In a Young double slit experiment, the wavelength of incident light is 6000 Å, the separation between slits S1 and S2 is 5 cm and the distance between slits plane and screen is 50 cm, as shown in the figure below. If the resultant intensity at P is equal to the intensity due to individual slits, the path difference between interfering waves is Å. [figure]
- 4000
- 3000
- 2000
- 1000
Answer
(3) 2000
Solution
Identical slits: intensity of each =I0. For phase difference at P:
I=I0+I0+2I0 =4I0 2 2
Given I=I0 ⇒ 4I0 2 2=I0 ⇒ 2= 12
⇒ 2= 3 ⇒ = 2 3 (least phase difference)
Path difference: x= 2 = 3
∴ x= 60003 Å =2000 Å
(The slit separation and screen distance are not needed.)
Answer: (3)