Answer:
[tex]\lambda = 420 nm[/tex]
Explanation:
As we know that position of maximum on the screen in diffraction pattern is given as
[tex]y = (\frac{2N + 1}{2})\frac{\lambda L}{d}[/tex]
here we know that
position of third maximum due to red color light and position of fifth maximum of unknown wavelength are same
so we have
[tex]\frac{7 \lambda_r L}{2d} = \frac{11 \lambda L}{2d}[/tex]
so we have
[tex]\lambda = \frac{7}{11} \lambda_r[/tex]
[tex]\lambda = \frac{7}{11}(660 nm)[/tex]
[tex]\lambda = 420 nm[/tex]
