Question

A beam of light of wavelength $6000 \,\mathring{A}$ 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 fringe is:

• A. $1.2 \, cm$
• B. $1.2 \, mm$
• C. $2.4 \, cm$
• D. $2.4 \, mm$

Answer 1

Answer: (D)

In single slit diffraction pattern, angular separation of first dark fringe from central bright fringe is given as
$b \sin \theta=\lambda$
For small angle we can use
$\sin \theta=\theta$
Separation distance of first dark fringe from central maxima is given as
$x =D \theta $
$\Rightarrow x =\frac{D \lambda}{b}=\frac{600 \times 10^{-9} \times 2}{1 \times 10^{-3}}$
$=12 \times 10^{-4} m$
So, distance between the first dark fringes on either side of the central bright fringe
$s=2 x $
$\Rightarrow s=2 \times 12 \times 10^{-4}\, m$
$\Rightarrow s=24 \times 10^{-4} m =2.4\, mm .$