Question

A parallel beam of light of wavelength 5.5 × 10^-7m in air is incident on the slits in Young's double slit experiment. A thin film of transparent plastic of refractive index 1.48 and thickness

5.2µm is placed over one of the slits.

(i.) Determine the increase in the optical path of the light passing through the thin film.

(ii.) Determine the number of fringes by which the central fringe is displaced.

(iii.) Explain how the fringe spacing would change if:

     • The slit separations were increased.

     • The slit-screen distances were increased.

Answer 1

Given,

Wavelength of the parallel beam of light $(\lambda) =$ $5.5 × 10^{-7}m$

Refractive index $(\mu) = 1.48$

Thickness of the film $(t)=5.2\mu m=5.2\times 10^{-6}m$

a) $\Delta x = 2\mu t$

Now, substituting the values,

$\Delta x=2\times 1.48\times 5.2\times 10^{-6}m$

$=15.39\mu m$

b) Total number of fringe shift $=$ total fringe shift/fringe width

$(n)=\frac{(\mu -1)t}{\lambda}$

Now, substituting the values,

$n=\frac{(1.48-1)\times 5.2\times 10^{-6}}{5.5\times 10^{-7}}$

$=\frac{0.48\times 5.2\times 10^{6}}{5.5\times 10^{-7}}$

$=4.53$

$\simeq =4$

c)

As the separation between the slit gets increase then the fringe width will get decreased because less interference will occur.

As the slit screen distance will get increase then the fringe width will get increase because the light would h ave more space to diffract outwards and the interference will increase.