When waves from two herent sources, having amplitudes `a` and `b` superimpose, the amplitude `R` of the resultant wave is given by `R = sqrt(a^(2) + b^(2) + 2ab cos phi)` where `phi` is the constant phase angle between the two waves. The resultant intensity `I` is directly proportional to the square of the amplitude of the resultant wave, i.e., `I prop R^(2)`,
i.e., `I prop (a^(2) + b^(2) + 2 ab cos phi)`
For constructive interference, `phi = 2 n pi`,
`I_(max) = (a + b)^(2)`
For destructive interference, `phi = (2 n - 1)pi`
`I_(min) = (a - b)^(2)`
If `I_(1), I_(2)` are intensities of light from two slits of widths `omega_(1)` and `omega_(2)`, then `(I_(1))/(I_(2)) = (omega_(1))/(omega_(2)) = (a^(2))/(b^(2))`
Light waves from two coherent sources of intensity ratio `81 : 1` produce interference. With the help of the passsage given above, choose the most appropriate alternative for each of the following questions :
The ratio of slit widths of the two sources is

Options

A. `1 : 4`
B. `1 : 16`
C. `9 : 1`
D. `9 : 16`

Answer

C

Explanation

Here, `(w_(1))/(w_(2)) = (1)/(4) = (a^(2))/(b^(2)) :. (a)/(b) = (2)/(1)` or `b = 2 a`
`(I_(max))/(I_(min)) = ((a + b)^(2))/((a - b)^(2)) = ((a + 2 a)^(2))/((a - 2 a)^(2)) = (9)/(1)`