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In Young's double slit experiment the slits are $3 \mathrm{~mm}$ apart and are illuminated by light of two wavelengths $3750 Å$ and $7500 Å$. The screen is placed at $4 \mathrm{~m}$ from the slits. The minimum distance from the common central bright fringe on the screen at which the bright fringe of one interference pattern due to one wavelength coincide with the bright fringe of the other is
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Verified Answer
The correct answer is:
$1 \mathrm{~mm}$
Given, $\lambda_1=3750 Å$
$\begin{aligned} \lambda_2 & =7500 Å \\ D & =4 \mathrm{~m} \\ d & =3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}\end{aligned}$
since, $\frac{\lambda_1}{\lambda_2}=\frac{3750}{7500}=\frac{1}{2}$
We know that,
$\Rightarrow \quad n_1 \lambda_1=n_2 \lambda_2 \Rightarrow \frac{n_2}{n_1}=\frac{\lambda_1}{\lambda_2}=\frac{1}{2}$
$\therefore$ Required minimum distance,
$x=\frac{n_1 \lambda_1 D}{d}$
$\begin{aligned} & =\frac{2 \times 3750 \times 10^{-10} \times 4}{3 \times 10^{-3}}=\frac{3 \times 10^4 \times 10^{-10}}{3 \times 10^{-3}} \\ & =10^{-3} \mathrm{~m}=1 \mathrm{~mm}\end{aligned}$
$\begin{aligned} \lambda_2 & =7500 Å \\ D & =4 \mathrm{~m} \\ d & =3 \mathrm{~mm}=3 \times 10^{-3} \mathrm{~m}\end{aligned}$
since, $\frac{\lambda_1}{\lambda_2}=\frac{3750}{7500}=\frac{1}{2}$
We know that,
$\Rightarrow \quad n_1 \lambda_1=n_2 \lambda_2 \Rightarrow \frac{n_2}{n_1}=\frac{\lambda_1}{\lambda_2}=\frac{1}{2}$
$\therefore$ Required minimum distance,
$x=\frac{n_1 \lambda_1 D}{d}$
$\begin{aligned} & =\frac{2 \times 3750 \times 10^{-10} \times 4}{3 \times 10^{-3}}=\frac{3 \times 10^4 \times 10^{-10}}{3 \times 10^{-3}} \\ & =10^{-3} \mathrm{~m}=1 \mathrm{~mm}\end{aligned}$
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