Question

A 460 nm monochromatic light is split into two beams. The two beams arrive at a point on a screen with a path difference of 920nm. Calculate the phase difference.

Answer 1

Answer:

phase difference $\Delta \Phi = 12.566$

Explanation:

Phase difference is expressed as $\Delta \Phi$

$\Delta \Phi =\frac {2\pi}{\lambda} \Delta X$

Where,

$\Delta \Phi$= phase difference

$\lambda$  wavelength = 460nm

$\Delta X$  path difference = 920 nm

$\Delta \Phi =\frac {2\pi}{460} *920 = 4\pi$

therefore

phase difference $\Delta \Phi = 12.566$