Question

In a double-slit interference experiment, the fringe width obtained with a light of wavelength $5900\mathrm{A}\mathrm{was}1.2$ mm for parallel narrow slits placed $2\mathrm{mm}$ apart. In this arrangement, if the slit separation is increased by one-and-half times the previous value, then the fringe width is?

• A. $0.9\mathrm{mm}$
• B. $0.8\mathrm{mm}$
• C. $1.8\mathrm{mm}$
• D. $1.6\mathrm{mm}$

Answer 1

The correct option is B

$0.8\mathrm{mm}$

Step 1: Given data:

Wavelength in a double-slit interference fringe experiment, $\lambda =5900A^{\circ }$
Fringe width obtained, $\beta =1.2mm$
Distance between the parallel narrow slits, $d=2mm$

The separation is given to be increased by one and half times the previous value,
$d^1=d+{\displaystyle \frac{d}{2}}={\displaystyle \frac{3}{2}}d$

Where ${\mathrm{d}}^1=$the new distance between the parallel narrow slits.
Step 2: Formula for calculating fringe width:

Fringe width is known as the difference between two light fringes in succession or two dark fringes in succession.

The fringe distance for all the fringes is constant in the interference pattern, which is directly proportional to the wavelength of the light employed.
The fringe width can be given as:
$\beta ={\displaystyle \frac{\lambda D}{d}}$

where, $\beta =$Fringe width, $\lambda =$ Wavelength,
$D=$Distance between the slit and the screen,$d=$ The distance between the slits.
Also, Fringe width is found out to be inversely proportional to the distance between the slits.
Therefore,

$\beta \alpha {\displaystyle \frac{1}{d}}$
${\displaystyle \frac{{\beta }^1}{\beta }}={\displaystyle \frac{d}{d^1}}$ ...... (1)

Where, ${\mathrm{\beta }}^1=$New fringe width

Step 3: Calculation of new fringe width:
Now we can find the fringe width when the separation is increased by one-and-half times the previous value as :
From equation (1), we get,
$$\begin{gathered} {\displaystyle \frac{{\beta }^1}{\beta }}={\displaystyle \frac{d}{d^1}} \\ \Rightarrow {\displaystyle \frac{{\beta }^1}{1.2}}={\displaystyle \frac{2}{3}} \\ \therefore {\beta }^1=0.8mm \end{gathered}$$
Hence, if the slit separation is increased by one-and-half times the previous value, then the fringe width is $0.8mm$, Hence, option B is correct.