Answer:
a
  [tex]\lambda = 1.667 nm[/tex]
b
   [tex]\theta = 0.8681^o[/tex]
Explanation:
From the question we are told that
  The distance of separation is [tex]d = 0.220 \ mm = 0.00022 \ m[/tex]
  The  is distance of the screen from the slit is  [tex]D = 2.60 \ m[/tex]
  The distance between the central bright fringe and either of the adjacent bright  [tex]y = 1.97 cm = 1.97 *10^{-2}\ m[/tex]
Generally  the condition for constructive interference is Â
   [tex]d sin \tha(\theta ) = n \lambda[/tex]
From the question we are told that small-angle approximation is valid here.
So   [tex]sin (\theta ) = \theta[/tex]
=> Â Â Â Â [tex]d \theta = n \lambda[/tex]
=> Â Â Â Â [tex]\theta = \frac{n * \lambda }{d }[/tex]
Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes
Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as
     [tex]y = D * sin (\theta )[/tex]
From the question we are told that small-angle approximation is valid here.
So
    [tex]y = D * \theta[/tex]
=> Â [tex]\theta = \frac{ y}{D}[/tex]
So
   [tex]\frac{n * \lambda }{d } = \frac{y}{D}[/tex]
   [tex]\lambda =\frac{d * y }{n * D}[/tex]
substituting values
    [tex]\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }[/tex]
    [tex]\lambda = 1.667 *10^{-6}[/tex]
    [tex]\lambda = 1.667 nm[/tex]
In the b part of the question we are considering the next set of bright fringe so  n=  2
  Hence
   [tex]dsin (\theta ) = n \lambda[/tex]
  [tex]\theta = sin^{-1}[\frac{ n * \lambda }{d} ][/tex]
  [tex]\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ][/tex]
  [tex]\theta = 0.8681^o[/tex]
