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: Compute the extra optical path from the plate.
\[ \Delta x = (\mu-1)t = \left(\frac{13}{7}-1\right)(1.4\times10^{-6}) = \frac{6}{7}\times1.4\times10^{-6} = 1.2\times10^{-6}\,\text{m} \]

Step 2: Convert to phase difference.
\[ \delta = \frac{2\pi}{\lambda}\Delta x = \frac{2\pi}{480\times10^{-9}}\times1.2\times10^{-6} = \frac{2\pi\times1.2}{0.48} = 5\pi \] \[ \boxed{5\pi} \]