The maximum intensity in a Young's double slit experiment is I0. Distance between the slits (d) is 5 , where is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at D=10d is .
- I04
- I02
- I0
- 3I04
Answer
(2) I02
Solution
[figure]
Point P opposite slit S1: y= d2 from the central point O
Path difference: x= ydD= (d/2) d10d= d20
d=5 x= 5 20= 4
Phase difference: = 2 4= 2
I=I0 2 2=I0 2 4
I= I02
Answer: (2)