Question

In a Young's double slit experiment, $m$ th order and $n$ th order of bright fringes are formed at point $P$ on a distant screen, if monochromatic source of wavelength $400nm$ and $600nm$ are used respectively. The minimum value of $m$ and $n$ are respectively,

• A. 4,6
• B. 3,2
• C. 2,3
• D. 4,2

Answer 1

Answer: (B)

Let $m^{\text{th }}$ bright fringe of wavelength ${\lambda }_m$ and $n^{\text{th }}$ bright fringe of wave length ${\lambda }_n$ formed at a distance $x_n$ from central maxima at same point $P$,
i.e., they coincide, therefore, $x_m=x_n$.
$\frac{m{\lambda }_{m}D}{d}=\frac{n{\lambda }_{n}D}{d}$
$\Rightarrow \frac{m}{n}=\frac{{\lambda }_n}{{\lambda }_m}$
Putting the given values, we get
$=\frac{600\times 10^{-9}}{400\times 10^{-9}}$
$[$ Given, ${\lambda }_m=400\times 10^{-9}m,{\lambda }_n=600\times 10^{-9}m]$
$\therefore \frac{m}{n}=\frac{3}{2}$
So, least integral values of $m$ and $n$ satisfying above requirement are $m=3$ and $n=2$.