Q. In a double slit arrangement fringes are produced using light of wavelength $4800\ \mathring{\mathrm{A}}$. One slit is covered by a thin plate of glass of refractive index 1.4 and the other with another glass plate of same thickness but of refractive index 1.7. By doing so the central bright shifts to original fifth bright fringe from centre. Thickness of glass plate is
A) $8\ \mu m$
B) $6\ \mu m$
C) $4\ \mu m$
D) $10\ \mu m$
Explanation
(a) Shift $\Delta x = \frac{\beta}{\lambda}(\mu - 1)t$
Shift due to one plate $\Delta x_{1} = \frac{\beta}{\lambda}(\mu_{1} - 1)t$
Shift due to another path $\Delta x_{2} = \frac{\beta}{\lambda}(\mu_{2} - 1)t$
Net shift $\Delta x = \Delta x_{2} - \Delta x_{1} = \frac{\beta}{\lambda}(\mu_{2} - \mu_{1})t$ … (i)
Also it is given that $\Delta x = 5\beta$ ….(ii)
Hence $5\beta = \frac{\beta}{\lambda}(\mu_{1} - \mu_{2})t$
$\Rightarrow t = \frac{5\lambda}{(\mu_{2} - \mu_{1})} = \frac{5 \times 4800 \times 10^{- 10}}{(1.7 - 1.4)} = 8 \times 10^{- 6}m = 8\mu m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.