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A light beam of wavelength $800 \mathrm{~nm}$ passes through a single slit and projected on a screen lsept at $5 \mathrm{~m}$ away from the slit. What should be the slit width for the ray optics approximation to be valid?
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Verified Answer
The correct answer is:
$2 \mathrm{~mm}$
Ray optics approximation can be taken upto FresneIs' distance which is given by
$$
z_f=\frac{a^2}{\lambda}
$$
Given, $\quad Z_f=5 \mathrm{~m}$, wavelength, $\lambda=800 \mathrm{~nm}=800 \times 10^{-9} \mathrm{~m}$, Slit width $a=$ ?
$$
\begin{aligned}
a & =\sqrt{Z_f \cdot \lambda} \\
& =\left(\sqrt{5 \times 800 \times 10^{-9} \times 10^3 \times 10^3}\right) \mathrm{mm} \\
a & =2 \mathrm{~mm}
\end{aligned}
$$
$$
z_f=\frac{a^2}{\lambda}
$$
Given, $\quad Z_f=5 \mathrm{~m}$, wavelength, $\lambda=800 \mathrm{~nm}=800 \times 10^{-9} \mathrm{~m}$, Slit width $a=$ ?
$$
\begin{aligned}
a & =\sqrt{Z_f \cdot \lambda} \\
& =\left(\sqrt{5 \times 800 \times 10^{-9} \times 10^3 \times 10^3}\right) \mathrm{mm} \\
a & =2 \mathrm{~mm}
\end{aligned}
$$
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