Q. In a YDSE bi-chromatic light of wavelengths $400\ nm$ and $560\ nm$ are used. The distance between the slits is $0.1\ mm$ 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) $4\ mm$
B) $5.6\ mm$
C) $14\ mm$
D) $28\ mm$
Explanation
(d) Let $n$th minima of $400\ nm$ coincides with $m$th minima of $560\ nm$ then
$(2n - 1)400 = (2m - 1)560 \Rightarrow \frac{2n - 1}{2m - 1} = \frac{7}{5} = \frac{14}{10} = \frac{21}{15}$
$i.e.,$ $4th$ minima of $400\ nm$ coincides with $3rd$ minima of $560\ nm$
The location of this minima is
$= \frac{7(1000)(400 \times 10^{- 6})}{2 \times 0.1} = 14\ mm$
Next, $11th$ minima of $400nm$ will coincide with $8th$ minima of $560\ nm$
Location of this minima is
$= \frac{21(1000)(400 \times 10^{- 6})}{2 \times 0.1} = 42\ mm$
$\therefore$ Required distance $= 28\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.