Question

In Young's double slit experiment, light of wavelength 480 nm is incident on two slits separated by a distance of \(4 \times 10^{-7}\) m. If a thin plate of thickness \(1.4 \times 10^{-6}\) 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

• A. \(\frac{7\pi}{3}\)
• B. \(\frac{7\pi}{4}\)
• C. \(\frac{7\pi}{2}\)
• D. \(\frac{7\pi}{6}\)

Hint:

Phase difference formula: \(\Delta\phi = \frac{2\pi}{\lambda}(\mu - 1)t\)
Central maxima shifts when path difference changes
Watch units - convert all lengths to same units (meters recommended)

Answer 1

The Correct Option is A

The phase difference \(\Delta\phi\) introduced by the thin plate is given by: \[ \Delta\phi = \frac{2\pi}{\lambda}(\mu - 1)t \] Where: \[ \lambda = 480 \text{ nm} = 480 \times 10^{-9} \text{ m}, \quad \mu = \frac{13}{7}, \quad t = 1.4 \times 10^{-6} \text{ m} \] \[ \Delta\phi = \frac{2\pi}{480 \times 10^{-9}}\left(\frac{13}{7} - 1\right)1.4 \times 10^{-6} \] \[ = \frac{2\pi}{480 \times 10^{-9}}\left(\frac{6}{7}\right)1.4 \times 10^{-6} \] \[ = \frac{2\pi \times 6 \times 1.4}{480 \times 7} \times 10^3 = \frac{7\pi}{3} \]