In $$YDSE$$ a light containing two wavelengths $$500\,nm$$ and $$700\,nm$$ are used. Find the minimum distance where maxima of two wavelengths coincide. Given $$\frac{D}{d} = {10^3},$$ where $$D$$ is the distance between the slits and the screen and $$d$$ is the distance between the slits.
A.
$$1.2\,m$$
B.
$$3.5\,mm$$
C.
$$2.8\,mm$$
D.
$$8.1\,mm$$
Answer :
$$3.5\,mm$$
Solution :
At the place where maxima for both the wavelengths coincide, $$y$$ will be same for both the maxima, i.e.,
$$\eqalign{
& \frac{{{n_1}{\lambda _1}D}}{d} = \frac{{{n_2}{\lambda _2}D}}{d} \cr
& \Rightarrow \frac{{{n_1}}}{{{n_2}}} = \frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{700}}{{500}} = \frac{7}{5} \cr} $$
Minimum integral value of $${{n_2}}$$ is 5.
∴ Minimum distance of maxima of the two wavelengths from central fringe $$ = \frac{{{n_2}{\lambda _2}D}}{d} = 5 \times 700 \times {10^{ - 9}} \times {10^3} = 3.5\,mm.$$
Releted MCQ Question on Optics and Wave >> Wave Optics
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