In Young's double slit experiment, the two slits are illuminated by a light beam consisting of wavelengths…
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
0.7 mm
1.4 mm
2.1 mm
2.8 mm
Solution
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}
$