Question

Consider the arrangement shown in the figure (17-E7). By some mechanism, the separation between the slits S3 and S4 can be changed. The intensity is measured at the point P, which is at the common perpendicular bisector of S1S2 and S2S4. When

z=Dλ2d, the intensity measured at P is I. Find the intensity when z is equal to

a Dλd, b 3Dλ2d and c 2Dλd

Answer 1

Given:

Fours slits S1, S2, S3 and S4.

The separation between slits S3 and S4 can be changed.

Point P is the common perpendicular bisector of S1 S2 and S3 S4.

(a)

For z=λD/d:

The position of the slits from the central point of the first screen is given by

y=OS3=OS4=z2=λD/2d

The corresponding path difference in wave fronts reaching S3 is given by

∆x=yd/D=[λD/2d ]×dD=λ2

Similarly at S4, path difference,

∆x=yd/D=λD/2d ×dD=λ2

i.e. dark fringes are formed at S3 and S4.

So, the intensity of light at S3 and S4 is zero. Hence, the intensity at P is also zero.

Answer 2

Answer:

Given:

Fours slits S1, S2, S3 and S4.

The separation between slits S3 and S4 can be changed.

Point P is the common perpendicular bisector of S1 S2 and S3 S4.

(a)

For z=λD/d:

The position of the slits from the central point of the first screen is given by

y=OS3=OS4=z2=λD/2d

The corresponding path difference