on a...

A parallel beam of light of wavelength

\lambda

is incident normally on a narrow slit. A diffraction pattern formed on a screen placed perpendicular to the direction of the incident beam. At the second minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of slit is:

Options

A $2 \pi$
B $3 \pi$
C $4 \pi$
D $\pi \pi$

Correct answer

C. $4 \pi$

Step-by-step solution

We know that path difference for second minimum

=2 \lambda

\begin{aligned} \therefore \text{ Phase difference } & =\frac{2 \pi}{\lambda} \times 2 x \\ & =4 \pi \end{aligned}

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