Question

In a Young's double slit experiment, a thin sheet of refractive index $1.6$ is used to cover one slit while a thin the sheet of refractive index $1.3$ is used to cover the second slit. The thickness of both the sheets are same and the wavelength of light used is $600nm$. If the central point on the screen is now occupied by what had been the $10$ th bright fringe $(m=10)$, then the thickness of covering sheets is

• A. $50\mu m$
• B. $8\mu m$
• C. $20\mu m$
• D. $40\mu m$

Answer 1

Answer: (C)

Given, refractive index of first sheet ${\mu }_1=1.6$ refractive index of second sheet ${\mu }_2=1.3$ and wavelength of light, $\lambda =600nm=600\times 10^{-9}m$
A Young's double slit experiment is shown below, in which two thin sheets are covered the slits. So, the path difference introduced in slit $1.$
$\Delta x_1=\left({\mu }_1-1\right)t=(1.6-1)t=0.6t$

Similarly, path difference introduced in slit 2 ,
$\Delta x_2=\left({\mu }_2-1\right)t=(1.3-1)t=0.3t$
So, the net path difference introduced in central maxima,
$\Delta x_{\text{central maxima }}=\Delta x_1-\Delta x_2=0.6t-0.3t=0.3t$
For central maxima, which occupied the $10^{\text{th }}$ bright fringe,
$\Delta x_{\text{cen }}=10\lambda$
$\Rightarrow t=\frac{10\times 600\times 10^{-9}}{0.3}=20\mu m$