Question
In an experiment, sodium light $$\left( {\lambda = 5890\,\mathop {\text{A}}\limits^ \circ } \right)$$ is employed and interference fringes are obtained in which 20 fringes equally spaced occupy $$2.30\,cm$$ on the screen. When sodium light is replaced by blue light, the setup remaining the same otherwise, 30 fringes occupy $$2.80\,cm.$$ The wavelength of blue light is
A.
$$4780\,\mathop {\text{A}}\limits^ \circ $$
B.
$$5760\,\mathop {\text{A}}\limits^ \circ $$
C.
$$9720\,\mathop {\text{A}}\limits^ \circ $$
D.
$$6390\,\mathop {\text{A}}\limits^ \circ $$
Answer :
$$4780\,\mathop {\text{A}}\limits^ \circ $$
Solution :
$$\eqalign{
& {\beta _1} = \frac{{2.30}}{{20}} = \frac{{D{\lambda _1}}}{d} \cr
& {\text{and}}\,\,{\beta _2} = \frac{{2.80}}{{30}} = \frac{{D{\lambda _2}}}{d} \cr
& {\text{or}}\,\,\frac{{{\beta _1}}}{{{\beta _2}}} = \frac{{2.30 \times 30}}{{20 \times 2.80}} = \frac{{{\lambda _1}}}{{{\lambda _2}}} \cr
& \therefore {\lambda _2} = 0.81\,{\lambda _1} = 0.81 \times 5890 = 4780\mathop {\text{A}}\limits^ \circ . \cr} $$