Answer:
The wavelength of visible light = 489.33 nm
Explanation:
In an [tex]m^{th}[/tex] order bright fringe;
[tex]y_m = m \frac{ \lambda R}{d}[/tex]
For first order ; m = 1
∴ [tex]y_1 = \frac{\lambda R}{d}[/tex]
For [tex]m^{th}[/tex] order dark fringe ;
[tex]y_m = (m+ \frac{1}{2}) \frac{\lambda R}{d}[/tex]
first order [tex]y_1 = (1+ \frac{1}{2}) \frac{\lambda R}{d}[/tex]
[tex]y_1 = (\frac{3}{2}) \frac{\lambda R}{d}[/tex]
Assuming the wavelength of second light is [tex]\lambda'[/tex]
Thus:
[tex]\phi \ y_1 = \frac{3}{2} \frac{\lambda' R}{d}[/tex]
Given that : [tex](y_1) bright =(b_1)dark[/tex]
[tex]\frac{\lambda R}{d}= \frac{3}{2}\frac{\lambda' R}{d}[/tex]
[tex]\lambda' = \frac{2}{3} \lambda[/tex]
[tex]\lambda' = \frac{2}{3} *734[/tex]
[tex]\lambda' =489.33 \ nm[/tex]
