Question

A transparent paper (refractive index = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed?
(a)$14$
(b)$15$
(c)\[18\]
(d)\[44\]

Answer 1

Hint:In order to solve this question, we need to use the formula for wave optics. The fringes created can be calculated using formula for diffraction.

Complete solution
Refractive index of transparent paper $=\mu =1.45$
Thickness $=t=0.2mm=0.02\times {{10}^{-3}}m$
$\lambda =620nm=620\times {{10}^{-9}}m$
$\lambda $ is wavelength,
If we paste a transparent paper in front of one of the slits, we realize that the optical path changes by
$=\left( \mu -1 \right)t$
And for the one fringe shift, the optical direction should be modified by\[\gamma \].
Then,
The number of fringes which cross the centre is
$n=\dfrac{\left( \mu -1 \right)t}{\lambda }$
So,
On substituting the values, we get,
$=\dfrac{\left( 1.45-1 \right)\times 0.02\times {{10}^{-3}}}{620\times {{10}^{-9}}}$
$=\dfrac{\left( 0.45 \right)\times 0.02\times {{10}^{-3}}}{620\times {{10}^{-9}}}$
\[=\dfrac{9\times {{10}^{-3}}\times {{10}^{-3}}}{620\times {{10}^{-9}}}\]
$=\dfrac{0.0145\times {{10}^{-3}}\times {{10}^{-3}}}{{{10}^{-9}}}$
$=14.5$
Hence, When the paper is cut, $14.5$(approx.) fringes pass through the centre.

The correct answer to the question is option (a).

Additional Information:Diffraction refers to various phenomena that occur when a wave encounters an obstacle or a slit. It is defined as the bending of waves around the corners of an obstacle or through an aperture into the region of geometrical shadow of the obstacle/aperture.

Note:While solving this question, we should be aware of the different types of formula used here. Especially how the different values of the variable of the formula is used in the question. The formula is modified and used here to take out the required solution for the problem given here.