Question

A fringe width of a certain interference pattern is $\beta =0.002\text{cm}$. What is the distance of ${{5}^{\text{th}}}$ dark fringe from the centre?
$\begin{align}
  & \text{A}.\text{ }1\times {{10}^{-2}}cm \\
 & \text{B}.\text{ }11\times {{10}^{-2}}cm \\
 & \text{C}.\text{ 1}.1\times {{10}^{-2}}cm \\
 & \text{D}.\text{ }3.28\times {{10}^{-2}}cm \\
\end{align}$

Answer 1

Hint: Firstly to find the distance of the distance of ${{5}^{\text{th}}}$ dark fringe from the centre we will write the formula. After that we will build a relation between $\beta $ and distance of ${{5}^{\text{th}}}$ dark fringe from the centre and by solving the equation we will obtain our result.
Formula used:
${{x}_{n}}=\dfrac{\left( 2n+1 \right)\lambda D}{2d}$

Complete answer:
Distance of ${{5}^{\text{th}}}$ dark fringe from the centre, since it is dark fringe
 ${{x}_{n}}=\dfrac{\left( 2n+1 \right)\lambda D}{2d}$
For ${{5}^{\text{th}}}$ dark fringe we will put n=5

$\begin{align}
  & {{x}_{n}}=\dfrac{11\lambda D}{2d} \\
 & \beta =\dfrac{\lambda D}{d} \\
\end{align}$

We will put $\beta $ in the above equation.
$\begin{align}
  & {{x}_{n}}=\dfrac{11\lambda D}{2d} \\
 & =\dfrac{11}{2}\times \beta \\
 & {{x}_{n}}=\dfrac{11}{2}\times 0.002 \\
 & {{x}_{n}}=1.1\times {{10}^{-2}}cm \\
\end{align}$
The distance of ${{5}^{\text{th}}}$ dark fringe from the centre is $1.1\times {{10}^{-2}}cm$.

So, the correct answer is “Option A”.

Additional Information:
Interference is the result of superposition of secondary waves starting from two different wavefronts originating from two different coherent sources. All bright and dark fringes are of equal width. All bright fringes are of the same intensity. Regions of dark fringes are perfectly dark. So there is good contrast between bright and dark fringes. At an angle of $\dfrac{\lambda }{d}$ , we get a bright fringe in the interference pattern of two narrow slits separated by distance d .

Condition for sustained interference:

The two sources should continuously emit waves of the same frequency and wavelength. The interfering waves should be in the same state of polarisation. The interfering waves must nearly travel along the same direction. The sources should be monochromatic, otherwise fringes of different colours will overlap just to give a few observable fringes.

Note:
In Young’s double slit experiment, the width of the central bright fringe is equal to the distance between the first dark fringes on the two sides of the central bright fringe. So the width of the central fringe is given by
${{\beta }_{{}^\circ }}=\dfrac{D\lambda }{d}$
As all the bright and dark fringes are of the same width the angular width of a fringe is given by $\theta =\dfrac{\beta }{d}$.