In a young's double slit arrangement frings are produced using light of wavelength $4000 Å$. 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 sixth fringe from centre. Thickness of glass plate is

• A

  $2 \mu \mathrm{m}$
• Correct answer
  B

  $8 \mu \mathrm{m}$
• C

  $11 \mu \mathrm{m}$
• D

  $16 \mu \mathrm{m}$

Solution

As, shift $\Delta x=\frac{\beta}{\lambda}(\mu-1) t$
Shift due to one plate $\Delta x_1=\frac{\beta}{\lambda}\left(\mu_1-1\right) t$ Shift due to another path $\Delta x_2=\frac{\beta}{\lambda}\left(\mu_2-1\right) t$ Net shift $\Delta x=\Delta x_2-\Delta x_1=\frac{\beta}{\lambda}\left(\mu_2-\mu_1\right) t$ Also, it is given that $\Delta x=6 \beta$ $$ \begin{aligned} & \text { Hence, } 6 \beta=\frac{\beta}{\lambda}\left(\mu_1-\mu_2\right) t \\ & \Rightarrow \quad t=\frac{6 \lambda}{\left(\mu_1-\mu_2\right)}=\frac{6 \times 4000}{(1.7-1.4)}-80000 Å \\ & \Rightarrow \quad t=8 \times 10^{-6}=8 \mu \mathrm{m} \end{aligned} $$

Hence, the correct answer is option: 'B'.