Question

In Young’s double-slit experiment two coherent sources are placed 0.90mm apart and fringe are observed 1m away. If it produces a second dark fringe, at a distance of 1mm from the central fringe. The wavelength of monochromatic light is used would be,
A) $60 \times {10^{ - 4}}cm$
B) $10 \times {10^{ - 4}}cm$
C) $10 \times {10^{ - 5}}cm$
D) $6 \times {10^{ - 5}}cm$

Answer 1

Hint: Two coherent sources must be close to each other; if not then the difference between the light waves reaching a particular point will be very large. As a result, the separation between the fringes will decrease and it is just possible that the fringe may not be seen at all.

Formula used:
The distance of the second dark fringe from the central fringe, ${x_n} = \left( {2n - 1} \right)\dfrac{{\lambda D}}{{2d}}$
Where, d is the distance between two coherent source or two slits \[\left( m \right)\]
D is the distance between slit and screen \[\left( m \right)\]
$\lambda $ is wavelength of light used \[\left( m \right)\]
\[n\]is the ${n^{th}}$fringe.


Complete step by step answer:
Given, the distance between two coherent source or two slits,$d = 0.90mm \Rightarrow 0.9 \times {10^{ - 3}}m$
Let the distance between slit and screen $D = 1m$ and $n = 2$
We know that the distance of the second dark fringe from the central fringe,${x_2} = 1mm \Rightarrow 1 \times {10^{ - 3}}m$
Now we need to calculate the wavelength of light used.
We have,
${x_n} = \left( {2n - 1} \right)\dfrac{{\lambda D}}{{2d}}$
Wavelength of light used is given by,
$\lambda = \dfrac{{2{x_2}d}}{{\left( {2n - 1} \right)D}}$
Substituting all the values we get,
$\lambda = \dfrac{{2 \times 1 \times {{10}^{ - 3}} \times 0.9 \times {{10}^{ - 3}}}}{{\left( {2 \times 2 - 1} \right) \times 1}}$
$\lambda = \dfrac{{1.8 \times {{10}^{ - 6}}}}{{3 \times 1}}$
Simplifying the given step, we get,
$\lambda = 0.6 \times {10^{ - 6}}m$
$ \Rightarrow 6 \times {10^{ - 7}}m$
$\therefore \lambda = 6 \times {10^{ - 5}}cm$

The correct option is (D).

Additional information:
Two sources must be coherent sources. A coherent source means two sources must emit waves of the same frequency or wavelength having zero phase difference.
Fringe width can be defined as the distance between any two consecutive dark or bright fringes.
From the young’ double-slit experiment, we can conclude that all bright and dark fringes are equally spaced.

Note:
Condition for constructive interference (bright fringe): When the crest of one superimposes with the crest of another or the trough of one wave superimposes with a trough of another then maximum amplitude is formed and there is a formation of bright fringe is called constructive interference.
Phase difference $\phi = 2n\pi $ where \[n = 0,1,2,3 \ldots .\]
Path difference, $\Delta = n\lambda $
Destructive interference (dark fringe): When the crest of one wave is superimposed with the through of another wave then the resultant amplitude is minimum, and there is dark fringe is formed. This is called destructive interference.
Phase difference $\phi = (2n - 1)\pi $ where \[n = 0,1,2,3 \ldots \]
Path difference, $\Delta = (2n - 1)\dfrac{\lambda }{2}$