Question

Laser light of wavelength $630\,nm$ incident on a pair of slits produces an interference pattern in which the bright fringes are separated by $8.3\,mm$ . A second light produces an interference pattern in which the bright fringes are separated by $7.6\,mm$. Find the wavelength of the second light.

Answer 1

In order to solve this question we need to understand Young’s Double Slit Experiment. In Young’s Double Slit Experiment, incident rays from a monochromatic source of light fall perpendicularly on two slits which are separated by some distance. These rays after passing through slits are made to fall on a screen which is at some distance from slits. These rays form an interference pattern of bright and dark fringes continuously. Also these rays on reaching a single point on screen suffer path difference and hence the phase difference.

Complete step by step answer:
Let the slit separation be “d” and Slits and screen separation be “D”. Also let the linear fringe width be denoted as $\beta$ , so we know from Young’s Double slit formula that fringe width is defined as, $\beta = \dfrac{{\lambda D}}{d} \to (i)$
Let the wavelength in first case be denoted as, ${\lambda _1}$
So for case $1$ we have, ${\beta _1} = 8.3mm$ and ${\lambda _1} = 630nm$
So from equation (i) we have, ${\beta _1} = \dfrac{{{\lambda _1}D}}{d} \to (ii)$

Let the wavelength in the second case be denoted as, ${\lambda _2}$. So for case $2$ we have, ${\beta _2} = 7.6\,mm$ and ${\lambda _2}$ be wavelengths.So from equation (i) we have,
${\beta _2} = \dfrac{{{\lambda _2}D}}{d} \to (iii)$
Dividing equation (iii) by equation (ii) we get,
$\dfrac{{{\beta _2}}}{{{\beta _1}}} = \dfrac{{(\dfrac{{{\lambda _2}D}}{d})}}{{(\dfrac{{{\lambda _1}D}}{d})}}$
$\Rightarrow \dfrac{{{\beta _2}}}{{{\beta _1}}} = \dfrac{{{\lambda _2}}}{{{\lambda _1}}}$
$\Rightarrow {\lambda _2} = {\lambda _1}\dfrac{{{\beta _2}}}{{{\beta _1}}}$
$\Rightarrow {\lambda _2} = (630\,nm)(\dfrac{{7.6\,mm}}{{8.3\,mm}})$
$\therefore {\lambda _2} = 576.87\,nm$

Therefore, the wavelength of the second light is ${\lambda _2} = 576.87\,nm$.

Note: It should be remembered that, here we have assumed that a monochromatic beam falls on slits and also there are no filters like glass is placed before any slit. We have also assumed that the experiment is conducted in an air medium having refractive index $n = 1$ . Also we know the fringe width between two maxima and minima is the same in the YDSE experiment.