Question

Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (Assume the small-angle approximation is valid here.)
(a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.97 cm, find the wavelength of the incident light.
(b) At what angle does the next set of bright fringes appear?

Answer 1

Answer:

a

   $\lambda = 1.667 nm$

b

     $\theta = 0.8681^o$

Explanation:

From the question we are told that

   The distance of separation is $d = 0.220 \ mm = 0.00022 \ m$

    The  is distance of the screen from the slit is  $D = 2.60 \ m$

    The distance between the central bright fringe and either of the adjacent bright   $y = 1.97 cm = 1.97 *10^{-2}\ m$

Generally  the condition for constructive interference is  

      $d sin \tha(\theta ) = n \lambda$

From the question we are told that small-angle approximation is valid here.

So    $sin (\theta ) = \theta$

=>        $d \theta = n \lambda$

=>        $\theta = \frac{n * \lambda }{d }$

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

         $y = D * sin (\theta )$

From the question we are told that small-angle approximation is valid here.

So

       $y = D * \theta$

=>   $\theta = \frac{ y}{D}$

So

     $\frac{n * \lambda }{d } = \frac{y}{D}$

     $\lambda =\frac{d * y }{n * D}$

substituting values

       $\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }$

        $\lambda = 1.667 *10^{-6}$

        $\lambda = 1.667 nm$

In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

     $dsin (\theta ) = n \lambda$

    $\theta = sin^{-1}[\frac{ n * \lambda }{d} ]$

    $\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ]$

    $\theta = 0.8681^o$