Question

A two-slit Young's interference experiment is done with monochromatic light of wavelength $6000\,\mathring{A}$ . The slits are $2\,mm$ apart. The fringes are observed on a screen placed $10\,cm$ away from the slits. Now a transparent plate of thickness $0.5\,mm$ is placed in front of one of the slits and it is found that the interference pattern shifts by $5\,mm$ . The refractive index of the transparent plate is $\frac{k}{10}$ , find $k$ ?

Answer 1

Given,
Wavelength of light is, $\lambda =6000\,\mathring{A}$
Slit width is, $d=1\,mm$
Screen distance is, $D=10\,cm$
Thickness of plate is, $t=0.5\,mm$
Shift in position of interference is, $y=5\,mm$
Position of fringe from central axis is given by, $y=\frac{\left(\Delta x\right) D}{d}\Rightarrow \Delta x=\frac{y d}{D}...\left(1\right)$
From the expression of optical path difference, extra path length introduced is given by, $\Delta x=\left(\right.\mu -1\left.\right)t...\left(2\right)$
Equating $\left(1\right)$ and $\left(2\right)$ , we have
$\left(\right.\mu -1\left.\right)t=\frac{y d}{D}$
$\Rightarrow \mu -1=\frac{y d}{t D}$
On putting all the corresponding values, we have
$\Rightarrow \mu -1=\frac{5 \times 10^{- 3} \times 2 \times 10^{- 3}}{0 . 5 \times 10^{- 3} \times 0 . 10}$
$\Rightarrow \mu =0.2+1=1.2$