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Two slits are made one millimetre apart and the screen is placed one metre away from the slits. The fringe width when light of wavelength $500 \mathrm{~nm}$ is used is
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$0.5 \mathrm{~m}$
Separation between two slit, $\mathrm{d}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}$
Distance between screen and the slit, $\mathrm{D}=1 \mathrm{~m}$
Wavelength of light, $\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}$
The fringe width will be
$$
\begin{aligned}
& \beta=\frac{D \lambda}{d} \\
& =\frac{1 \times 500 \times 10^{-9}}{10^{-3}}=0.5 \mathrm{~m} .
\end{aligned}
$$
Distance between screen and the slit, $\mathrm{D}=1 \mathrm{~m}$
Wavelength of light, $\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}$
The fringe width will be
$$
\begin{aligned}
& \beta=\frac{D \lambda}{d} \\
& =\frac{1 \times 500 \times 10^{-9}}{10^{-3}}=0.5 \mathrm{~m} .
\end{aligned}
$$
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