Question
A monochromatic light is used in Young's double slit experiment when one of the slits is covered by a transparent sheet of thickness $$1.8\,mm,$$ made of material of refractive index $${\mu _1}$$ number of fringes which shift is 18, when another sheet of thickness $$3.6\,mm,$$ made of material of refractive index $${\mu _2}$$ is used, number of fringes which shift is 9. Relation between $${\mu _1}$$ and $${\mu _2}$$ is given by
A.
$$4{\mu _2} - {\mu _1} = 3$$
B.
$$4{\mu _1} - {\mu _2} = 3$$
C.
$$3{\mu _2} - {\mu _1} = 4$$
D.
$$2{\mu _1} - {\mu _2} = 4$$
Answer :
$$4{\mu _2} - {\mu _1} = 3$$
Solution :
$$\eqalign{
& \left( {\mu - 1} \right)t = n\beta \cr
& \frac{{\left( {{\mu _1} - 1} \right) \times 1.8 \times {{10}^{ - 5}}}}{{\left( {{\mu _2} - 1} \right) \times 3.6 \times {{10}^5}}} = \frac{{18\beta }}{{9\beta }} \cr
& \left( {{\mu _1} - 1} \right) = 4\left( {{\mu _2} - 1} \right) \cr
& 4{\mu _2} - {\mu _1} = 3 \cr} $$