In a interference pattern with two identical coherent sources, the intensity on the screen at a point where path difference is λ is I₀. The intensity at a point where path difference is λ/4, will be

Concept: 

• When two waves of the same frequency, same wavelength, same velocity (nearly equal amplitude) move in the same direction. Then their superimposition results in interference.
• In young's double-slit experiment, lightwave produces an interference pattern of alternate bright and dark fringes or interference bands.

• For two coherent sources s1 and s2, the resultant intensity at some point p is given by

\(I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}cos\phi }\)

Calculation:

• The relationship between phase difference and path difference is 

\(Phase\;difference\;\left( {{\rm{\Delta \;\Phi }}} \right) = \frac{{2π }}{λ }\;Path\;difference\;\left( {{\rm{\Delta \;}}x} \right)\)

\( ⇒ {\rm{\Delta \;\Phi }} = \frac{{2π \;{\rm{\Delta }}x}}{λ }\)

• When the path difference is λ, then the phase difference will be 

\( ⇒ {\rm{\Delta \;\Phi }} = \frac{2π}{\lambda}\times\lambda =2π\)

In the first case, when the path difference is λ, the phase difference is  2π 

\(⇒ I_0 = {I} + {I} + 2{I}\cos 2π = 4{I}\)      ----- (1)

• When the path difference is λ/4, then the phase difference will be 

\( ⇒ {\rm{\Delta \;\Phi }} = \frac{2π}{\lambda}\times\frac{\lambda}{4} =\frac{\pi}{2}\)

In the second case, when the path difference is λ/3, the phase difference is  π/2

\(⇒ I' = {I} + {I} + 2{I}\cos \frac{{π }}{2} = {2I}\)      ----- (2)

On dividing equations 1 and 2, we get

\(⇒ \frac{{I_o}}{I'} = \frac{{{4I}}}{{2{I}}} =2\)

⇒ I' = Io/2