Question

In a YDSE bi-chromatic light of wavelengths, $400nm$ and $560nm$ are used. The distance between the plane of the slits is $0.1mm$ and the distance between 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

Answer: (D)

Let $n\text{th}$ minimum of $400nm$ coincides with $m\text{th}$ minimum of $560nm$, then
For $n^{th}$ minima $\delta =\left(2n-1\right)\frac{\lambda }{2}$
$\left(2n-1\right)\left(\frac{400}{2}\right)=\left(2\text{m}-1\right)\left(\frac{560}{2}\right)$
$\frac{2n-1}{2\text{m}-1}=\frac{7}{5}=\frac{14}{10}=\frac{21}{15}$
i.e., $4\text{th}$ minimum of $400nm$ coindes with $3$rd minimum of $560nm$.
Distance of $n^{th}$ minima from center $=\frac{\left(2n-1\right)\lambda D}{2d}$
Location of this minimum is,
${\left(\text{Y}\right)}_1=\frac{\left(2\times 4-1\right)\left(1000\right)\left(400\times 10^{-6}\right)}{2\times 0\cdot 1}=14\text{mm}$
Next $11$th minimum of $400nm$ will coincide with $8$th
minimum of $560nm$.
Location of this minimum is,
${\left(\text{Y}\right)}_2=\frac{\left(2\times 11-1\right)\left(1000\right)\left(400\times 10^{-6}\right)}{2\times 0\cdot 1}=42\text{mm}$
$\therefore$ Required distance $={\text{Y}}_2-{\text{Y}}_1=28\text{mm}$
Hence, the correct option is ($28mm$)