Question

A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?

Answer 1

Given:-

Refractive index of the paper, μ = 1.45

The thickness of the plate,

\[t = 0 . 02  mm = 0 . 02 \times  {10}^{- 3}   m\]

Wavelength of the light,

\[\lambda = 620  nm = 620 \times  {10}^{- 9}   m\]

We know that when we paste a transparent paper in front of one of the slits, then the optical path changes by \[\left( \mu - 1 \right)t.\]

And optical path should be changed by λ for the shift of one fringe.

∴ Number of fringes crossing through the centre is

\[n = \frac{\left( \mu - 1 \right)t}{\lambda}\]

\[     = \frac{\left( 1 . 45 - 1 \right) \times 0 . 02 \times {10}^{- 3}}{620 \times {10}^{- 9}}\]

\[     = 14 . 5\]

Hence, 14.5 fringes will cross through the centre if the paper is removed.