Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is:
- 3 : 1
- (2 2 + 1) : (2 2 - 1)
- 9 : 1
- (3 + 2 2) : (3 - 2 2)
Answer
(4) (3 + 2 2) : (3 - 2 2)
Solution
Intensity from a slit its width ⇒ I1I2 = 21
⇒ a1a2 = 2
ImaxImin = ( 2 + 1)2( 2 - 1)2
= 3 + 2 23 - 2 2
Answer: (4)