Question

A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0⋅50 mm and the separation between the slits is 0⋅12 cm. The refractive index of mica and polystyrene are 1.58 and 1.55, respectively, for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen at a distance one metre away. (a) What would be the fringe-width? (b) At what distance from the centre will the first maximum be located?

Answer 1

Given:
The thickness of the strips = $t_1=t_2=t=0.5\mathrm{mm}=0.5\times {10}^{-3}\mathrm{m}$
Separation between the two slits,$\mathit{}d=0.12\mathrm{cm}=12\times {10}^{-4}\mathrm{m}$
The refractive index of mica, μ_m = 1.58 and of polystyrene, μ_p = 1.58
Wavelength of the light, $\mathrm{\lambda }=590\mathrm{nm}=590\times {10}^{-9}\mathrm{m},$
Distance between screen and slit, D = 1 m

(a)
We know that fringe width is given by
$\beta =\frac{\lambda D}{d}$
$$\begin{gathered} \Rightarrow \beta =\frac{590\times {10}^{-9}\times 1}{12\times {10}^{-4}} \\ =4.9\times {10}^{-4}\mathrm{m} \end{gathered}$$

(b) When both the mica and polystyrene strips are fitted before the slits, the optical path changes by
$$\begin{gathered} ∆x=\left({\mathrm{\mu }}_{\mathrm{m}}-1\right)t-\left({\mathrm{\mu }}_{\mathrm{p}}-1\right)t \\ =\left({\mathrm{\mu }}_{\mathrm{m}}-{\mathrm{\mu }}_{\mathrm{p}}\right)t \\ =\left(1.58-1.55\right)\times \left(0.5\right)\left({10}^{-3}\right) \\ =\left(0.015\right)\times {10}^{-3}\mathrm{m} \end{gathered}$$
∴ Number of fringes shifted, n = $\frac{∆x}{\lambda }$.
$\Rightarrow n=\frac{0.015\times {10}^{-3}}{590\times {10}^{-9}}=25.43$
∴ 25 fringes and 0.43^th of a fringe.
⇒ In which 13 bright fringes and 12 dark fringes and 0.43^th of a dark fringe.
So, position of first maximum on both sides is given by
On one side,
$$\begin{gathered} x=\left(0.43\right)\times 4.91\times {10}^{-4}\left(\because \beta =4.91\times {10}^{-4}\mathrm{m}\right) \\ =0.021\mathrm{cm} \end{gathered}$$
On the other side,
$$\begin{gathered} x\text{'}=\left(1-0.43\right)\times 4.91\times {10}^{-4} \\ =0.028\mathrm{cm} \end{gathered}$$