Q: In young’s double slit experiment intensity at a point is (1/4) of the maximum intensity. Angular position of this points is

Sol: $\large I = I_{max} cos^2 (\phi/2)$

$\large I_{max}/4 = I_{max} cos^2 (\phi/2)$

$\large cos\frac{\phi}{2} = \frac{1}{2} = cos\frac{\pi}{3}$

φ = 2π/3

$\large \phi = \frac{2\pi}{\lambda}. \Delta x $ ; where ∆x = d sin θ

$\large \frac{2\pi}{3} = \frac{2\pi}{\lambda}. d sin\theta $

λ/3 = d sin θ , sin θ = λ/3d ;

$\large \theta = sin^{-1}(\frac{\lambda}{3 d}) $