In YDSE for wavelength `lambda = 589nm`, the interference fringe have angular separation of `3050 xx 10^(-3)` rad. For what wavelength would the angular separation be 10.0% greater?

Explanation

The angular position of maxima is given by
`d sin theta = n lambda`
d theta = n lambda`
[For small theta, sin theta = theta]`
The angular separation of two adjacent maxima is
`Delta theta = (lambda)/(d)`
Let angular separation be 10% greater for wavelength `lambda'`.
`:. (1.10 lambda)/(d) = (lambda ' )/(d)`
Then, `lambda' = 1.10 lambda = (1.10 xx 589) = 648 nm`