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?

Answer 1

Answer:

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Answer 2

Fringes that will cross through the centre if the paper is removed is  $14.5$

Explanation:

Step 1:

Refractive index of  transparent paper $=\mu=1 \cdot 45$

Thickness $=\mathrm{t}=0 \cdot 02 \mathrm{mm}=0.02 \times 10^{-3} \mathrm{m}$

$\lambda=620 \mathrm{nm}=620 \times 10^{-9} \mathrm{m}$

λ is wavelength  

If we paste a transparent paper in front of one of the slits, we realize that the optical path changes by  

$=(\mu-1) t$

And for the one fringe shift, the optical direction should be modified by γ.

Step 2:

The number of fringes which cross the center is

$n=\frac{(\mu-1) t}{\lambda}$

Step 3:

On substituting the values, we get  

$=\frac{(1.45-1) \times 0.02 \times 10^{-3}}{620 \times 10^{-9}}$

$=\frac{(0.45) \times 0.02 \times 10^{-3}}{620 \times 10^{-9}}$

$=\frac{9 \times 10^{-3} \times 10^{-3}}{620 \times 10^{-9}}$

$=\frac{0.0145 \times 10^{-3} \times 10^{-3}}{10^{-9}}$

$=14.5$

When the paper is cut, $14.5$ fringes pass through the centre.