Problem : One of the slits in a Young’s experiment is wider than the other, so that amplitudes of light reaching the central point of the screen from one slit, acting alone, is twice that from the other slit, acting alone . Find the expression for resultant intensity I_{θ} at a direction θ on the screen .

Solution: We know that

$ \displaystyle I = a_1^2 + a_2^2 + 2 a_1 a_2 cos\phi$

At central point φ = 0, if I_{0} be the intensity, then

$ \displaystyle I_0 = a_1^2 + a_2^2 + 2 a_1 a_2 cos0 $

$ \displaystyle I_0 = a_1^2 + a_2^2 + 2 a_1 a_2 $

$ \displaystyle I_0 = (a_1 + a_2)^2 $

$ \displaystyle I_0 = 9a^2 $

$ \displaystyle a^2 = \frac{I_0}{9} $

For any other direction

Path difference = d sinθ

Phase difference $ \displaystyle \phi =\frac{2\pi}{\lambda}\times d sin\theta $

$ \displaystyle I = a_1^2 + a_2^2 + 2 a_1 a_2 cos\phi$

$ \displaystyle I = a^2 + (2a)^2 + 2 a \times 2a cos(\frac{2\pi}{\lambda}\times d sin\theta)$

$ \displaystyle I = 5a^2 + 4 a^2 cos(\frac{2\pi}{\lambda}\times d sin\theta)$

$ \displaystyle I = a^2 (5 + 4 cos(\frac{2\pi}{\lambda}\times d sin\theta) )$

$ \displaystyle I = \frac{I_0}{9} (5 + 4 cos(\frac{2\pi}{\lambda}\times d sin\theta) )$