Question

Light of wavelength $6000 \,\mathring{A}$ is normally incident on a slit. Angular position of second minimum from central maximum us $30^{\circ}$. Width of the slit should be :

• A. $12 \times 10^{-5} \, cm$
• B. $18 \times 10^{-5} \, cm$
• C. $24 \times 10^{-5} \, cm$
• D. $36 \times 10^{-5} \, cm$

Answer 1

Answer: (C)

For diffraction minima due to a single slit of width $b$, we use
$ b \sin \theta=n \lambda $
$\Rightarrow b=\frac{n \lambda}{\sin \theta}=\frac{2 \times 6000 \times 10^{-10}}{\sin 30^{\circ}} $
$\Rightarrow b=\frac{2 \times 6000 \times 10^{-10}}{1 / 2}$
$=4 \times 6000 \times 10^{-10} $
$\Rightarrow b=24 \times 10^{-7} m $
$=24 \times 10^{-5} \,cm $