The width of one of the two slits in Young's double slit experiment is d while that of the other slit is x d. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is 9:4 then what is the value of x ? (Assume that the field strength varies according to the slit width.)
- 2
- 3
- 4
- 5
Answer
(4) 5
Solution
I A2
Amplitude is proportional to the slit width: A1=A, A2=xA.
I I = (x+1)2(x-1)2= 94 x+1x-1= 32
2x+2=3x-3
x=5
Answer: (4)