Question

A thin paper of thickness 0.02 mm having a refractive index 1.45 is pasted across one of the slits in a Young's double slit experiment. The paper transmits 4/9 of the light energy falling on it. (a) Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern. (b) How many fringes will cross through the centre if an identical paper piece is pasted on the other slit also? The wavelength of the light used is 600 nm.

Answer 1

Given:-

The thickness of the thin paper,

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

Refractive index of the paper,

\[\mu = 1 . 45\]

Wavelength of the light,

\[\lambda = 600  nm = 600 \times  {10}^{- 9}   m\]

(a)

Let the intensity of the source without paper = I₁

and intensity of source with paper =I₂

Let a₁ and a₂ be corresponding amplitudes.

As per the question,

\[I_2  = \frac{4}{9} I_1\]

We know that

\[\frac{I_1}{I_2} = \frac{{a_1}^2}{{a_2}^2}............\left( \because I \propto a^2 \right)\]

\[ \Rightarrow \frac{a_1}{a_2} = \frac{3}{2}\]

Here, a is the amplitude.

We know that \[\frac{I_\max}{I_\min} = \frac{\left( a_1 + a_2 \right)^2}{\left( a_1 - a_2 \right)^2}. \]

\[ \Rightarrow   \frac{I_\max}{I_\min} = \frac{\left( 3 + 2 \right)^2}{\left( 3 - 2 \right)^2}\]

\[= \frac{25}{1}\]

\[ \Rightarrow  I_\max :  I_\min  = 25  :   1\]

(b)

Number of fringes that will cross through the centre is given by \[n = \frac{\left( \mu - 1 \right)t}{\lambda}\]

\[\Rightarrow n = \frac{\left( 1 . 45 - 1 \right) \times 0 . 02 \times {10}^{- 3}}{600 \times {10}^{- 9}}\]

\[= \frac{0 . 45 \times 0 . 02 \times {10}^4}{6} = 15\]