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A screen is placed $2 \mathrm{~m}$ away from a narrow slit. It the first minimum lies 5 mm from either side of the central maxima, when plane waves of wavelength $5 \times 10^{-7} \mathrm{~m}$ are used, the slit width is
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Verified Answer
The correct answer is:
$2 \times 10^{-4} \mathrm{~m}$
Given, distance between source and screen, $D=2 \mathrm{~m}$
Distance of first minimum from the central maximum, $y=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m}$
Wavelength, $\lambda=5 \times 10^{-7} \mathrm{~m}$
We know that,
$$
y_{n}=\frac{n \lambda D}{d}
$$
For first minimum, $n=1$
$$
\begin{aligned}
& & y_{1} &=\frac{\lambda D}{d} \\
\Rightarrow & & d &=\frac{\lambda D}{y_{1}}=\frac{5 \times 10^{-7} \times 2}{5 \times 10^{-3}}=2 \times 10^{-4} \mathrm{~m}
\end{aligned}
$$
Distance of first minimum from the central maximum, $y=5 \mathrm{~mm}=5 \times 10^{-3} \mathrm{~m}$
Wavelength, $\lambda=5 \times 10^{-7} \mathrm{~m}$
We know that,
$$
y_{n}=\frac{n \lambda D}{d}
$$
For first minimum, $n=1$
$$
\begin{aligned}
& & y_{1} &=\frac{\lambda D}{d} \\
\Rightarrow & & d &=\frac{\lambda D}{y_{1}}=\frac{5 \times 10^{-7} \times 2}{5 \times 10^{-3}}=2 \times 10^{-4} \mathrm{~m}
\end{aligned}
$$
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