Question

In Young’s double slit experiment, the fringe width is $1 \times {10^{ - 4}}m$, if the distance between the slit and screen is doubled and the distance between two slits is reduced to half and the wavelength is changed from $6.4 \times {10^{ - 7}}m$ to $4.0 \times {10^{ - 7}}m$, then the value of new fringe width will be
$
  {\text{A}}{\text{. 2}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 4}}m \\
  {\text{B}}{\text{. 2}}{\text{.0}} \times {\text{1}}{{\text{0}}^{ - 4}}m \\
  {\text{C}}{\text{. 1}} \times {\text{1}}{{\text{0}}^{ - 4}}m \\
  {\text{D}}{\text{. 0}}{\text{.4}} \times {\text{1}}{{\text{0}}^{ - 4}}m \\
$

Answer 1

Hint: We are given fringe width and we know the formula for fringe width. By comparing the new and old values of distance between slits and the distance between slit and screen, we can find the value of new fringe width.

Formula used:
In Young’s double slit experiment, the fringe width of the interference pattern is given as
$\beta = \dfrac{{\lambda D}}{d}{\text{ }}...{\text{(i)}}$
where $\beta $ is called the fringe width of the pattern obtained on the screen, $\lambda $ signifies the wavelength of light used in the YDSE, D is the distance between the slits and the screen while d denotes the distance between the two slits.

Detailed step by step solution:
In Young’s double slit experiment, we obtain an interference pattern on a screen when two coherent sources of light are allowed to overlap each other. The pattern obtained on the screen consists of maximum and minimum intensities of light. The distance between two maxima is called the fringe width. The expression of fringe width is given in equation (i).
We are given the following parameters for the fringe width in a YDSE. The fringe width is given as:

$\beta = 1 \times {10^{ - 4}}m$

The wavelength of light used is

$\lambda = 6.4 \times {10^{ - 7}}m$

Therefore, we can write:

$1 \times {10^{ - 4}} = \dfrac{{6.4 \times {{10}^{ - 7}} \times D}}{d}{\text{ }}...{\text{(ii)}}$
Let new fringe width be $\beta '$ , wavelength be $\lambda '$, distance between slits be d’ and distance between slit and screen be d’ then we can write

$\beta ' = \dfrac{{\lambda 'D'}}{{d'}}$

We are given that new parameters are modified as follows:

$
  \lambda ' = 4 \times {10^{ - 7}}m \\
  D' = 2D \\
  d' = \dfrac{d}{2} \\
  \therefore \beta ' = \dfrac{{4 \times {{10}^{ - 7}} \times 2D}}{{\dfrac{d}{2}}} \\
   = \dfrac{{4 \times {{10}^{ - 7}} \times 2D \times 2}}{d}{\text{ }}...{\text{(iii)}} \\
$

Now, by dividing equation (iii) by (ii), we get

$
  \dfrac{{\beta '}}{{1 \times {{10}^{ - 4}}}} = \dfrac{{4 \times {{10}^{ - 7}} \times 2D \times 2}}{d} \times \dfrac{d}{{6.4 \times {{10}^{ - 7}} \times D}} \\
   \Rightarrow \dfrac{{\beta '}}{{1 \times {{10}^{ - 4}}}} = \dfrac{5}{2} \\
   \Rightarrow \beta ' = \dfrac{5}{2} \times {10^{ - 4}} \\
   \Rightarrow \beta ' = 2.5 \times {10^{ - 4}}m \\
$

This is the required value of fringe width. Hence, the correct answer is option A.

Note: The fringe pattern is formed only when coherent sources of light are used. Coherent sources have the same frequency and constant phase difference. They can be obtained from a single source by passing the same light from two slits.