Q: A screen is placed 50 cm from a single slit, which is illuminated with 6000 A° light, If distance between the first and third minima in the diffraction pattern is 3.00 mm, what is the width of the slit ?

Sol: in case of diffraction at single slit, the position of minima is given by d sin θ = nλ. Where d is the the

aperture size and for small θ:

$\large sin\theta = \theta = \frac{y}{D} $

$\large d sin\theta = n \lambda $

$\large d \frac{y}{D} = n \lambda $

$\large y = \frac{n \lambda D}{d} $

$\large y_3 -y_1 = \frac{2 \lambda D}{d} $

$\large 3 \times 10^{-3} = \frac{2 \times 6 \times 10^{-7} \times 0.5}{d} $

Hence, d = 2 × 10^{-4} m

= 0.2 mm