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In a certain double slit experimental arrangement, interference fringes of width $1 \mathrm{~mm}$ each are observed when light of wavelength $5000 Å$ is used. Keeping the setup unaltered, if the source is replaced by another of wavelength $6000 Å$, the fringe width will be
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$1.2 \mathrm{~mm}$
Fringe width, $\beta=\frac{\lambda D}{d}$ Since $D$ and $d$ are unaltered, $\beta \propto \lambda$.
$$
\begin{aligned}
& \therefore \frac{\beta^{\prime}}{\beta}=\frac{\lambda^{\prime}}{\lambda} \\
& \text { or } \beta^{\prime}=\beta \times \frac{\lambda^{\prime}}{\lambda}=1 \times \frac{6000}{5000}=1.2 \mathrm{~mm}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \frac{\beta^{\prime}}{\beta}=\frac{\lambda^{\prime}}{\lambda} \\
& \text { or } \beta^{\prime}=\beta \times \frac{\lambda^{\prime}}{\lambda}=1 \times \frac{6000}{5000}=1.2 \mathrm{~mm}
\end{aligned}
$$
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