Question

In YDSE, bichromatic 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 distance of the second region of complete darkness from central maximum is :

• A. $14\text{mm}$
• B. $28\text{mm}$
• C. $42\text{mm}$
• D. $56\text{mm}$

Answer 1

The correct option is C $42\text{mm}$

Compelet darkness is formed when minima of both wavelenghts coincides.
Distance of $n$th minima from the central maximum is given as,
$y=\frac{(2n-1)\lambda D}{2d}$

Let us say $x$th minima of $400\text{nm}$ coincides with $y$th minima of $560\text{nm}$. Their distance from the central maxima will be same.
So,
$\frac{(2x-1)\times 400\times D}{2d}=\frac{(2y-1)\times 560\times D}{2d}$

$\Rightarrow \frac{2x-1}{2y-1}=\frac{560}{400}$

$\Rightarrow \frac{2x-1}{2y-1}=\frac{7}{5}=\frac{14}{10}=\frac{21}{15}=.......$

Suppose, $\frac{2x-1}{2y-1}=\frac{7}{5}$

Then $x=4$ and $y=3$
[First compelete darkness]

Now suppose,
$\frac{2x-1}{2y-1}=\frac{14}{10}$

$\Rightarrow x=\frac{15}{2}$ and $y=\frac{11}{2}$
It is not possible because of fraction.

Now suppose,
$\frac{2x-1}{2y-1}=\frac{21}{15}$

$\Rightarrow x=11$ and $y=8$
[Second compelete darkness]

So, distance of second compelete darkness,
$y=\frac{(2\times 11-1)\times 400\times D}{2d}$

$y=\frac{21\times 400\times {10}^{-9}}{2\times 0.1\times {10}^{-3}}=42\times {10}^{-3}\text{m}=42\text{mm}$