Question

The slits in a double-slit interference experiment are illuminated by orange light ($\lambda =60nm$). A thin transparent plastic of thickness $t$ is placed in front of one of the slits. The number of fringes shifting on screen is plotted versus the refractive index $\mu$ of the plastic in graph shown in figure. The value of $t$ is:

• A. $4.8mm$
• B. $640\mu m$
• C. $2.4\mu m$
• D. none of these

Answer 1

Answer: (C)

$2.4\mu m$

Number of fringes shifted $N=40$, refractive index of plastic $\mu =2$

Lateral shift $\overline{X}=\frac{\beta }{\lambda }(\mu -1)t$ and $\beta =\frac{\overline{X}}{N}$

$\therefore \beta N=\frac{\beta }{\lambda }(\mu -1)t\Rightarrow t=\frac{N\lambda }{(\mu -1)}=\frac{40\times 60\times {10}^{-9}}{2-1}=2.4\times {10}^{-6}=2.4\mu m$