Answer:
λ' = 379.22nm
Explanation:
In order to find the longest wavelength that allows one to dark fringe coincides with the bright fringe, you use the following formulas:
[tex]y_{bright}=\frac{m\lambda D}{d}[/tex] (1)
[tex]y_{dark}=(m+\frac{1}{2})\frac{\lambda' D}{d}[/tex] (2)
y-dark and y-bright are the positions of dark fringes and bright fringes respectively.
m: order of the fringe (dark or bright) = 4
D: distance to the screen
d: distance between slits
λ: light for y-bright= 427nm
λ': second light for y-dark = ?
By the information of the statement you know that y-bright = y-dark.
You divide equation (2) into the equation (1) and solve for λ':
[tex]\frac{y_{dark}}{y_{bright}}=1=\frac{(m+1/2)\lambda'}{m\lambda}\\\\\lambda'=\frac{m\lambda}{m+1/2}[/tex] (3)
Finally, you solve the equation (3) by replacing the values of m and the wavelength:
[tex]\lambda'=\frac{4(427nm)}{4+1/2}=379.55nm[/tex]
The longest wavelength which produces the fourth dark fringe in the same location for the fourth bright fringe of the first wavelength is 379.22nm
