Question

In $\text{YDSE}$, dichromatic light of wavelengths $400\text{nm}$ and $560\text{nm}$ are used. The distance between the slits is $0.1\text{mm}$ and the distance between the plane of the slits and the screen is $1\text{m}$. The minimum distance between two successive regions of complete darkness is

• A. $4\text{mm}$
• B. $5.6\text{mm}$
• C. $14\text{mm}$
• D. $28\text{mm}$

Answer 1

The correct option is D $28\text{mm}$

Let $n^{th}$ minima of $400\text{nm}\left({\lambda }_1\right)$ coincides with $m^{th}$ minima of 560 $\text{nm}\left({\lambda }_2\right)$, then

$\frac{(2n-1){\lambda }_{1}D}{2d}=\frac{(2m-1){\lambda }_{2}D}{2d}$

$\Rightarrow (2n-1)\left(\frac{400}{2}\right)=(2m-1)\left(\frac{560}{2}\right)$

$\Rightarrow \frac{2n-1}{2m-1}=\frac{7}{5}=\frac{14}{10}=\frac{21}{15}....=\frac{7k}{5k}$

If we take the first ratio, then $n=4$ and $m=3$ (Acceptable)

If we take the second ratio, then $n=7.5$ and $m=5.5$ (This is not acceptable, as $n$ and $m$ should be integer.)

If we take the third ratio, then $n=11$ and $m=8$ (Acceptable)

By first ratio, $4^{th}$ minima of $400\text{mm}$ coincides with $3^{rd}$ minima of $560\text{nm}$ coincide.

The general formula for the location of minima is given by,

$y=\frac{(2n-1)\lambda D}{d}$

Then, location of these minima is

$y_1=\frac{\left(\right(2\times 4)-1)(400\times {10}^{-9})}{2\times 0.1\times {10}^{-3}}=14\text{mm}$

Next, ${11}^{th}$ minima of $400\text{nm}$ will coincide with $8^{th}$ minima of $560\text{nm}$.

Location of these minima is,

$y_2=\frac{\left(\right(2\times 11)-1)(400\times {10}^{-9})}{2\times 0.1\times {10}^{-3}}=42\text{mm}$

$\therefore$ Required distance $=y_2-y_1=42-14=28\text{mm}$

Hence, the correct option is $\left(D\right)$