Question

In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900{A^ \circ }$ was $1.2\,mm$ for parallel narrow slits placed $2\,mm$ apart. In this arrangement, if the salt separation is increased by one-and –half times the previous value, then the fringe width is:
(A) $0.9\,mm$
(B) $0.8\,mm$
(C) $1.8\,mm$
(D) $1.6\,mm$

Answer 1

Here first we have to establish the relationship between fringe width and distance between the slits.Fringe width is 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. The distance of the fringe is directly proportional to the wavelength of the light employed.Fringe width depends on wavelength of light, distance between the slits and the screen or slit

Complete step by step answer:
The fringe width is given by –
$\beta = \dfrac{{\lambda D}}{d}$, where $\beta=$fringe width,$\lambda=$ wavelength,
$D =$distance between the slit and the screen,$d =$distance between the slits.
From the above expression we can see that,
Fringe width is inversely proportional to the distance between the slits. If the separation between the slits is increased the fringe width decreases and vice versa.
Therefore, $\beta \alpha \dfrac{1}{d}$
$\dfrac{{{\beta ^1}}}{\beta } = \dfrac{d}{{{d^1}}}$ ...... (1)
Given,
Wavelength in a double slit interference fringe experiment, $\lambda = 5000\,{A^ \circ }$
Fringe width obtained, $\beta = 1.2\,mm$
Distance between the parallel narrow slits, $d = 2\,mm$
Also the separation is increased by one and half times the previous value,
${d^1} = d + \dfrac{d}{2} = \dfrac{3}{2}d$
Now we have to find the fringe width when the separation is increased as above-
From equation (1), we get-
$\dfrac{{{\beta ^1}}}{\beta } = \dfrac{d}{{{d^1}}} \\\ \Rightarrow\dfrac{{{\beta ^1}}}{{1.2}} = \dfrac{2}{3} \\\ \therefore {\beta ^1} = 0.8\,mm \\\$
Hence, the fringe width is $0.8\,mm$.So, option B is correct.

Additional information:
Good interference bands can be produced if the fringe width becomes very small. In this case, the maxima and minima will be spaced so closely that it will appear like a uniform pressure pattern.

Note: Here we have to pay attention to the fact that fringe width is inversely proportional to the distance between the slits.
So, $\dfrac{{{\beta ^1}}}{\beta } = \dfrac{d}{{{d^1}}}$. If we happen to upturn the numerator and the denominator in this expression, the answer would be wrong. So, we have to pay attention to this carefully.