Discussion Forum

In the Young's double-slit experiment using a monochromatic ;light of wavelength λ the path difference (in terms of an integer n)  corresponding  to any point having half the peak intensity is 

Answer:

Explanation:

$I= I_{max} \cos^{2}\frac{\phi}{2}$    ...............(i)

  Given,         $I= \frac{I_{max}}{2}$.........(ii)

 From Eqs( i) and (ii) , we have

   $\phi= \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}$  

  Or path difference ,  $\triangle x=\left(\frac{\lambda}{2\pi}\right).\phi$

  $\therefore$     $\triangle x=\frac{\lambda}{4},\frac{3\lambda}{4},\frac{5\lambda}{4},.....(\frac{2n+1}{4})\lambda$