Question

A beam of light consisting of wavelength $6000{A^\circ }$ and $4500{A^\circ }$ is used in a YDSE with $D = 1m$ and $d = 1mm$. Find the least distance from central maxima, where bright fringes due to the two wavelengths coincide:

Answer 1

Hint: To solve this question, at first we need to know what happens when the bright fringes meet. Using the condition of least distance, we will then consider the fringes that will meet such that the least distance can be obtained. Using the formula for central fringe, we can obtain the required distance.

Complete Step-By-Step Solution:
We know, in case of Young’s Double Slit interference, light waves are allowed to pass through the two slits places, The light waves due to wave nature of light , diffracts and they pass through the slit and form different fringes on the screen. The fringes forms are bright fringes or dark fringes. The central line is the brightest of all other fringes present and is called the central fringe.
Fringe width, as we know, is defined as the distance between two consecutive dark fringes or bright fringes.
In the case of YDSE, all fringes are of the same width.
Now, we are given the wavelength of one light is and the wavelength of other light is
Let us consider that the ${n^{th}}$fringe of one wavelength light meets with ${m^{th}}$fringe of another wavelength light.
Let ${\beta _1}$ be the fringe width of ${n^{th}}$fringe and ${\beta _2}$ be the fringe width of\[{m^{th}}\].
Since the fringes meet at a point, we can say:
\[m{\beta _1} = n{\beta _2}\]
Now, we know the formula for fringe width is given by:
\[\beta = \dfrac{{\lambda D}}{d}\]
Where,
\[\beta \] is the fringe width
\[\lambda \] is the wavelength of the light
\[D\] is the distance between source and screen \[1m\]
\[d\] is the distance between the slits.
Thus putting the value of \[{\beta _1}\] and\[{\beta _2}\], we get:
\[m\dfrac{{{\lambda _1}D}}{d} = n\dfrac{{{\lambda _2}D}}{d}\]
Now putting the values as given in the question:
$m\dfrac{{6000 \times 1}}{{1 \times {{10}^{ - 2}}}} = n\dfrac{{4500 \times 1}}{{1 \times {{10}^{ - 2}}}}$
On solving, we get:
$m = 3$ and $n = 4$
On putting the values, we obtain the least distance from central maxima as: $m{\beta _1}$
We get:
$ = 3\dfrac{{6000 \times {{10}^{10}} \times 1}}{{1 \times {{10}^{ - 3}}}}$
Thus, we obtain:
$m{\beta _1} = 1.8m$
This is our required solution.

Note:
The fringes observed in YDSE are due to interference of monochromatic light waves as they pass through slits. The conditions that are required for the interference to occur is that the waves must have constant phase difference and they must be monochromatic which means emit waves of one colour only and be coherent, that means emit waves continuously.