Question

In Young’s double slit interference experiment, separation is made $3$ fold. The fringe width becomes
a. Six fold times
b. Three fold times
c. $\dfrac{3}{6}$ fold times
d. $\dfrac{1}{3}$ fold times.

Answer 1

In young’s double slit experiment, fringe width depends on the wavelength of light used, distance of the screen from the silts and the distance between the screen and slit. Here, distance of the screen from the silts is inversely proportional to the fringe width. The distance between any two consecutive bright or dark fringes is called fringe width.

Complete step by step answer:
Consider two coherent sources A and B separated by a distance $d$ and emitting waves of wavelength $\lambda$ let $d$is the distance of the screen from the silts A and B as shown in the figure. Also, D is the distance between the screen and slit.

The waves from the coherent sources A and B interfere to produce interference pattern on the screen. Since point O on the screen is equidistant from A and B, the waves from A and B arriving at point O are in phase. Therefore, constructive interference occurs and a bright fringe is formed at O. This is called central maximum.
The expression for the fringe width in Young’s double slit experiment is given by,
Fringe width, $\beta = \dfrac{{\lambda D}}{d}....\left( 1 \right)$
Given, separation is made $3$ fold means separation between slits $d$ is increased $3$ times. That is,
$d' = 3d$
Remaining parameter as it is.
Then fringe width becomes, $\beta ' = \dfrac{{\lambda D}}{{d'}}$
$\Rightarrow \beta ' = \dfrac{{\lambda D}}{{3d}}$
This can be written as,
$\Rightarrow \beta ' = \dfrac{1}{3}\beta$
From equation $\left( 1 \right)$ we have,
.$\Rightarrow \beta ' = \dfrac{1}{3}\left( {\dfrac{{\lambda D}}{d}} \right)$.
Which tells us that, as separation between slits d is increasing $3$ times then value of fringe width decreases 3 times.
$\therefore$ The fringe width becomes $\dfrac{1}{3}$ fold times.

Hence, the correct answer is option (D).

Additional information:
Condition for constructive interference (bright fringe): when crest of one superimpose with crest of another or trough of one wave superimpose with 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 $n = 0,{\text{ }}1,{\text{ }}2,{\text{ }}3 \ldots ..$
Path difference, $\Delta = n\lambda$
Destructive interference (dark fringe): When crest of one wave superimposed with trough of another wave then resultant amplitude is minimum, and there is dark fringe is formed. This is called Destructive interference.
Phase difference $\phi = (2n - 1)\pi$ where $n = 0,{\text{ }}1,{\text{ }}2,{\text{ }}3 \ldots ..$
Path difference, $\Delta = (2n - 1)\dfrac{\lambda }{2}$

Note:
• All the bright and dark fringes are of equal width.
• When the apparatus is immersed in a liquid of refractive index n, then wavelength of light decreases, and hence fringe width decreases.
• Two sources must be coherent sources means two sources must emit waves of same frequency or wavelength having zero phase difference.