Question

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

• A. \(5\pi\)
• B. \(\dfrac{7}{3}\pi\)
• C. \(\dfrac{7}{4}\pi\)
• D. \(4\pi\)

Hint:

A thin plate introduced in the path of one ray produces an extra optical path difference \((\mu-1)t\). The corresponding phase difference is \[ \Delta \phi=\frac{2\pi}{\lambda}(\mu-1)t \]

Answer 1

The Correct Option is A

Step 1: Write the formula for optical path difference introduced by a thin plate.
When a thin plate of thickness \(t\) and refractive index \(\mu\) is introduced in the path of one ray, the extra optical path difference introduced is \[ \Delta x=(\mu-1)t \] Given, \[ \mu=\frac{13}{7} \] and \[ t=1.4\times 10^{-6}\,\text{m} \]

Step 2: Calculate the optical path difference.
\[ \Delta x=\left(\frac{13}{7}-1\right)(1.4\times 10^{-6}) \] \[ \Delta x=\left(\frac{6}{7}\right)(1.4\times 10^{-6}) \] \[ \Delta x=1.2\times 10^{-6}\,\text{m} \]

Step 3: Convert wavelength into SI unit.
Given wavelength is \[ \lambda=480\,\text{nm} \] Since, \[ 1\,\text{nm}=10^{-9}\,\text{m} \] Therefore, \[ \lambda=480\times 10^{-9}\,\text{m} \] \[ \lambda=4.8\times 10^{-7}\,\text{m} \]

Step 4: Find the phase difference.
Phase difference introduced is \[ \Delta \phi=\frac{2\pi}{\lambda}\Delta x \] Substituting the values, \[ \Delta \phi=\frac{2\pi}{4.8\times 10^{-7}}\times 1.2\times 10^{-6} \] \[ \Delta \phi=2\pi\times \frac{1.2\times 10^{-6}}{4.8\times 10^{-7}} \] \[ \Delta \phi=2\pi\times 2.5 \] \[ \Delta \phi=5\pi \]

Step 5: Final conclusion.
Therefore, the phase difference introduced at the position of central maxima is \[ \boxed{5\pi} \]