The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is:
- 1 : 1
- 4 : 1
- 16 : 1
- 9 : 1
Answer
(4) 9 : 1
Solution
Intensity slit width ⇒ I1 = I0, I2 = 4I0
ImaxImin = ( I2 + I1 I2 - I1 )2 = ( 2 + 12 - 1 )2
= 9 : 1
Answer: (4)