Question

Find the intensity at a point on a screen in Young's double slit experiment where the interfering waves have a path difference of (i) λ/6, and (ii) λ/2. 

Answer 1

Resultant intensity of the interfering waves can be given by

       I' = I₁ + I₂ +2√I₁I₂ cosδ

where I₁ and I₂  are the intensities of light emitted by first and second source respectively and δ is the phase difference between two waves emitted by light sources.

In Young's double slit experiment since a single light source is used. Therefore I₁ = I₂ 
This implies,   I' = 2I + 2Icosδ  

(i) If path difference is λ/6

\[\text { Then phase difference }, \delta = ( \frac{2\pi}{\lambda} ) \times \text { path difference } \]

\[ = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} \]

\[ = \frac{\pi}{3}\]

∴ I' = 2I +2I cos(π/3) 
      = 2I + 2I(1/2) 
      = 2I + I
      = 3I

(ii) If path difference is λ/2

\[\text { Then phase difference }, \delta = ( \frac{2\pi}{\lambda} ) \times \text { path difference } \]

\[ = \frac{2\pi}{\lambda} \times \frac{\lambda}{2} \]

\[ = \pi\]

I' = 2I +2I cos(π)
      = 2I + 2I (-1) 
      = 2I - 2I
      = 0