Question

In a YDSE bichromatic light of wavelength $400nm$ and $560nm$ are used. The distance between the slits is $0.1mm$ and the distance between the plane of the slits and the screen is $1m$. The minimum distance between two successive regions of complete darkness is

• A. $4mm$
• B. $5.6mm$
• C. $14mm$
• D. $28mm$

Answer 1

The correct option is D $28mm$

$\begin{array}{c}Letnthminimaof400nmcoincideswithmthminimaof560nm\\ then,\left(2n-1\right)400=\left(2m-1\right)560\\ \Rightarrow \frac{2n-1}{2m-1}=\frac{7}{5}=\frac{14}{10}=\frac{21}{15}\\ i.e.4thminimaof400nmcoincideswith3rdminimaof560nm.\\ =\frac{7\left(1000\right)\left(400\times {10}^{-6}\right)}{2\times 0.1}=14mm\\ next,11thminimaof400nmwillcoincidewith8thminimaof560nm\\ locationofthisminimais\\ =\frac{21\left(1000\right)\left(400\times {10}^{-6}\right)}{2\times 0.1}=42mm]\\ \therefore requireddis\tan ce=28mm\end{array}$

Hence,

option $\left(D\right)$ is correct answer.