Two slits separated by a distance of 0.5 mm are illuminated by light of wavelength 5000 Å. The interference fringes are obtained on a screen at a distance of 1.2 m. what is the phase difference between two interfering waves at a point 3 mm from the central bbright fringe?

Options

A. `7pi` rad
B. `3pi` rad
C. `5pi` rad
D. `4pi` rad

Answer

C

Explanation

`X=(lamdaD)/(d)=(5000xx10^(-7)xx1.2)/(5xx10^(-4))=1.2mm`
and `5/2 X=(5)/(2)xx1.2=3mm`
This corresponds to third dark band.
for the third d.b, phase difference`=5pi` radian.