In Young's double-slit experiment using monochromatic light, the … - Vidyadip

Question

In Young's double-slit experiment using monochromatic light, the fringe pattern shifts by certain distance on the screen when a mica sheet of refractive index 1.6 and thickness 1.964 microns is introduced in the path of one of the interfering waves. The mica sheet is then removed and the slits-to-screen distance is doubled. It is found that the distance between successive maxima now is the same as the observed fringe shift with the mica sheet. Calculate the wavelength of the monochromatic light used.

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Answer

Data $: n _{ m }=1.6, b =1.964$ microns $=1.964 \times 10^{-6} m$,
$
D _2=2 D _1, W _2= y _0
$
The fringe shift with the mica sheet,
$
y _0=\frac{D_1}{d}\left( n _{ m }-1\right) b
$
Subsequent to the removal of the mica sheet and doubling the slits-toscreen distance, the new fringe width is,
$
W_2=\frac{\lambda D_2}{d}=\frac{\lambda\left(2 D_1\right)}{d}
$
Since, $W_2=y_0$,
$
\frac{2 \lambda D_1}{d}=\frac{D_1}{d}\left(n_{ m }-1\right) b
$
$\therefore$ The wavelength of the light used,
$
\begin{aligned}
\lambda & =\frac{n_{ m }-1}{2} \cdot b=\frac{1.6-1}{2} \cdot\left(1.964 \times 10^{-6}\right) \\
& =0.3 \times 1.964 \times 10^{-6} \\
& = 5 . 8 9 2 \times 1 0 ^{-7} ~ = 5 8 9 2 Å
\end{aligned}
$

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