QJEEM17Ph9363: JEE Main 2017 Physics - Class 12 JEEM17Ph9363 Solution: A single slit of width 0.1 mm is illuminated by a parallel beam …

The slit width is a = 0.1 mm = 10 ^{$-$ 4} m.

The wavelength of the light is $\lambda$ = 6000 $\times$ 10 ^{$-$ 10} = 6 $\times$ 10 ^{$-$ 7}.

The distance from the slit to diffraction bands is D = 0.5 m.

We calculate the third dark band from the central band as follows:

$A\sin \theta = 3\lambda \Rightarrow \sin \theta = {{3\lambda } \over a} = {x \over D}$

$\Rightarrow x = {{3\lambda D} \over a} = {{3 \times 6 \times {{10}^{ - 7}} \times 0.5} \over {{{10}^{ - 4}}}} = 9$ mm