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In a Young's double-slit experiment, the slits are separated by $0.28 \mathrm{~mm}$ and the screen is placed $1.4 \mathrm{~m}$ away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2 \mathrm{~cm}$. Determine the wavelength of light used in the experiment.
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Given : $\mathrm{d}=0.28 \mathrm{~mm}=0.28 \times 10^{-3} \mathrm{~m}$ $\mathrm{D}=1.4 \mathrm{~m}, \mathrm{x}=1.2 \mathrm{~cm}=1.2 \times 10^{-2} \mathrm{~m}, \mathrm{n}=4$,
$$
\begin{aligned}
\text { As, } \quad x=& \frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}} \\
\therefore \quad \lambda &=\frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4}=6 \times 10^{-7} \mathrm{~m} \\
&=600 \times 10^{-9} \mathrm{~m}=600 \mathrm{~nm}
\end{aligned}
$$
$$
\begin{aligned}
\text { As, } \quad x=& \frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}} \\
\therefore \quad \lambda &=\frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4}=6 \times 10^{-7} \mathrm{~m} \\
&=600 \times 10^{-9} \mathrm{~m}=600 \mathrm{~nm}
\end{aligned}
$$
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