Question

A thin paper of thickness 0.02mm having a refractive index 1.45 is placed 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 600nm.
A. 25:1, 15
B. 35:1, 15
C. 25:3, 16
D. 25:2, 18

Answer 1

Hint: For the first part, you could recall the expression for the ratio of maximum intensity to that of minimum intensity and substitute accordingly to get the required ratio. Now for the second part, you could recall the expression for shift and also the fringe width and ratio of these two will give the number of fringes shifted and hence, the answer.

Formula used:
Ratio of maximum and minimum intensities,
$\dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{{{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}}{{{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}}$
Fringe width,
$\beta =\dfrac{D\lambda }{d}$
Shift,
$s=\dfrac{\left( \mu -1 \right)tD}{d}$

Complete step-by-step answer:
In the question, we are given a thin paper with thickness $t=0.02mm=2\times {{10}^{-5}}m$and refractive index 1.45. This paper when kept in one of the slits in a Young’s double slit experiment, the paper transmits $\dfrac{4}{9}$ of the light energy falling on it. This information is followed by a set of questions.
(a) Here we are asked to find the ratio of the maximum to the minimum intensity in the fringe pattern.
Initially the intensity through each slit was,
${{I}_{1}}={{I}_{2}}={{I}_{0}}$
After keeping paper on one of the slits, say${{S}_{2}}$, the intensity of light through first slit will remain the same while, the intensity of the light through that slight is given to be,
${{I}_{2}}=\dfrac{4}{9}{{I}_{0}}$
So the ratio of intensities through the two slits will be,
$\dfrac{{{I}_{1}}}{{{I}_{2}}}=\dfrac{{{I}_{0}}}{\dfrac{4}{9}{{I}_{0}}}=\dfrac{9}{4}$
Taking square root on both sides,
$\sqrt{\dfrac{{{I}_{1}}}{{{I}_{2}}}}=\dfrac{3}{2}$ ……………………………………………… (1)
We know that the ratio of maximum and minimum intensities in YDSE is given by,
$\dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{{{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}}{{{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}}$
Dividing both numerator and denominator by$\sqrt{{{I}_{2}}}$, we get,
$\dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{{{\left( \dfrac{\sqrt{{{I}_{1}}}}{\sqrt{{{I}_{2}}}}+1 \right)}^{2}}}{{{\left( \dfrac{\sqrt{{{I}_{1}}}}{\sqrt{{{I}_{2}}}}-1 \right)}^{2}}}$
Substituting (1) we get,
$\dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{{{\left( \dfrac{3}{2}+1 \right)}^{2}}}{{{\left( \dfrac{3}{2}-1 \right)}^{2}}}$
$\Rightarrow \dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{{{\left( 5 \right)}^{2}}}{{{\left( 1 \right)}^{2}}}$
$\therefore \dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{25}{1}$
Therefore, the ratio of the maximum intensity to the minimum intensity in the fringe pattern is found to be, 25:1.
(b) Now, for the second part, we know that the shift in fringes caused due to the paper is given by the expression,
$s=\dfrac{\left( \mu -1 \right)tD}{d}$ …………………………………………. (2)
Where, $\mu $ is the refractive index of the paper, t is its thickness, D is the distance between the slits and the screen and d is the separation of the slits.
Also, we have the fringe width given by,
$\beta =\dfrac{D\lambda }{d}$ ………………………………………….. (3)
Where, ‘$\lambda $’ is the wavelength of light.
We know that the number of fringes shifted is given by,
$n=\dfrac{s}{\beta }$
$\Rightarrow n=\dfrac{\left( \mu -1 \right)tD}{d}\times \dfrac{d}{D\lambda }$
$\Rightarrow n=\dfrac{\left( \mu -1 \right)t}{\lambda }$
Substituting the given values,
$n=\dfrac{\left( 1.45-1 \right)2\times {{10}^{-5}}}{600\times {{10}^{-9}}}$
$\therefore n=15$
Therefore, the number of fringes that will cross through the centre if an identical paper piece is pasted on the other slit will be 15.
Hence, option A is found to be the right answer.

So, the correct answer is “Option A”.

Note: In a Young’s double slit experiment coherent sources of light are used. They are normally kept in at a very small distance apart, slightly greater than the order of wavelength of light being used. The fringes formed as the result of this experiment is found to be hyperbolic with a central section.