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In Young's double slit experiment, fringes of width $\beta$ are produced on a screen kept at a distance of $1 \mathrm{~m}$ from the slit. When the screen is moved away by $5 \times 10^{-2} \mathrm{~m}$, fringe width changes by $3 \times 10^{-5} \mathrm{~m}$. The separation between the slits is $1 \times 10^{-3} \mathrm{~m}$. The wavelength of the light used is
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Verified Answer
The correct answer is:
$600 \mathrm{~nm}$
In YDSE, we have
$$
\begin{aligned}
&\text { Wavelength } \begin{aligned}
\lambda &=\frac{\Delta \beta \mathrm{d}}{\Delta \mathrm{D}}=\frac{3 \times 10^{-5} \times 1 \times 10^{-3}}{5 \times 10^{-2}} \\
&=6 \times 10^{-7} \mathrm{~m}
\end{aligned} \\
&\text { or } \lambda=600 \mathrm{~nm}
\end{aligned}
$$
$$
\begin{aligned}
&\text { Wavelength } \begin{aligned}
\lambda &=\frac{\Delta \beta \mathrm{d}}{\Delta \mathrm{D}}=\frac{3 \times 10^{-5} \times 1 \times 10^{-3}}{5 \times 10^{-2}} \\
&=6 \times 10^{-7} \mathrm{~m}
\end{aligned} \\
&\text { or } \lambda=600 \mathrm{~nm}
\end{aligned}
$$
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