Question

In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phase difference (in degrees) between the waves from the two slits arriving at a point 5 mm from the central maximum when the wavelength is 400 nm? (Convert your result so the angle is between 0 and 360°.)

Answer 1

Answer:

The phase difference is  $\Delta \phi = 180^o$

Explanation:

From the question we are told that

     The distance between the slits is $d = 0.2 \ mm = \frac{0.2}{1000} = 0.2 *10^{-3} \ m$

     The distance to the screen is $D = 100 cm = \frac{100}{100} = 1 \ m$

      The wavelength is  $\lambda = 400nm$

 The distance of the wave from the  central maximum is $L = 5mm = 5*10^{-3} m$

   

Generally the path difference of this  waves is mathematically represented as

              $y = d sin \theta$

Here $\theta$ is the angle between the the line connecting the mid-point of the slits with  the screen and the line  connecting the mid-point of the slits to the central maximum

  This implies that

              $tan \theta = \frac{L}{D}$

     =>     $\theta = tan ^{-1} \frac{L}{D}$

             $\theta = tan ^{-1} [\frac{5*10^{-3}}{1}]$

           $\theta =0.2865$

Substituting values into the formula for path difference

       $y = 0.2 *10^{-3} sin(0.2864)$  

       $y = 9.997*10^{-7} \ m$  

The phase difference is mathematically represented as

          $\Delta \phi = \frac{2 \pi }{\lambda } * y$

Substituting values        

         $\Delta \phi = \frac{2 \pi }{400 *10^{-9} } \ * 9.997*10^{-7}$

         $\Delta \phi =5 \pi$

Converting to degree

         $\Delta \phi =5 \pi radians = 5 (180^o) = 180^o$  

the solution is subtracted by 360° in order to get the actual angle