Question

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

Answer 1

Answer: 12

Given,
Wavelength of light is, $\lambda =6000\mathring{A}$
Slit width is, $d=1mm$
Screen distance is, $D=10cm$
Thickness of plate is, $t=0.5mm$
Shift in position of interference is, $y=5mm$
Position of fringe from central axis is given by, $y=\frac{\left(\Delta x\right)D}{d}\Rightarrow \Delta x=\frac{yd}{D}...\left(1\right)$
From the expression of optical path difference, extra path length introduced is given by, $\Delta x=(\mu -1)t...\left(2\right)$
Equating $\left(1\right)$ and $\left(2\right)$ , we have
$(\mu -1)t=\frac{yd}{D}$
$\Rightarrow \mu -1=\frac{yd}{tD}$
On putting all the corresponding values, we have
$\Rightarrow \mu -1=\frac{5\times 10^{-3}\times 2\times 10^{-3}}{0.5\times 10^{-3}\times 0.10}$
$\Rightarrow \mu =0.2+1=1.2$