In Young's double slit interference experiment the wavelength of light used is $6000 Ã…$. If the path…
In Young's double slit interference experiment the wavelength of light used is $6000 Ã…$. If the path difference between waves reaching a point $P$ on the screen is 1.5 microns, then at that point $P$.
Second bright band occurs
Second dark band occurs
Third dark band occurs
Third bright band occurs
Solution
$
\begin{aligned}
\lambda & =6000 Ã…, \Delta x=1.5 \mu \\
& =6 \times 10^{-7} \mathrm{~m},=1.5 \times 10^{-6} \mathrm{~m}
\end{aligned}
$
Path difference for maxima
$
\begin{aligned}
\Delta x & =n \lambda \\
1.5 \times 10^{-6} & =n \times 6 \times 10^{-7} \\
n & =\frac{1.5 \times 10^{-6}}{6 \times 10^{-7}}=\frac{15}{6}=2.5=\frac{5}{2}
\end{aligned}
$
Value of $n$ is always an integer for maxima, so point $P$ is not a maxima.
$\therefore$ For maxima
$
\begin{aligned}
\Delta x & =(2 n+1) \frac{\lambda}{2} \\
1.5 \times 10^{-6} & =(2 n+1) \times \frac{6 \times 10^{-7}}{2}
\end{aligned}
$
$
\begin{aligned}
(2 n+1) & =\frac{1.5 \times 10^{-6} \times 2}{6 \times 10^{-7}} \\
2 n & =5-1 \\
2 n & =4 \Rightarrow n=2
\end{aligned}
$
So, at $P$ third dark band occurs