Question

In a YDSE experiment, if the position of the first minima is given as ${{Y}_{0}}$, then find out the wavelength of light used if distance between slits is $'d'$ and screen is $D$.

Answer 1

Hint: This problem can be solved by using the formula for the path difference in the YDSE in terms of the separation between the slits, the separation between the slits and the screen and the vertical distance of the point on the screen from the midpoint on the separation of the two slits, on which the light rays meet and the formula for the path difference for minima in YDSE.
Formula used:
$\Delta x=d\dfrac{y}{D}$
$\Delta {{x}_{\text{minima}}}=\left( n+\dfrac{1}{2} \right)\lambda $

Complete step by step answer:
We will use the formula for the path difference in the YDSE.
The path difference $\Delta x$ between the rays emanating from the slits and meeting at a point on the screen is given by
$\Delta x=d\dfrac{y}{D}$ ---(1)
where $d$ is the separation between the slits, $y$ is the vertical distance of the point where the light rays meet on the screen from the midpoint of the separation between the slits and $D$ is the separation of the screen from the slits.
Now, dark fringes are formed in YDSE on the screen whenever there is a minima or destructive interference between the two interfering light rays emanating from the slits.
The path difference $\Delta {{x}_{\text{minima}}}$ is the path difference between two light rays of wavelength $\lambda $ which interfere destructively and produce minima in YDSE is given by
$\Delta {{x}_{\text{minima}}}=\left( n+\dfrac{1}{2} \right)\lambda $ $\left( n=0,1,2,3,4..... \right)$ --(2)
Now for the first minima, $n=0$. Therefore, the path difference $\Delta {{x}_{\min 1}}$ will be
$\Delta {{x}_{\text{min1}}}=\left( 0+\dfrac{1}{2} \right)\lambda =\dfrac{1}{2}\lambda $ --(3)
Now, to get the required wavelength, we will equate (1) and (3).
$\therefore \Delta x=\Delta {{x}_{\min 1}}$
$\therefore d\dfrac{y}{D}=\dfrac{1}{2}\lambda $ --(4)
Now, according to the question, for the first minima, $y={{Y}_{0}}$.
Putting this value in (4), we get
$\therefore d\dfrac{{{Y}_{0}}}{D}=\dfrac{1}{2}\lambda $
$\therefore \lambda =2d\dfrac{{{Y}_{0}}}{D}$

Hence, we have got the required expression for the wavelength.

Note: Students must note that the wavelength of the light used in the YDSE experiment depends upon the refractive index of the medium in which the apparatus is put in and the experiment is conducted. Therefore, a change in the wavelength will mean a change in all the different observations of the experiment such as the position of the fringes.
Students must also keep in mind that there is a presence of a central maxima or the zeroth maxima. However, there is nothing such as the central minima in YDSE. The minima next to the central maxima is the first minima.