Answer:
The wavelength is 503 nm Â
Explanation:
Considering constructive interference , this means that route(path) difference is equal to the product of order of fringe and wavelength of the light
 i.e  dsinθ  = m[tex]\lambda[/tex]
Where [tex]\lambda[/tex]  is the wavelength of light  and m is the order of the fringe
  Looking at θ to be very small , sin θ can be approximated to  θ
       and  [tex]\theta \approx \frac{x}{l}[/tex]
Substituting this into the above equation
      [tex]d[\frac{x}{l} ] =m\lambda[/tex]
   making x the subject
          [tex]x =\frac{m\lambda l}{d}[/tex]
   This above equation will give the value of the distance of the [tex]m^{th}[/tex] order fringe of the wavelength [tex]\lambda[/tex] from the central fringe
    Replacing with the value given in the question we have
 [tex]\lambda[/tex] = 710 nm  m = 2  d =0.66 mm , l = 2.0 m
         [tex]x = \frac{(2)(710nm)(2.0m)[\frac{10^9}{1m} ]}{(0.66mm)(\frac{10^6}{1mm} )}[/tex]
          [tex]=(4.303*10^6nm)[\frac{\frac{1}{10^6}mm }{1nm} ][/tex]
          [tex]=4.303mm[/tex]
 The separation of the second fringe from central maximum is 4,303 mm  Â
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To obtain the separation of the second order fringe of the unknown light from central maximum
    [tex]x' = 4.303mm - 1.25 mm = 3.053mm[/tex]
Now to obtain the wavelength of this second source
          from [tex]x = \frac{m\lambda l}{d}[/tex]
            [tex]\lambda' = \frac{x'd}{ml}[/tex]
Now substituting 3,053 mm for [tex]x'[/tex] 2.0 mm for l , 0.66 mm  for d and 2 for m in the above formula
      [tex]\lambda' =\frac{(3.053mm)(0.66mm)}{(2)(2.0)(\frac{10^3mm}{1m} )}[/tex]
         [tex]= (503.7*10^{-6}mm)(\frac{10^6nm}{1mm} )[/tex]
          [tex]=503.7nm[/tex]
