In young's double slit experiment, light of wavelength $480 \mathrm{~nm}$ is incident on two slits separated by a distance of $4 \times 10^{-4} \mathrm{~m}$. If a thin plate of thickness $1.4 \times 10^{-6} \mathrm{~m}$ and refractive index $\frac{13}{7}$ is placed between one of the slits and screen, the phase difference introduced at the position of central maxima is

• Correct answer
  A

  $5 \pi$
• B

  $\frac{7}{3} \pi$
• C

  $\frac{7}{4} \pi$
• D

  $4 \pi$

Solution

The YDSE setup is shown below
Additional path difference introduced by medium plate, $\Delta L=(\mu-1) t$ As, a path difference of one wavelength is equals to a phase difference $\Delta \phi=\frac{\Delta L \times 2 \pi}{\lambda}$ so, $\quad \Delta \phi=\frac{(\mu-1) t \times 2 \pi}{\lambda}$ Here, $\mu=\frac{13}{7}, t=1.4 \times 10^{-6}, \lambda=480 \times 10^{-9} \mathrm{~m}$ $\therefore \quad \Delta \phi=\frac{\left(\frac{13}{7}-1\right) \times 1.4 \times 10^{-6} \times 2 \pi}{480 \times 10^{-9}}$ $=\frac{6 \times 1.4 \times 2 \pi \times 10^{-6}}{7 \times 480 \times 10^{-9}}=5 \pi \mathrm{rad}$

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