Question
Monochromatic light of wavelength $$400\,nm$$ and $$560\,nm$$ are incident simultaneously and normally on double slits apparatus whose slits separation is $$0.1\,mm$$ and screen distance is $$1\,m.$$ Distance between areas of total darkness will be
A.
$$4\,mm$$
B.
$$5.6\,mm$$
C.
$$14\,mm$$
D.
$$28\,mm$$
Answer :
$$28\,mm$$
Solution :
At the area of total darkness minima will occur for both the wavelengths.
$$\eqalign{
& \therefore \frac{{\left( {2n + 1} \right)}}{2}{\lambda _1} = \frac{{\left( {2m + 1} \right)}}{2}{\lambda _2} \cr
& \Rightarrow \left( {2n + 1} \right){\lambda _1} = \left( {2m + 1} \right){\lambda _2} \cr
& {\text{or}}\,\,\frac{{\left( {2n + 1} \right)}}{{\left( {2m + 1} \right)}} = \frac{{560}}{{400}} = \frac{7}{5} \cr
& {\text{or}}\,\,10n = 14m + 2 \cr} $$
by inspection for $$m= 2,n =3$$ and for $$m=7,n= 10,$$ the distance between them will be the distance between such points.
i.e., $$\Delta s = \frac{{D{\lambda _1}}}{d}\left\{ {\frac{{\left( {2{n_2} + 1} \right) - \left( {2{n_1} + 1} \right)}}{2}} \right\}$$
put $${n_2} = 10,{n_1} = 3$$
On solving we get, $$\Delta s = 28\,\,mm.$$