Discussion Forum

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

Answer:

Explanation:

Let $m ^{th} $ bright fringe of wavelength  $\lambda_{m}$ and $n ^{th}$  bright fringe  of wavelength $\lambda_{n}$ 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.