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A beam of light of wavelength $600 \mathrm{~nm}$ from a source falls on a single slit $1 \mathrm{~mm}$ wide and the resulting diffraction pattern is observed on a screen $2 \mathrm{~m}$ away. The distance between the first dark fringes on either side of the central bright fringe is
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Verified Answer
The correct answer is:
$2.4 \mathrm{~mm}$
Given $\lambda=600 \mathrm{~nm}=600 \times 10^{-9} \mathrm{~m}$
$$
D=2 \mathrm{~m}, d=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}
$$
The distance of first dark fringe from central bright fringe,
$$
\begin{aligned}
y &=\frac{D \lambda}{d} \\
&=\frac{2 \times 600 \times 10^{-9}}{1 \times 10^{-3}}=1.2 \mathrm{~mm}
\end{aligned}
$$
$\therefore$ Distance between first dark fringe on either side of central bright fringe $=2 y=1.2 \times 2=2.4 \mathrm{~mm}$
$$
D=2 \mathrm{~m}, d=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}
$$
The distance of first dark fringe from central bright fringe,
$$
\begin{aligned}
y &=\frac{D \lambda}{d} \\
&=\frac{2 \times 600 \times 10^{-9}}{1 \times 10^{-3}}=1.2 \mathrm{~mm}
\end{aligned}
$$
$\therefore$ Distance between first dark fringe on either side of central bright fringe $=2 y=1.2 \times 2=2.4 \mathrm{~mm}$
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