A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000 \mathop A\limits^ \circ and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is :

Wave Optics PYQ — JEE Main 2017 · 9 Apr · Shift 1 · Q55 with Solution

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Correct answer: B

Slit width: $a = 0.1\text{ mm} = 1\times 10^{-4}\text{ m}$
Wavelength: $\lambda = 6000\,\mathring{A} = 6000\times 10^{-10}\text{ m} = 6\times 10^{-7}\text{ m}$
Screen distance: $D = 0.5\text{ m}$

For minima (dark bands),

a\sin\theta = n\lambda, \qquad n=1,2,3,\dots

For small angles,

\sin\theta \approx \tan\theta = \frac{y_n}{D}

so,

y_n = \frac{n\lambda D}{a}

where $y_n$ is the distance of the $n$ th dark band from the central bright band.

Here $n=3$ :

y_3 = \frac{3\lambda D}{a}

Substitute the values:

y_3 = \frac{3(6\times 10^{-7})(0.5)}{1\times 10^{-4}}

y_3 = \frac{9\times 10^{-7}}{10^{-4}} = 9\times 10^{-3}\text{ m}

y_3 = 9\text{ mm}

$9\text{ mm}$ corresponds to Option B.

Stored correct answer: B

My derived answer: B

So, the derived answer agrees with the stored answer.