Sol. From the given condition,

OS 3 = = = 2 × 10 –4 m

Let light reaching from S 1 and S 2 to S 4 has phase difference φ and intensity of incident light is I 0 .

Resultant intensity at S 4 , I = 4I 0 cos 2

As I = I 0 ,

Hence = cos 2 or cos = = cos 60º

or φ =

For φ = , OS 4 =

Therefore, S 3 S 4 = OS 3 + OS 4 = = × 10 –4 m

Now resultant wave coming out of S 3 has intensity 4I 0 and waves coming out of S 4 have intensity I 0 .

Phase difference at S 3 = 2 π Phase difference at S 4 = 2 π /3. These phase differences are relative to the light incident on slits S 1 and S 2 .

Now S 3 and S 4 are secondary sources of light.

Phase difference at O′ = , equal to initial phase difference between the light reaching at O′ = 2 π – = . Let intensity at O′ be I′.

I′ = I 0 + 4I 0 + 2 cos

= 5I 0 + 4I 0 cos = 3I 0

For brightest fringe, phase difference = 2n π , n = 0, ±1, ±2, …

Let I′′ be the intensity of brightest fringe.

I′′ = I 0 + 4I 0 + 2 cos φ (where cos φ = 1)

= 9I 0 .