Question

A transparent paper (refractive index $= 1.45$ ) of thickness $0.02\,mm$ is pasted on one of slits of a Young’s double experiment which uses monochromatic light of wavelength $620\,nm$ How many fringes will cross the center if the paper is removed?

Answer 1

In order to solve this problem we need to understand interference. Interference is the physical phenomena in which light is considered to be wave and these waves can add together by crest to crest and through to through creating constructive interference while they may join crest to trough and trough to crest creating destructive interference . Constructive interference leads to brightness and destructive interference leads to darkness as waves here cancel each other out.

Complete step by step answer:
Young’s double slit experiment is one in which two slits are placed in front of monochromatic light and light after falling on two slits divides into two different waves which add on after emanating from slits either constructively or destructively creating bright and dark fringes on screen.

If any material having some refractive index is placed before one of the slits then an additional path difference is created between them. Given, material refractive index is $\mu = 1.45$ and wavelength $\lambda = 620\,nm$. And thickness of material is $t = 0.02 \times {10^{ - 3}}\,m$.

Let the slit separation distance be $d$ and let an angular position of bright fringe be $\theta$.So due to paper in front of the slit additional path difference created is $\Delta = (\mu - 1)t$. So for a bright fringe we know,
${\Delta _T} = n\lambda$
Where, “n” is the number of bright fringes and $\lambda$ is wavelength and ${\Delta _T}$ is total path difference.
So from formula of Young’s double slit we get, ${\Delta _T} = d\sin \theta - (\mu - 1)t$
Putting values we get $d\sin \theta - \left( {\mu - 1} \right)t = n\lambda \to (i)$

So when the paper is removed let the new number of bright fringes be ${n_1}$
So using formula of young double slit we get, $d\sin \theta = {n_1}\lambda$
put this value in equation (i)
${n_1}\lambda - (\mu - 1)t = n\lambda$
Since number of fringes cross at center is $({n_1} - n) = N$
So, $N = \dfrac{{(\mu - 1)t}}{\lambda }$
Putting values we get,
$N = \dfrac{{(1.45 - 1) \times (0.02 \times {{10}^{ - 3}}m)}}{{620 \times {{10}^{ - 9}}m}}$
$\therefore N = 14.5$

So the number of complete fringes that cross the center is $14$.

Note: It should be remembered that using a material having some refractive index in front of the slit actually created a path different between the two waves emanating from two slits. Actually what happens is that due to the material's refractive index one wave velocity is altered and hence a phase difference is created between two waves of two slits which in turn creates a path difference between them.