Question

In a Young's double slit experiment, using monochromatic light, the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1⋅6 and thickness 1⋅964 micron (1 micron = 10⁻⁶ m) is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the screen and the slits is doubled. It is found that the distance between the successive maxima now is the same as the observed fringe-shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment.

Answer 1

Given:
Refractive index of the mica sheet,μ = 1.6
Thickness of the plate, $t=1.964\mathrm{micron}=1.964\times {10}^{-6}\mathrm{m}$
Let the wavelength of the light used = λ.
Number of fringes shifted is given by
$n=\frac{\left(\mathrm{\mu }-1\right)t}{\lambda }$
So, the corresponding shift in the fringe width equals the number of fringes multiplied by the width of one fringe.
$$\begin{gathered} \mathrm{Shift}=n\times \beta \\ =\frac{\left(\mu -1\right)t}{\lambda }\times \frac{\lambda D}{d} \\ =\frac{\left(\mu -1\right)t\times D}{d}...\left(\mathrm{i}\right) \end{gathered}$$
As per the question, when the distance between the screen and the slits is doubled,
i.e. $D\text{'}=2D$,
fringe width, $\beta =\frac{\lambda D\text{'}}{d}=\frac{\lambda 2D}{d}$
According to the question, fringe shift in first case = fringe width in second case.
$$\begin{gathered} \mathrm{So},\frac{\left(\mathrm{\mu }-1\right)t\times \mathrm{D}}{d}=\frac{\lambda 2\mathrm{D}}{d} \\ \Rightarrow \lambda =\frac{\left(\mu -1\right)t}{2} \\ =\frac{\left(1.6-1\right)\times \left(1.964\right)\times {10}^{-6}}{2} \\ =589.2\times {10}^{-9}=589.2\mathrm{nm} \end{gathered}$$

Hence, the required wavelength of the monochromatic light is 589.2 nm.