Question

Light from a 600 nm source goes through two slits 0.080 mm apart. What is the angular separation of the two first order maxima occurring on a screen 2.0 m from the slits

Answer 1

Answer:

The angular separation is  $k = 0.8594^o$

Explanation:

From the question we are told that

   The  wavelength of the light is $\lambda = 600 \ nm = 600*10^{-9} \ m$

   The  distance of separation between the slit is  $d = 0.080 \ mm = 0.080 *10^{-3} \ m$

    The distance from the screen is

Generally the condition for  constructive interference is mathematically represented as

        $d \ sin(\theta) = n \lambda$

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

    here $\theta$ is the angular separation between the central maxima and one side of the first order maxima

given that we are considering the first order of maxima n =  1  

        =>   $\theta = sin ^{-1} [ \frac{1 * 600*10^{-9} }{ 2.0 } ]$

        =>    $\theta = sin ^{-1} [ 0.0075 ]$

        =>   $\theta = 0.4297^o$

So the angular separation of the two first order maxima  is  

     $k = 2 * \theta$

     $k = 2 * 0.4297$

      $k = 0.8594^o$