Answer to Question #164548 in Optics for Halyufer

Question

When a thin film of transparent plastic is placed over one of the slits in Young's double slit

     experiment, the central fringe is displaced by 4.5 fringes. The refractive index of the material is 1.48

    and the wavelength of the light in air is 5.5 × 10^-7 m.

    (i.) By how much does the film increase the optical path?

    (ii.) What is the thickness of the film?

    (iii.) What would be observed if the material had a thickness of 1.00mm? Explain. 

Answer 1

Answers

Thickness of the width for maximum shift

(I) optical path will be increased by

$4.5\lambda=4.5\times5.5\times10^{-7}\\=24.75\times10^{-7}m$

ii)

$t=\frac{4.5\lambda}{\mu-1}$

$t=\frac{4.5(5 .5\times10^{-7})}{1.48-1}=5.15\mu m$

$t=11.6\mu m$

(Iii) if thickness is t then

optical path=$(\mu-1) t=(1.48-1) (10^{-3}) =0.48mm$