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A monochromatic light source $\mathrm{S}$ of wavelength $440 \mathrm{~nm}$ is placed slightly above a plane mirror $\mathrm{M}$ as shown. Image of $\mathrm{S}$ in $\mathrm{M}$ can be used as a virtual source to produce interference fringes on the screen. The distance of source $S$ from $O$ is $20.0 \mathrm{~cm}$, and the distance of screen from $\mathrm{O}$ is $100.0 \mathrm{~cm}$ (figure is not to scale). If the angle $\theta=0.50 \times 10^{-3}$ radians, the width of the interference fringes observed on the screen is $-$
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Verified Answer
The correct answer is:
$2.64 \mathrm{~mm}$
$\mathrm{S}$ and $\mathrm{S}_{\mathrm{I}}$ are source of $\mathrm{YDSE}$ $\theta=0.5 \times 10^{-3}$ radian (very small)
$$
\begin{array}{l}
\mathrm{D}=\mathrm{SO} \cos \theta+100 \\
=20 \times 1+100 \\
=120 \mathrm{~cm} \\
\mathrm{~d}=2 \times \mathrm{SO} \sin \theta \\
\Rightarrow \quad 2 \times 20 \times \theta \\
\Rightarrow \quad 40 \times 0.5 \times 10^{-3} \mathrm{~cm} \\
2 \times 10^{-2} \mathrm{~cm} \\
\beta=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{440 \times 10^{-6} \times 120 \times 10^{2}}{2 \times 10^{-2} \times 10^{2}} \\
264 \times 10^{-2} \\
\Rightarrow 2.64 \mathrm{~mm}
\end{array}
$$
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