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 $600\, nm$. 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, $\lambda=600\, nm =600 \times 10^{-9} m$
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 $1 .$
$\Delta x_{1}=\left(\mu_{1}-1\right) t=(1.6-1) t=0.6\, t$

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