Question

Maximum intensity in YDSE is I_0. Find the intensity at a point on the screen where (a) The phase difference between the two interfering beams is pi/3. (b) the path difference between them is lambda/4.

Answer 1

Answer:

For path difference λ, phase difference =2πrad.

For path difference

4

λ

, phase difference =

2

π

rad.

As K=4I

0

so intensity at given point where path difference is

4

λ

K

′

=4I

0

cos

2

(

4

π

) (cos

4

π

=cos45

∘

)

K

′

=2I

0

=

2

K

.

Answer 2

Answer:

hence proved..............