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In Young's double slit experiment, the two slits are separated by $0.2 \mathrm{~mm}$ and they are $1 \mathrm{~m}$ from the screen. The wavelength of the light used is $500 \mathrm{~nm}$. The distance between 6th maxima and 10 th minima on the screen is closest to
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Verified Answer
The correct answer is:
$8 \mathrm{~mm}$
Given, $d=0.2 \mathrm{~mm}=2 \times 10^{-4} \mathrm{~m}$
$$
D=1 \mathrm{~m}, \lambda=500 \mathrm{~nm}=5 \times 10^{-7} \mathrm{~m}
$$
The distance between 6 th maxima and 10 th minima is given as
$$
\begin{aligned}
\Delta x & =\left(x_{10}\right)_{\text {dark }}-\left(x_6\right)_{\text {bright }} \\
& =\frac{(2 \times 10-1) D \lambda}{2 d}-\frac{6 D \lambda}{d} \\
& =\frac{D \lambda}{d}\left[\frac{19}{2}-6\right) \\
& =\frac{1 \times 5 \times 10^{-7}}{2 \times 10^{-4}}\left[\frac{7}{2}\right]=\frac{35}{4} \times 10^{-3} \mathrm{~m} \\
& =8.75 \times 10^{-3} \mathrm{~m} \\
& =8.75 \mathrm{~mm}
\end{aligned}
$$
Hence, $8 \mathrm{~mm}$ is closest to $8 \mathrm{~mm}$.
$$
D=1 \mathrm{~m}, \lambda=500 \mathrm{~nm}=5 \times 10^{-7} \mathrm{~m}
$$
The distance between 6 th maxima and 10 th minima is given as
$$
\begin{aligned}
\Delta x & =\left(x_{10}\right)_{\text {dark }}-\left(x_6\right)_{\text {bright }} \\
& =\frac{(2 \times 10-1) D \lambda}{2 d}-\frac{6 D \lambda}{d} \\
& =\frac{D \lambda}{d}\left[\frac{19}{2}-6\right) \\
& =\frac{1 \times 5 \times 10^{-7}}{2 \times 10^{-4}}\left[\frac{7}{2}\right]=\frac{35}{4} \times 10^{-3} \mathrm{~m} \\
& =8.75 \times 10^{-3} \mathrm{~m} \\
& =8.75 \mathrm{~mm}
\end{aligned}
$$
Hence, $8 \mathrm{~mm}$ is closest to $8 \mathrm{~mm}$.
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