Question

Q20. In Young’s double slit experiment, a maximum is obtained when the path difference between the interfering waves is (n ∈ l): (a) nλ (b) n λ/2 (c) (2n + 1) λ/2 (d) (2n – 1) λ/4

Answer 1

Answer:

(a)

Explanation:

Answer 2

Answer:

In the young's double-slit experiment, maximum intensity is obtained when the path difference is (a) n$\lambda$ where n∈I

Step by Step Explanation:

In the young's double-slit experiment, for the maximum intensity or bright fringe.

Path difference, $\Delta z=n\lambda$  n = 0, ±1, ±2, ±3, ...

That is $\frac{xd}{D}=n\lambda$

Here d = slit separation,

D = screen distance

Therefore in the young's double-slit experiment, maximum intensity is obtained when the path difference is (a) n$\lambda$ where n∈I