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In Young's double slit experiment, the two slits are illuminated by a light beam consisting of wavelengths $4200 Å$ and $5040 Å$. If the distance between the slits is $2.4 \mathrm{~mm}$ and the distance between the slits and the screen is $200 \mathrm{~cm}$, the minimum distance from the central bright fringe to the point where the bright fringes due to both the wavelengths coincide is
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Verified Answer
The correct answer is:
2.1 mm
Let $n_1$ fringes of $\lambda_1=4200 Å$ and $n_2$ fringes of $\lambda_2=5040 Å$ wavelength are formed in a fixed distance on screen.
So, $n \lambda=$ constant or $n_1 \lambda_1=n_2 \lambda_2$
For minimum distance to be coincide,
$$
\begin{gathered}
n_1=n+1, n_2=n \\
(n+1) 4200=n(5040) \\
4200 n+4200=5040 n \\
n \\
\text { Required distance, } x=\frac{n_1 \lambda_1 D}{d} \\
\text { Given, } D=200 \mathrm{~cm}=2 \mathrm{~m}, d=2 \cdot 4 \times 10^{-3} \mathrm{~m}, \\
n_1=5+1=6, \lambda_1=4200 Å \\
x=\frac{6 \times 4200 \times 10^{-10} \times 2}{24 \times 10^{-3}} \\
x=2.1 \times 10^{-3} \mathrm{~m}=2.1 \mathrm{~mm}
\end{gathered}
$$
So, $n \lambda=$ constant or $n_1 \lambda_1=n_2 \lambda_2$
For minimum distance to be coincide,
$$
\begin{gathered}
n_1=n+1, n_2=n \\
(n+1) 4200=n(5040) \\
4200 n+4200=5040 n \\
n \\
\text { Required distance, } x=\frac{n_1 \lambda_1 D}{d} \\
\text { Given, } D=200 \mathrm{~cm}=2 \mathrm{~m}, d=2 \cdot 4 \times 10^{-3} \mathrm{~m}, \\
n_1=5+1=6, \lambda_1=4200 Å \\
x=\frac{6 \times 4200 \times 10^{-10} \times 2}{24 \times 10^{-3}} \\
x=2.1 \times 10^{-3} \mathrm{~m}=2.1 \mathrm{~mm}
\end{gathered}
$$
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