Question

In a YDSE fringes are observed by using light of wavelength 4800 $A^{\circ }$, if a glass plate ($\mu$ =1.5) is introduced in the path of one of the wave and another plate of same thickness is introduced in the path of the ($\mu$ = 1.8) other wave. The central fringe takes the position of fifth bright fringe. The thickness of plate will be

• A. 8 micron
• B. 80 micron
• C. 0.8 micron
• D. None of these

Answer 1

The correct option is A 8 micron

Shift due to the first plate $x_1=\frac{\beta }{\lambda }({\mu }_1-1)t\left(Upward\right)$
and shift due to the second $x_2=\frac{\beta }{\lambda }({\mu }_2-1)t\left(Downward\right)$
Hence net shift $=x_2-x_1=\frac{\beta }{\lambda }({\mu }_2-{\mu }_1)t$
$\Rightarrow 5\beta =\frac{\beta }{\lambda }(1.8-1.5)t\Rightarrow t=\frac{5\lambda }{0.3}=\frac{5\times 4800\times {10}^{-10}}{0.3}=8\times {10}^{-6}m=8micron$