Question

A parallel beam of light of wavelength a 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
(a) 2π (b) 3π (c) 4π (d)
πlamba
​

Answer 1

Answer:

c) 4π

Explanation:

second minimum path difference is 2π.

therefore, value of phase difference is

∆l=2π\wavelength ×path difference

therefore , =2π/ lambda ×2 lambda

= 4π