In the Young's double slit experiment, the interference pattern is found to have as intensity ratio between the bright and dark fringes as 9. This implies that

Options

A. the intensities at the screen due to the two slits are 5 units and 4 units, respectively
B. the intersities at the screen due to the two slits are 4 units and 1 units, respectively
C. the amplitude ratio is 3
D. the amplitude ratio is 2

Answer

b.,d

Explanation

We know that
`(I_(max))/(I_(min)) = (a + b)^(2)/((a - b)^(2)`
Where `I_(1) prop a^(2)` (a is amplitude of wave 1) and
`I_(2) prop b^(2)`.
Hence,
`(I_(max))/(I_(min)) = (9)/(1)`
`(a + b)/(a -b) = (3)/(1) implies (a)/(b) = (1)/(2)`
`(I_(1)^(2))/(I_(2)^(2)) = (a^(2))/(b^(2)) = (1)/(4)`