Question
In Young’s double slit experiment using monochromatic light the fringe pattern shifts by a 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 distance between the slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.
A.
$$38.2\,\mathop {\text{A}}\limits^ \circ $$
B.
$$68.32\,\mathop {\text{A}}\limits^ \circ $$
C.
$$5892\,\mathop {\text{A}}\limits^ \circ $$
D.
$$528.32\,\mathop {\text{A}}\limits^ \circ $$
Answer :
$$5892\,\mathop {\text{A}}\limits^ \circ $$
Solution :
Fringe shift when a sheet of thickness $$t$$ and refractive index $$\mu $$ is introduced in path of one of interfering waves is
$$\Delta x = \frac{{Dt\left( {\mu - 1} \right)}}{d} = \frac{{D \times 1.964 \times {{10}^{ - 6}}\left( {1.6 - 1} \right)}}{d}\,......\left( {\text{i}} \right)$$
The distance between two maxima where mica sheet is remove and the distance between the slits and the screen is doubled $$ = \frac{{\lambda \left( {2D} \right)}}{d}\,.......\left( {{\text{ii}}} \right)$$
Given that the value in eq. (i) and eq. (ii) are equal
$$\eqalign{
& \therefore \frac{{D \times 1.964 \times {{10}^{ - 6}} \times 0.6}}{d} = \frac{{\lambda \times 2D}}{d} \cr
& \Rightarrow \lambda \frac{{1.964 \times {{10}^{ - 6}} \times 0.6}}{2} \cr
& = 0.5892 \times {10^{ - 6}}m \cr
& = 5892\,\mathop {\text{A}}\limits^ \circ \cr} $$