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In Young's double slit experiment, the spacing between the slits is $d$ and wavelength of light used is $6000 Å$. If the angular width of a fringe formed on a distance screen is $1^{\circ}$, then value of $d$. is
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$0.03 \mathrm{~mm}$
Here, $\sin \theta=\left(\frac{Y}{D}\right) \quad \therefore \Delta \theta=\frac{\Delta Y}{D}$
Angular fringe width $\theta_{0}=\Delta \theta$
$$
\text { (width } \Delta \mathrm{Y}=\beta \text { ) }
$$
$$
\begin{array}{l}
\theta_{0}=1^{\circ}=\frac{\pi}{180} \mathrm{rad} \text { and } \lambda=6 \times 10^{-7} \mathrm{~m} \\
d=\frac{\lambda}{\theta_{0}}=\frac{180}{\pi} \times 6 \times 10^{-7}=0.03 \mathrm{~mm}
\end{array}
$$
Angular fringe width $\theta_{0}=\Delta \theta$
$$
\text { (width } \Delta \mathrm{Y}=\beta \text { ) }
$$
$$
\begin{array}{l}
\theta_{0}=1^{\circ}=\frac{\pi}{180} \mathrm{rad} \text { and } \lambda=6 \times 10^{-7} \mathrm{~m} \\
d=\frac{\lambda}{\theta_{0}}=\frac{180}{\pi} \times 6 \times 10^{-7}=0.03 \mathrm{~mm}
\end{array}
$$
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