A beam of light of wave length 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright is
Answer & explanation
Correct answer: option 4
For first minimum putting n = 1 in the condition for minima given as $d \sin θ = nλ$
we obtain $d \sin θ = λ$ …(1)
where $\sin θ ≅ θ ≅ \frac{2/x}{D}$
$⇒ sin θ = \frac{x}{2D}$ …(2)
using (1) and (2) we obtain,
$\frac{λ}{d}=\frac{x}{2D}$
$x=\frac{2λD}{d}=\frac{2(600×10^{-9})(2)}{10^{-3}}$
$= 2.4 × 10^{-3} m$
= 2.4 mm