Question

Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits

Answer 1

Answer:

The maximum number of bright spot is $n_{max} =5001$

Explanation:

From the question we are told that

     The  slit distance is $d = 1 \ mm = 0.001 \ m$

      The  wavelength is  $\lambda = 400 \ nm = 400*10^{-9 } \ m$

       

Generally the condition for interference is  

        $n * \lambda = d * sin \theta$

Where n is the number of fringe(bright spots) for the number of bright spots to be maximum  $\theta = 90$

=>     $sin( 90 )= 1$

So

     $n = \frac{d }{\lambda }$

substituting values

     $n = \frac{ 1 *10^{-3} }{ 400 *10^{-9} }$

     $n = 2500$

given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as

        $n_{max} = 2 * n + 1$

The  1  here represented the central bright spot

So  

      $n_{max} = 2 * 2500 + 1$

     $n_{max} =5001$