In a Young's double slit experiment using monochromatic light of wavelength 620 nm , a transparent paper (refractive index = 1.45 ) with a thickness of 0.02 mm is pasted over one of the slits. Determine the number of

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Correct answer: B. 14.5

The path difference introduced by the transparent paper is given by

\Delta x = (\mu - 1)t

.

The number of fringes shifted across the centre upon removal of the paper is

n = \dfrac{\Delta x}{\lambda} = \dfrac{(\mu - 1)t}{\lambda}

.

Substituting the given values:

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

n = \dfrac{0.45 \times 2 \times 10^{-5}}{6.2 \times 10^{-7}}

n = \dfrac{0.9 \times 10^{-5}}{6.2 \times 10^{-7}}

n = \dfrac{90}{6.2} \approx 14.5

Thus, approximately

14.5

fringes will cross through the centre.

In a Young's double slit experiment using monochromatic light of...

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