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A beam of light of wavelength $600 \mathrm{~nm}$ from a distant 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, wavelength of light beam,
$$
\begin{aligned}
\lambda &=600 \mathrm{~nm} \\
&=600 \times 10^{-9} \mathrm{~m}
\end{aligned}
$$
Distance between slits. $d=1 \mathrm{~mm}=10^{-3} \mathrm{~m}$
Distance between slits and screen, $D=2 \mathrm{~m}$
Distance between first dark fringes from the central bright fringe, $2 \beta=$ ?
As we know, fringe width, $\beta=\frac{\lambda D}{d}$
$$
\begin{aligned}
&=\frac{600 \times 10^{-9} \times 2}{10^{-3}} \\
&=1.2 \times 10^{-3} \mathrm{~m}
\end{aligned}
$$
Distance between first dark fringes on either side of central bright fringe $(2 \beta)=2 \times \beta=2 \times 1.2 \times 10^{-3} \mathrm{~m}$
$$
\Rightarrow \quad 2 \beta=2.4 \mathrm{~mm}
$$
$$
\begin{aligned}
\lambda &=600 \mathrm{~nm} \\
&=600 \times 10^{-9} \mathrm{~m}
\end{aligned}
$$
Distance between slits. $d=1 \mathrm{~mm}=10^{-3} \mathrm{~m}$
Distance between slits and screen, $D=2 \mathrm{~m}$
Distance between first dark fringes from the central bright fringe, $2 \beta=$ ?
As we know, fringe width, $\beta=\frac{\lambda D}{d}$
$$
\begin{aligned}
&=\frac{600 \times 10^{-9} \times 2}{10^{-3}} \\
&=1.2 \times 10^{-3} \mathrm{~m}
\end{aligned}
$$
Distance between first dark fringes on either side of central bright fringe $(2 \beta)=2 \times \beta=2 \times 1.2 \times 10^{-3} \mathrm{~m}$
$$
\Rightarrow \quad 2 \beta=2.4 \mathrm{~mm}
$$
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