Question

Given a double slit apparatus with slit distance 2 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 500 nm on the slits

Answer 1

Answer:

The values is  $m_{max} = 8001 \ bright \ spots$

Explanation:

From the question we are told that

    The slit distance is  $d = 2 \ mm = 2*10^{-3} \ m$

    The  wavelength is  $\lambda = 500 \ nm = 500 *10^{-9} \ m$

At the first half of the screen from the central maxima

   The number of bright spot according to the condition for constructive interference is  

          $n = \frac{d * sin (\theta )}{\lambda}$

For maximum number of spot $\theta = 90^o$

So  

       $n = \frac{2*10^{-3} * sin (90 )}{500 *10^{-9}}$

        $n =4000$

Now for the both sides plus the central maxima  we have

      $m_{max} = 2 * n + 1$

substituting values

       $m_{max} = 2 * 4000 + 1$

       $m_{max} = 8001 \ bright \ spots$